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Is the heat capacity infinite for a process at constant temperature?

The heat capacity of $\ce{H2}$ at constant $P$ and temperature is ______.

It was explained to me that the answer is infinity with the following reasoning:

Since heat capacity = $\mathrm{d}H/\mathrm{d}T$ and $\mathrm{d}T$ is zero since constant temperature is part of the question, the answer is infinity.

My attempt: Since $\ce{H2}$ is a diatomic gas, its $C_p$ is $7R/2$ which gives $7~\mathrm{cal \cdot mol^{-1} \cdot K^{-1}}$.
Which method is correct?

I had an elementary school teacher who loved to flip A/THE to make true false questions. This is one of those gotcha type questions. // Look at it another way. No amount of hydrogen was defined in the problem, yet your answer has units of $mol^{-1}$.  // So with an infinite amount of hydrogen you could dump an infinite amount of heat and still not get a temperature or pressure rise. There are a lot of these sorts of "fuzzy problems" that get weird if you dissect them too much.

To me I thought of the hydrogen like a brass cylinder which gets rid of pressure.  "The heat capacity of a brass cylinder at constant temperature is ___". Since I don't know how big the brass cylinder is, I can't determine its heat capacity (change in temperature per gram of brass).  But if I add any heat then the temperature has to change. So to keep the brass cylinder at the same temperature it can't lose or gain heat. So dH =0.

If you assume ideal behaviour and a closed system, then since the ideal gas law holds, with pressure and temperature remaining constant volume must also be constant.So that means that nothing is happening to the system, everything is as it is with $dq=0$ and $dw=0$.Since there is no process involved there is no meaning for the term heat capacity.

I'd answered 0. If you add heat then either (1) system does work to keep P and T constant (2) P rises (3) T rises (4) some combination of P and T rising. I never would have deduced infinity.

Heat capacity is not a property of a process. It is a property of a system: $C_{p,\mathrm{m}} = (\partial H_\mathrm m / \partial T)_p$.

@orthocresol but Heat capacity is not independent of the path.Here though the pressure is constant, since temperature is also the same,how should one define heat capacity-we could even claim it to be C_v instead of using C_p.That is what seems ambiguous to me.

In my humble opinion, your answer is correct and your teacher's answer is incorrect.  Heat capacity is an intensive property of a material, defined as you expressed it.  If the gas is ideal, then, if dT is equal to zero, dH is also equal to zero, so the ratio dH/dT appears to be indeterminate.  However the limit of $\frac{\Delta H}{\Delta T}$ as $\Delta T$ approaches zero is the heat capacity, and that is not indeterminate.

I hate problems like this because you have "read between the lines." "Heat capacity" is ill defined. Does it mean (1) "molar heat capacity" or (2) "specific heat capacity" or (3) "heat capacity" of some undefined system (amount of hydrogen.)  I took it to mean (3). So it doesn't matter how much hydrogen gas that the system has, if you add heat then at least one of the two parameters of pressure and temperature must change.

I agree.  The most precise definition of  $C_p$ that I am familiar with is $C_p=\left(\frac{\partial H}{\partial T}\right)_P$, but this doesn't specify whether if is per mole or per unit mass.  Anyway, in terms of the problem at hand, with a change of phase such as vaporization or melting, the enthalpy can change discontinuously at constant temperature and pressure, in which case the heat capacity of the material at such a point can be regarded as infinite (a Dirac delta function in temperature).

I think the answer infinity is correct.
We know that $c=\frac{\partial H}{\partial T}$. But, in your case, $\partial T$ is not $\to0$, in fact, $\partial T$ is said to be exactly zero. You might be tempted now to say that $c=\frac{\partial H}{\partial T}=\frac{\partial H}0=\infty$ but this is NOT the correct way to interpret this question (and neither is it mathematically valid). The correct way to interpret this question is that "a very large change in enthalpy causes a negligible change in temperature". Hence, $c\to\infty$ (a mathematical way to express "enormously large").
PS: This process is commonly taken to be the phase change process at the boiling point or the melting point. However, it is more complicated at that point than it seems, because the mass of the system also changes. At this point, we may have to thus use the specific heat capacity, $c=\frac 1m \frac{\partial H}{\partial T}$ which is defined per unit mass. But again, the specific heat isn't defined for processes in which mass changes. So, it is really tough to do any reasonable calorimetry at this point.

The question never says that there is a change in enthalpy, let alone  "a very large change in enthalpy ".

@DavePhD Indeed. I am not saying that there is necessarily an exchange in enthalpy. Certainly, the hydrogen gas can remain wherever it is without gaining or losing any enthalpy. I am saying that if there was a certain change in enthalpy, then the change in temperature wrt that enthalpy change would be almost negligible.

It seems like you're saying that if I have a glass of water sitting in my kitchen, at constant temperature and pressure, then the heat capacity of water becomes zero.

@DavePhD No. But if you do supply enthalpy to that glass of water, then its temperature will rise. By calculating $\left(\frac{\partial H}{\partial T}\right)$, you would be able to measure the heat capacity of that glass of water at that temperature. This is also what I am doing in my answer post above.

Heat capacity $C$ is not well-defined for a closed system undergoing constant temperature processes, at least with a reversible process. For a reversible process in a closed system,
$$\delta q_\pu{rev} = C\pu{d}T.$$
If temperature is constant, then $\mathrm{d}T=0$ gives $q_\pu{rev} = 0$ for any finite value of $C$. Other formulae for heat capacity, expressed through internal energy or enthalpy, require additional constraints such as constant volume or pressure, and only expansion work is allowed. Neither of these alternative depictions change the outcome.
This is not a problem because heat capacity does not make sense anyway if we assume a priori that temperature cannot change at all.

My attempt: Since $\ce{H2}$ is a diatomic gas, its $C_P$ is $7R/2$ which gives $\pu{7 cal mol-1 K^-1}$.

I would be fine with this answer as well. If heat is supplied at constant pressure to a closed non-reactive system, then the temperature must change. The alternative would be an change in volume to keep temperature constant but this would alter the pressure of the system.
In other words, in the net-constant pressure and net-constant temperature process dihydrogen had to warm up (according to its heat capacity at constant pressure), and then cool off again. It is not a good idea to equate the heat capacity here to infinity; that would make little physical and mathematical sense in my opinion.


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When calculating a dihedral angle for a chemical bond, how is the direction defined? What are the most common convention?
I understand the geometry and calculation of a dihedral angle, discussed in this question. It is not clear how the direction is defined for a chemical bond.  On the wiki page for dihedral angles, it states that, "the dihedral angle $\phi$ is the counterclockwise angle..." between the planes $P_{123}$ and $P_{234}$ (where $P_{123}$ is defined as the plane containing atoms 1, 2, and 3 in the diagram below, and likewise for $P_{234}$).
$\hspace{45 mm}$
This counterclockwise definition means that the direction of the dihedral angle is defined using the left-hand rule about the line connecting atom 2 to atom 3.  The left-hand rule is unusual in mathematics and geometry.  Alternatively, this example may be defining the directionality of $\phi$ simply so the result is a positive angle less than $180^\circ$.
I have not been able to find any chemistry standards on this.  Is there a common convention?

The IUPAC Gold Book states that,

In a Newman projection the torsion angle is the angle (having an absolute value between 0° and 180°) between bonds to two specified (fiducial) groups, one from the atom nearer (proximal) to the observer and the other from the further (distal) atom. The torsion angle between groups A and D is then considered to be positive if the bond A-B is rotated in a clockwise direction through less than 180° in order that it may eclipse the bond C-D: a negative torsion angle requires rotation in the opposite sense.

This implies a range of -180 to 180, which is by far the most typical range used, but I certainly have seen reports ranging from 0..360, e.g., 225 degree dihedrals.

When they report a negative dihedral angle, or an angle larger than 180, there has to be some way they are determining directionality.  Otherwise the results are ambiguous.

The link you provided does describe the orientation, "In a Newman projection the torsion angle is the angle (having an absolute value between 0° and 180°) between bonds to two specified (fiducial) groups, one from the atom nearer (proximal) to the observer and the other from the further (distal) atom. The torsion angle between groups A and D is then considered to be positive if the bond A-B is rotated in a clockwise direction through less than 180° in order that it may eclipse the bond C-D: a negative torsion angle requires rotation in the opposite sense."

I am not familiar with the IUPAC Gold Book but it seem valid.  According to this definition, the wiki page is incorrect and the torsion angle is defined using the right-hand rule along the line connecting atoms 2 and 3.

My point is that despite the Gold Book, different software (or people) report different references.

This is good to note.  I think the most important point is that they use the same directionality (but maybe this is not the case).  If a positive angle is always defined using the right-hand rule about the B-C line (simplified version of the IUPAC Gold Book), converting from -60 to +300 is straightforward.  Without a common orientation, anything but 0-180 is ambiguous.

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How do I calculate principal quantum numbers of the Lyman series
How do I calculate principal quantum numbers of the Lyman series?
I've been given that the lowest line is 121.6nm and the 5th line is 93.8nm
The formula of the wavenumber is   v=-R(1/n2^2 - 1/n1^2)
My taught was that as the wavenumber is known (1/wavelength) I could create an equation
v/-R = 1/(x+4)^2 - 1/x^2   and solve for x to get the principal quantum numbers
However this equation seems to be too hard to solve and I guess there is a different way to solve this.
Many thanks in advance

I know it's not a "good" answer, but since $n_1$ and $n_2$ are both small positive integers, trial and error works pretty well.

Lyman series, by definition, is a series with $n_2=1$, so you have to find just the $n_1$, which is not that hard.

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Why does soup turn purple when lemon juice and salt are added?
Well you have probably heard about salpeter. I don't know if it is a myth or not that it is used to reduce men's libido by adding to food especially in boarding schools and remote military campuses.
I didn't know that before but because I studied at boarding school, this topic and its alleged test intrigued me. According to the test, when you add lemon juice and salt to food and if the salt in the food turns to purple/blue, it means salpeter is added to the food. I don't know if it is scientifically proven.
Now I live with my parents but just for fun I wanted to try to test on my homemade food and I shocked because the salt I added to the soup, which was made by mom, turned to be  purple /blue. I've become really baffled and I checked if there is any salpeter in cupboards I found a bottle of white like flour thing. I presumed it was powdered salpeter and tested it with the same method, which I added lemon juice and salt and it turned blue/purple.
But when I asked what that white thing is, she said it was "powdered sugar".
and I tried the same test on new powdered sugar yeah it turned to be blue too.
I hope you are not confused. So far what I am sure about is that when I add salt and lemon juice to powdered sugar it turns to be blue.
My mom said there is neither salpeter nor powered sugar in the soup she made.
Buy how come does soup turn to be blue /purple when I added some lemon juice and a lot of salt?
Here is a picture.The soup is made of lentil, some spices and salt, some potato and carrot.

Your claims about saltpeter reducing libido are a myth according to this McGill University page.

A possible answer:
(1) Common table salt is usually "iodized" - i.e. it contains potassium iodide.
(2) Something in the soup could oxidize the iodide to elemental iodine.
(3) Iodine plus starch (from the potato in the soup) gives a deep blue color.

Could the lemon juice by itself oxidise potassium iodide to iodine?


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Conservation equation of reaction scheme
Consider the reaction scheme:
$$\ce{S + E ->[k_1] C1} \qquad \ce{C1 ->[k_2] E + P} \qquad \ce{S + C1 <=>[k_3][k_4] C2}$$
where $\ce{S}$ is the substrate, $\ce{E}$ is the enzyme, $\ce{P}$ is the product, $\ce{C1}$ and $\ce{C2}$ are enzyme substrate complexes. Let $[\ce{S}] = s$, $[\ce{E}] = e$, $[\ce{C1}] = c_1$, $[\ce{C2}] = c_2$ and $[\ce{P}] = p$ be the concentrations of each respective chemical. I have used the law of mass reaction to convert this reaction into a system of differential equations:
\begin{align}
\frac{\mathrm{d}s}{\mathrm{d}t} &= -k_1se-k_3sc_1+k_4c_2\\
\frac{\mathrm{d}e}{\mathrm{d}t} &= -k_1se+k_2c_1\\
\frac{\mathrm{d}c_1}{\mathrm{d}t} &= k_1se-k_2c_1-k_3sc_1+k_4c_2\\
\frac{\mathrm{d}c_2}{\mathrm{d}t} &= k_3sc_1-k_4c_2\\
\frac{\mathrm{d}p}{\mathrm{d}t} &= k_2c_1
\end{align}
I notice that
$$\frac{d}{dt}(e+c_1+c_2) = 0.$$
Hence $e+c_1+c_2 = const.$ Now is this a conservation equation? Also how do I use this equation to simplify the system?


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Why does a reaction stop when turning into a solid?
I recently was looking at a question about the volume of carbon dioxide produced when reacting calcium carbonate with dilute sulfuric acid versus using nitric acid. In the answer it says that calcium sulfate is produced, which is a solid, which stops the reaction, but I don't get why. Calcium nitrate produces an aqueous solution which causes the reaction to carry on reacting.
The reaction equations are:
$$\ce{CaCO3(s) + H2SO4(aq) -> CaSO4(s) + H2O(l) + CO2(g)}$$
$$\ce{CaCO3(s) + 2HNO3(aq) -> Ca(NO3)2(aq) + H2O(l) + CO2(g)}$$

The reason is due to the surface area of calcium carbonate available to react.
When reacting a block of solid calcium carbonate with dilute sulphuric acid, solid calcium sulphate is produced which is deposited on the surface of the calcium carbonate block (since that is where the reaction occurs). The calcium sulphate is insoluble in dilute sulphuric acid and so you end up with calcium carbonate coated in a layer of calcium sulphate, which stops the acid from getting at the carbonate and so stops the reaction.
In contrast, calcium nitrate, produced from the reaction with nitric acid, is highly soluble in dilute nitric acid and so no solid layer forms around the carbonate block and the reaction can continue until one of the reagents is used up.

Reactions such as these take place in an ionic form where Ca2+ and CO3 2- are formed in the aqueous medium and react with H2 2+ and SO4 2-. When CaSO4 is formed, it precipitates as it is insoluble. Thus it covers the surface of the CaCO3 and the reaction stops.
In the second reaction CaNO3 is not insoluble so the reaction continues and CaNO3 solution keeps being formed.
Hope this answers your question!

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Why can't Al3+ be reduced to solid, pure aluminum?
There are efforts underway to develop ionic liquids for the purpose of reducing bauxite using exotic intermediates such as aluminum fluoride. However, the aqueous $\ce{Al^3+}$ ion is tantalizingly close to pure aluminum. What prevents it from being reduced?

Nothing prevents it from being reduced other than the conditions of the aqueous solution (electric potential, pH, and so on): Are you asking about the more specific case of why aluminum metal is not produced this way commercially and is instead produced (usually) by the Hall-Heroult process?

@ToddMinehardt Yes, I am interested in why aluminum is not produced this way.

It is aluminium’s standard electrode potential that prevents it from easily being reduced. For the half-cell $\ce{Al^3+ + 3e- <=> Al(s)}$, the standard potential is $E^0 = -1.66~\mathrm{V}$. For comparison a few other values:

$\ce{Fe2O3 + 3 H2O + 2e- <=> 2 Fe(OH)2(s) + 2 OH-}$ and $\ce{Fe(OH)2 + 2 e- <=> Fe(s) + 2 OH-}$ have potentials of $E^0 = -0.86~\mathrm{V}$ and $-0,89~\mathrm{V}$, respectively;
$\ce{Zn^2+ + 2 e- <=> Zn(s)}$ has a potential of $E^0 = -0.76~\mathrm{V}$;
$\ce{Cr^3+ + 3 e- <=> Cr(s)}$ has a potential of $E^0 = -0.74~\mathrm{V}$;
$\ce{Co^2+ + 2 e- <=> Co(s)}$ has a potential of $E^0 = -0.28~\mathrm{V}$;
$\ce{Cu^2+ + 2 e- <=> Cu(s)}$ has a potential of $E^0 = +0.34~\mathrm{V}$.

(All values taken from Wikipedia)
The higher the standard potentials are, the easier the ion can be reduced. Hydrogen, of course, has a potential of $E^0 = 0~\mathrm{V}$ at pH 0 by definition. So in acidic media, hydrogen is reduced first. So let’s lower the pH to impede the formation of hydrogen. According to the Nernst equation:
$$E = E^0 + \frac{0.059~\mathrm{V}}{z} \lg \frac{a_\mathrm{Ox}}{a_\mathrm{red}}$$
Where $z$ is the number of electrons transferred in the half-cell, and $a$ is the activity of the reduced/oxidised species. $a_\mathrm{red}$ is $1$ by definition and $z = 2$ for the hydrogen half-cell. So we need to solve the following simplified equation to determine the acid concentration that will allow reduction of aluminium:
$$-1.66~\mathrm{V} = 0~\mathrm{V} + \frac{0.059~\mathrm{V}}{2} \lg \frac{[\ce{H+}]}{1}\\
\frac{0.059}{2} \mathrm{pH} = 1.66 \\
0.059 \times \mathrm{pH} = 3.32 \\
\mathrm{pH} = 3.32 / 0.059 = 56$$
Whoops, so we would need a pH of $56$ to allow the reduction of aluminium in the presence of water. This calculation was highly theoretical, because:

A pH of 56 is impossible in anything that once could have been water. Wikipedia gives the $\mathrm{p}K_\mathrm{b}(\ce{O^2-}) \approx -38$; because $\mathrm{p}K_\mathrm{a}(\ce{HA}) + \mathrm{p}K_\mathrm{b}(\ce{A-}) = 14$ in water that must mean that $\mathrm{p}K_\mathrm{a}(\ce{OH-}) \approx 52$ — thus we would be dealing with an ‘oxide solution’ rather than water.
Although it is nice that both equations (pH and Nernst) use acitivities, we cannot really say anything about the activities at this large a concentration range.
The redox system of both species would be totally different in these environments.

Note that you can reduce most air-stable metals in thermite reactions from their oxides thereby creating aluminium oxide. An example reaction would be:
$$\ce{Fe2O3 + 2Al -> 2Fe + Al2O3}$$
That alone should show the unwillingness of aluminium to be reduced in aquaeous solutions; think about the heat that the thermite reactions generate.

I thought that maybe the Al+3 ion would be not bonded to oxygen. Is Al+3 surrounded by water ligands or is it an oxide? MaxW's answer suggests that reducing aluminum may not be the problem, but preventing the reduced aluminum from being oxidized. This makes me wonder what the ionic liquid activity coefficients for the nernst equation must be for aluminum reduction.

@Dale For the standard potential, I believe we are dealing with $\ce{[Al(H2O)6]^3+}$. But it doesn’t really matter regardless, because the point is water (hydrogen) will be instantly reduced by aluminium metal.

This answer misses the fact that there is an https://en.wikipedia.org/wiki/Overpotential for the electrochemical evolution of a gas at a metal surface. I've not found a number for H2 at Al, though. The theoretically necessary pH level would however be quite a bit lower. Still impossible, of course.

Works for sodium, for example: https://commons.wikimedia.org/wiki/File:Einzelpotentiale_des_Amalgamverfahrens.svg , due to amalgamation AND overpotential.

@Karl Amalgamation introduces an entire additional layer of complexity to the question.

According to Wikipedia "elemental aluminium cannot be produced by the electrolysis of an aqueous aluminium salt because hydronium ions readily oxidize elemental aluminium." So the aluminium oxide is reduced via the Hall–Héroult process.
The alumina itself is produced from mined bauxite via the Bayer process.
So the gist is that industry spends many many millions of dollars converting bauxite too aluminium metal. No doubt they do it the cheapest way possible.

So it is not the Al+3 that can't be reduced, it is the solid aluminum oxide coating that forms AFTER the Al+3 has been reduced (assuming non-aluminum electrodes).?

In general it wouldn't matter what metal the "initial" electrode was made from. Once the electrode gets coated with Al metal then it would form an oxide coating which would "ruin" the electrochemical reaction. The only exception would be a Hg electrode. The Al would dissolve in the Hg rather than forming a coating. But you couldn't get a lot of Al in Hg, and cycling the Hg would be a problem. So that isn't practical.

Mercury does some damage to aluminum. https://www.youtube.com/watch?v=Z7Ilxsu-JlY   Also, see: https://en.wikipedia.org/wiki/Aluminium_amalgam

In addition to the Hall-Héroult process for electowinning aluminum, aluminum can be deposited from chloroaluminate melts, which are made by combining aluminum chloride with an organic or inorganic chloride. Such melts work at lower temperatures than the Hall-Héroult process. This reaction is used to electrochemically measure aluminum content in molten zinc during hot dip galvanizing (aluminum is added to control the reaction between the zinc coating and the steel).

authors: Саркома Капоши

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Dichloromethane cleaning
Is it safe to drink out of a bottle that use to contain Dichloromethane?  What should I use to clean a bottle of Dichloromethane?  Is there any chemicals that I could use to dissolve the Dichloromethane and make it safe to drink out of?
More info:
So if I had a glass bottle containing >= 99.5% Dichloromethane with amylene as a stabilizer and left out overnight is that safe? The bottle was used in a chemistry department so it is high quality, I would assume. Made from the company Sigma-Aldrich and in a Glass bottle.

Why on earth would you want to drink anything out of a bottle that used to contain Dichloromethane!?

To extend Philipp’s comment: Why on earth would you want to drink anything out of any bottle that used to contain any chemical?!?

No. Don't do that.

@Jan I regularly drink out of my bottle that used to contain various water-soluble organic compounds. ^^

@orthocresol Is that a coke bottle? xD I should have specified ‘a chemical not sold for food or drink’.

Perhaps OP can indicate what his motivation is, glass is not ideal for a portable lunch bottle but perhaps there is a good reason to inquire.

Safety regulations in every lab I have heard of strictly disallow mixing of containers of any sort that contain, have contained or will contain chemicals with containers of any sort that contain, have contained or will contain food, beverages or anything else intended for human consumption. You should never drink out of a bottle that had contained chemicals, you should never store food or drinks and chemicals together and you should not consume anything where there are chemicals. Your question has a premise that should never happen due to safety regulations.
With that gotten out of the way the answer is quick and dirty: It’s not the dichloromethane that is your prime point of concern. Dichloromethane boils at $40~\mathrm{^\circ C}$ and if you keep a bottle that contained it open and lying down overnight at room temperature, there are no traces of dichloromethane in harmful concentrations left. But unless you really go out of your way to buy $99.999~\%$ purity dichloromethane, it will contain impurities which generally do not evaporate as easily, i.e. which will stay in the bottle. Since no bottle of dichloromethane is designed for drinking out of it, there are no safety tests performed with these impurities or additives and thus you cannot assume it to be harmless even if dichloromethane is gone.
You should not read this answer to say that bottles of highest-purity dichloromethane be safe to drink out of. Never mix food or beverages and chemicals. Never, ever!

Amen. Such recklessness is just asking for trouble.

So if I had a glass bottle containing >= 99.5% Dichloromethane with amylene as a stabilizer and left out overnight is that safe?  The bottle was used in a chemistry department so it is high quality, I would assume.   Made from the company Sigma-Aldrich.

@CarlThon Really?!?!  I would politely suggest rereading the post you're commenting on.  It repeatedly states that this is a terrible idea.

@CarlThon Which part of ‘no!’ don’t you understand, the n or the o?

@Jan There have been safety tests done with the additives AND the dichloromethane still there, it does not kill on sight and if you remove the 99.9995% (see link in answer below) of the chemical mix it is unlikely to become more toxic.  OP is discussing an empty bottle and not large doses of unknown residue.

But what if I reeeeeeeally want to?!

@KalleMP "Does not kill on sight..." Are you for real? Cancer isn't usually an instantaneous death. Dimethyl mercury doesn't kill on sight either.

@long Yes I am for real. The cancer threat has not even been observed in humans who work with the stuff daily, only in force fed lab rats.  Dimethyl Mercury is a different chemical even though the first bit sounds the same.  If you read my answer below you will see what I base my comment on.  I am all for chemical safety but object to scaremongering and over regulation as a result that causes pointless social costs. We wash dishes in a dishwasher but would not dream of eating a detergent pill.

There are chemicals that kill and some that don't, learn to differentiate and how to read MSDSs.
Try not to mix random chemical bottles with food containers for a lot of reasons, mostly because it might kill someone (you) due to compounded mistakes.
As for this scenario I would say sure.
A typical residue on evaporation for this type of product is <0.0005% that is 5 parts per million. If you have an empty 1000ml bottle you might have 2ml of remaining product.  If you let the left over product in your empty bottle evaporate (bottle mouth down is better due to drainage and density of vapour) you would have 0.000005 x 2 = 0.000010g or 10ug or residue left. If you somehow were to dissolve in water and ingest all the residue and it was very toxic you would be getting 10ug total dose that is made of all the stuff that is non-volatile.
In any event the 2ml (near 2700mg) of product left in an empty bottle would be enough to kill half of a population of four 500g rats if NOT evaporated. By simple (not accurate) extrapolation a 70kg adult would need to drink 100g (75ml) of product to have a 50% chance of dying, instead we are talking about the residue from 2ml (2.7g) product with all the poisonous stuff gone.
You probably don't even need to rinse the bottle, though that is what I would do (using clean acetone if I was feeling paranoid, our bodies create acetone and know how to metabolise it in small quantities so don't fret) both solvents will evaporate to dryness (and be clean for almost any use), what is left may be less toxic than the residue from dish-washing liquid.
Also note it is used in some extraction processes in the food industry an produced in some way during the chlorination of municipal recycled drinking water.  It has been linked with lab rat cancer when administered in significant doses but not in humans working with it over long periods.
Please note that some chemicals are not this easy to avoid and may cause problems down the line even with low doses. I believe sugar and prescription drugs are a bigger concern than Dichloromethane residue.
Please down vote only if you have read the linked documents and have made an informed comment about Dichloromethane evaporation residue toxicity so people know where I have made errors, I will happily correct any brought to my attention.
EDIT:
Hmm, down votes without useful reasons.  If you look up the toxic doses for caffeine, ethyl alcohol, nicotine, dish washing liquid, dishwasher tablets, quinine (tonic water), cream of tartar, acetic acid and too many other 'acceptable' chemicals then you may find they are worse than dichloromethane and don't evaporate as well.
What gives, did you people not actually read this answer before down voting?
Is fear mongering so much fun that you have to remove your own judgement from your own actions. One day you may actually need to decide which of two chemical bottles you need to carry water across the desert to save your life, are you going to pick the right one or are you going to say, yuk, chemicals and walk to your death from dehydration.

"The bottle was used in a chemistry department so it is high quality, I would assume." I don't those aren't the words of someone who knows the minor contaminants of what they're using.

@DGS  My point is that with such very pure chemicals evaporated to dryness the original product is essentially gone and the fraction of minor contaminants is so small that it would be hard to detect.  You will also find that if you have toxic contaminants that are somehow worse that the main product they need to be listed in an MSDS as well even if very small fractions are somehow dangerous.

As for the high quality I believe OP is looking at the manufacturing quality of the Sigma-Aldrich bottle and wants it in his lunch box to make people leave his lunch alone.

And at the end of the day, someone needs to put away some dichloromethane and stumbles across your bottle … Sorry, $-1$ for unsafe working.

I feel the need to explain myself, since a) there was only a single downvote and b) it was me. I will agree fully that most of the evidence you gave makes drinking out of a well-vented DCM bottle rather safe (and so my lab experience would say). But out of general safety considerations especially when considering that part of the audience is chemically uneducated folk, I cannot approve of any answer that does not stress the safety aspect much much more. Your answer reads far too much in the vein of ‘yeah, go ahead.’

I also greatly disapprove of anecdotical desert scenarios that will never come true.

@Jan Thanks for explaining, there were two down votes so you were not alone, I up voted my own answer in protest when it was still at -1.  My edit was because I had rather politely asked people to substantiate their down votes and have to make the assumption that down voters don't bother to read in full, your clear (thought I disagree) explanation is all that was needed to justify your down vote.

You can't upvote your own answer.

@orthocresol it seemed to work ok for me.

I can see vote counts, and believe me: There is exactly one vote (down) on your question (my own). Also, what ortho said, you cannot vote on your own answer neither up nor down.

@Jan, ok, my memory is playing tricks with me then.

@Jan correct when he wrote his most recent comment.

Please read the original question before downvoting answers. Please keep Stack  Exchange factual and not nanny state.


text:
Finding Hybridization of Certain Atoms in Lewis Structure

So, I'm having trouble with this problem. We're given this incomplete lewis structure and asked to find the hybridization of 3 of the atoms.
I figured that the labelled Cl would be sp3 because of it's 1 single bond and the 3 lone pairs that would be on it.
I also thought the labelled C would be sp2 because of the 2 single bonds and 1 lone pair.
Now, I was confused at to what the labelled O would be. I thought maybe it would have 3 lone pairs but that doesn't seem to be correct, since my sp3 answer was rejected.
Any insight on what I'm doing wrong here?

A lone pair on C would be quite an unusual thing to find. Consider double bonds instead. Yes, some of these bonds are double, and they won't tell you which. But you may guess...

You should never consider halogens hybridised in any way.

Still, if confronted with a question like this one, go with $sp^3$ for Cl, because that's what they want from you.

Well, if you see valency of Oxygen is 2, but its making only 1 bond. Thats  enough for hint, and as @IvanNeretin said "some of these bonds are double, and they won't tell you which. But you may guess..."


text:
How to calculate the yield for an incomplete reaction?
I had a doubt in part b) of the following question.
a) When 0.527 g of platinum is heated in fluorine, 0.732 g of a dark red volatile solid forms. What is the empirical formula of this product?
b) Suppose fluorine was a limiting reagent, such that 0.155 g of platinum remains after all the fluorine has been consumed. Calculate the percent yield of this reaction. How many grams of product would be produced?
I've given part a) also because it was told that we should use data from a).
This is my attempt:

The answer for b) is given as 70.6% and 0.517 g, but I am unable to find where I am wrong. Is my logic correct? (I've double checked that my calculations at least are correct.)

Please format your question to include the information contained in the photo in a typeset format. That will enable others to find the information contained therein and make a more valuable contribution to the site.

Well the problem doesn't say anything about how to get the % yield. If you start with 0.527 grams of platinum, but have 0.155 grams left, then 0.372 grams reacted.
$\frac{0.372}{0.527} * 100$% = $70.6$%
As for the yield
$\frac{0.372}{0.527} * 0.732 = 0.517$ grams

OK, but where did I go wrong?

To start look at your yield calculation. Yes $0.732g - 0.155g = 0.577g$ but what makes you think that you can subtract $0.155 g$ $\ce{Pt}$ from $0.732 g$ $\ce{PTF4}$ ?!? // It like an orange weighs 8 ounces and is 3 inches in diameter. Therefore it has a measurement of 5 (8-3).

Yeah, I think the actual yield is the main issue. I used a reaction table and the law of mass conservation to arrive at that.

For the complete reaction you do use 0.00540 moles of fluorine gas ($\ce{F2}$) as a reactant, but you only get 0.00270 moles of platinum tetrafluoride ($\ce{PtF4}$) which does have a molecular weight of 271.1 g/mol. // How could you get 1.46 g of product for the 100% conversion when the problem tells you that you got 0.732 g of $\ce{PtF4}$ ?!?

So, the theoretical yield is supposed to be 0.732 g. Isn't the percent yield formula based on the amount of product formed? Is there no way of finding the actual yield first and then plugging it into the formula to find the percent yield?

Of course. If you start with 0.527 grams of platinum, but have 0.155 grams left, then 0.372 grams of platinum reacted. From that you can figure out how many moles, then use molecular weight of platinum tetrafluoride to get grams of product.

This question should have been clearer. It's so confusing what data from part a) we're supposed to use!

@Leponzo - This problem is quite clear and unambiguous. Keep working on understanding the chemistry and don't just focus on learning equations. You'll get the hang of it.


text:
How to rationalise the planar structre of I2Cl6?
$\ce{I2Cl6}$ is a dimer of $\ce{ICl3}$. $\ce{ICl3}$ is T-shaped, with $\ce{I}$ being $\mathrm{sp^3d}$ and having two lone pairs.
I read in Inorganic Reaction Mechanisms by R. K. Sharma that $\ce{I2Cl6}$ is formed by the extra two $\ce{Cl-I}$ bridge bonds. What I don't understand is, why does the bridge bond form? And is it a coordinate bond or something like a banana bond (as in $\ce{B2H6}$)?

Source: Sharma, R. K. Inorganic Reaction Mechanisms; Discovery Publishing House: 2007.

The bonding situation in the compound $\ce{(ICl3)2}$ is by far more complex than what is depicted in this book. The molecule itself has very high symmetry, i.e. $D_\mathrm{2h}$, that needs to be satisfied in the molecular orbital/ valence bond description.[1,2] That is why one cannot distinguish between the bridging chlorine bonds.
Analysing this bonding situation in the simple concept of Lewis-structures is not trustworthy and will lead to wrong conclusions. I have performed a short gas-phase optimisation at the DF-BP86/def2-SVP level of theory and analysed the resulting wave function with the Natural Bond Orbital theory. Covalent treatment of any of the bonds seems to be wrong. In this case I assumed all single bonds. Due to symmetry reasons, we only need to be concerned with two bonds. The outer $\ce{I-Cl}$ bond is polarised 93% towards the chlorine, but only occupied by 1.25 electrons. The remaining 0.75 electrons can be found in anti-bonding lone pair orbitals. This is not even close to expected Lewis structure behaviour. The bridging $\ce{I-\mu{}Cl}$ bond is still 81% polarised towards the chlorine, and has an occupation of 1.92 electrons. This only tells us, that this bond has a higher covalent character.
The analysis with the Quantum Theory of Atoms In Molecules (QTAIM) gives us more tools to analyse the electron density. In here the picture is quite obvious. All of the bonds in the molecule are of highly ionic character. The electron density at the bond critical point of the outer $\ce{I-Cl}$ is with $0.08~\mathrm{e\,a.u.^{-3}}$ very low and the Laplacian is positive at this point. The electron density at the bridging bond is even lower with $0.04~\mathrm{e\,a.u.^{-3}}$, again with a positive Laplacian.

(Laplacian distribution, solid blue lines indicate charge depletion $∇^2ρ<0$, dashed blue lines indicate charge accumulation $∇^2ρ>0$, red spheres are bond critical points, pink spheres are ring critical points, black lines are bond paths, dark red lines are zero flux surfaces.)
This finding is fairly consistent with the observation, that the melt of $\ce{(ICl3)2}$ conducts electricity, which according to Wikipedia is a hint for partial ionic dissociation.
$$\ce{(ICl3)2 <=> ICl2+ + ICl4- }$$
However, the planarity of this compound in solid state can be attributed to the possibility of having a somewhat quite extensive pi-bonding system as part of the small covalent bonding. For final conclusions about this, the results are not sufficient.
What can be said, judging from the molecular orbitals of $\ce{(ICl3)2}$, is that apart from the occupied core d-orbitals of iodine, no other d-orbitals are involved in any bonding - as it is expected. Shown here are the occupied non-core orbitals of $\ce{(ICl3)2}$. It is noteworthy, that the HOMO is fully antibonding.[3]

Finally a short warning:
While the assumption made by VSEPR may lead to the approximately correct conclusion for the interhalogenide compounds $\ce{XF3}$, where $\ce{X$=$Cl, Br, I}$, that the molecules are T-shaped, it is doubtfully the correct conclusion for the much less electronegative chlorine. This can be mainly explained by Bent's rule. Even in the monomeric compound the bonding situation is somewhat more complicated than you would expect.

Notes and References

The point group $D_\mathrm{2h}$ specifies that there is a main symmetry axis $C_2$ which passes through the centre of the ring. Additionally there are also two $C_2$ axis perpendicular to this and a horizontal mirror plane $\sigma_\mathrm{h}$. This means the molecule is planar. There are also non-planar molecules that can be described by this point group.)
As Max pointed out (Thank you), the structure is taken from the solid state. K. H. Boswijk and E. H. Wiebenga,  Acta Cryst. 1954, 7, 417-423. It mentions that it was purified by sublimation. This is no absolute proof of a gaseous dimer, but at least very suggestive. For our purposes we assume that the gas phase calculations are sufficient.
Orbitals obtained at the DF-BP86/def2-SVP level of theory with Gaussian09 rev. D. The energy increases row-wise from left to right and from top to bottom. The last shown structure is the molecule itself.

In the first paragraph I'd add (1)  the dimer is only in solid form. (2) $D_{2h}$ symmetry is planar for those who don't understand the point group symmetry notation. // "The bonding situation in the solid state of the compound...." and "... $D_{2h}$ (a planar symmetry) ..."

@Max There is no conclusive evidence, that the dimer only exists in solid state. During my literature query I did not come across an article which investigates other states of matter. It is probably highly unstable towards decomposition. The low level calculation, however, predicts a local minimum on the potential energy hypersurface, so it could exist in gas phase - if only kinetically stable. In any state of matter the bonding situation will be more complicated than it is suggested by the book. I need to think about this a bit more before I make any edits.


text:
Significant figures when standard deviation is 0
For analytical chem lab, our group did a titration in which we got the same exact value every single time. As a result, our standard deviation was 0 and our instructor says to report answer using real rule of sig figs (our last significant figure should be the first digit in the uncertainty measurement) I am not sure what to do as my STDev doesn't have any significant digits?
edit: Molarity is the measurement in question!

How many titrations? What readings did you get? How is burette marked? How many significant figures are you reading on burette? I wonder if  you technique is bad, or if you have so cheap student grade burettes.

I'm one of the top research universities (hint public) for chemistry in the country, the burets are surely calibrated as this is an upper level chemistry course. We did 4 trials, Moles CaCO3 Volume EDTA added corrected for blank molarity EDTA
0.0009975 33.7 33.39 0.029874214
0.0009975 33.7 33.39 0.029874214
0.0009975 33.7 33.39 0.029874214
0.0009975 33.7 33.39 0.029874214

You can be pretty sure that no matter where you go, how well funded or unfunded schools are, the burettes will be calibrated to something. That does not mean they are all accurate. The precision of the instrument should be printed on it (or you can find it in the manual). Your standard deviation cannot be smaller than the actual precision of the instrument. I would also like to encourage you to include your comment into the post with proper formatting. In the way it is presented now it is hardly comprehensible.

Related: Uncertainties after Averaging Trials

@camerobn - I think 33.7 were your burette volumes. Correct? Using 50 ml burette with 0.1 ml marks?

Various sources of error are inherent in measurements. The so-called Type A evaluation, i.e. the statistical analysis of series of observations (e.g. calculation of the standard deviation and the standard error of the mean), can be used to evaluate the uncertainty arising from random effects.
However, this evaluation does not include the uncertainty arising from systematic effects. For example, if the concentration of the titrant (the solution of a known concentration that is added during the titration) has a small unknown error and the same titrant is used for all titrations, each of the individual results includes the same systematic error. Or, if the burette has a small unknown error and the same burette is used for all titrations, again each of the individual results includes the same systematic error. You can neither eliminate such errors nor reduce the caused uncertainty by making additional trials, and you cannot evaluate the caused uncertainty by using statistical analysis (i.e. by calculating the standard deviation and the standard error of the mean) of the results. The uncertainty arising from such systematic effects has to be evaluated by other means. The so-called Type B evaluation is usually based on scientific judgment using all of the relevant information available on the possible variability of the results. Finally, you can combine all calculated or assumed uncertainties by using the rules for error propagation.
For example, the uncertainty arising from the unknown error of the burette may be estimated using the manufacturer’s specifications as follows:
Usually, you estimate (interpolate) the measurement between the graduation divisions of the burette. You might be able to estimate the measurement to one tenth of the smallest division. However, volumetric instruments, such as graduated cylinders, volumetric flasks, bulb pipettes, graduated pipettes, and burettes, – like any other measuring instrument – have a limited accuracy. The dimensions, graduating divisions, and accuracy limits of volumetric instruments are standardized. For example, the requirements of burettes are specified in the international standard ISO 385. Two classes of accuracy are specified: Class A or AS for the higher grade and Class B for the lower grade. The nominal capacity, graduation divisions, and accuracy limits are shown in the following tables.
$$
\textbf{Burettes, Classes A and AS} \\
\begin{array}{lll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} \\
\hline
\hphantom{00}1 & 0.01 & 0.006 \\
\hphantom{00}2 & 0.01 & 0.01 \\
\hphantom{00}5 & 0.01 & 0.01 \\
\hphantom{00}5 & 0.02 & 0.01 \\
\hphantom{0}10 & 0.02 & 0.02 \\
\hphantom{0}10 & 0.05 & 0.03 \\
\hphantom{0}25 & 0.05 & 0.03 \\
\hphantom{0}25 & 0.10 & 0.05 \\
\hphantom{0}50 & 0.10 & 0.05 \\
100 & 0.20 & 0.10 \\
\hline
\end{array}$$

$$
\textbf{Burettes, Class B}\\
\begin{array}{lll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} \\
\hline
\hphantom{00}1 & 0.01 & 0.01 \\
\hphantom{00}2 & 0.01 & 0.02 \\
\hphantom{00}5 & 0.01 & 0.02 \\
\hphantom{00}5 & 0.02 & 0.02 \\
\hphantom{0}10 & 0.02 & 0.05 \\
\hphantom{0}10 & 0.05 & 0.05 \\
\hphantom{0}25 & 0.05 & 0.05 \\
\hphantom{0}25 & 0.10 & 0.10 \\
\hphantom{0}50 & 0.10 & 0.10 \\
100 & 0.20 & 0.20 \\
\hline
\end{array}$$
For example, a burette of Class AS with a nominal capacity of 50 ml has graduation divisions of 0.10 ml. By interpolation, you might be able to estimate the measurement to one tenth of the division, i.e. to 0.01 ml. However, the accuracy limits of the burette are ±0.05 ml (at any point on the scale).

As I remember it the burettes were rated as "accuracy was at least" the quoted figure.


text:
How much CO2 is produced in burning of a candle?
I've been reading this book, page 149. In chapter 15 the following experiment is suggested,

Fix a candle in a container. Fill the container with water. Light the candle and then cover it with an inverted glass.

It is said that the oxygen in the space of inverted glass will be used up and to fill the space that was previously occupied with oxygen the water level will rise.
My question is, when the wax burns since $\ce{CO2}$ is emitted, won't this $\ce{CO2}$ fill up the space of used up $\ce{O2}$? What amount of $\ce{O2}$ is used up and what amount of $\ce{CO2}$ is released in the burning of wax of candle?

Surely $\ce{CO2}$ will fill some of the space of used up $\ce{O2}$. Candles these days are commonly made out of paraffine, which is basically $\mathrm{C_{many}H_{many}}$. The equation may look as follows:
$$\ce{C31H64 + 47O2 -> 31CO2 + 32H2O}$$
You see that we get approximately 2 moles of $\ce{CO2}$ for 3 moles of spent $\ce{O2}$, hence the decrease in volume.
Another factor is that $\ce{CO2}$ is much better soluble in water than other components of air, so over time (though not immediately) it will probably vanish from the glass and thus free some more space.
Yet another factor is that air in the glass is quite hot while the candle is burning. As soon as it cools down, it shrinks.

Are the 32 $\ce{H_2O] in liquid form or vapor form?

Initially in vapor form, but mostly turns to liquid pretty soon.

Another factor is that the combustion is incomplete, so there is even less $\ce{CO2}$ produced per mole of carbon atoms in the candle.

This factor does not necessarily work in favor of the effect in question. Incomplete combustion may end in C (soot) and produce less gases per 1 mole of used $\ce{O2}$, or it may go to $\ce{CO}$ and thus produce more gases.


text:
Gas Pressure and Molecular Mass
The ideal gas equation (or the van der Waals equation, for this matter) states that holding volume and temperature constant, increasing the number of moles of particles in the container will increase the total pressure exerted by the gas mixture.
That’s all well and good, but when it comes to looking at chemical reactions and their equilibria, Le Châtelier’s Principle states that when a gas mixture is pressurized, the equilibrium shifts to the side of the reaction that produces a lesser total number of moles. This would mean less particles, and therefore less pressure, as I see from the ideal gas equation.
What I do not understand is why the mass of each molecule does not affect the pressure exerted by the gas. If you have heavier molecules, they would each have more momentum, and would exert more force on the container’s walls—i.e. more pressure. So is this not a factor when considering Le Châtelier’s Principle? The equilibrium shifts toward the formation of more particles, but they are each lighter and exert less force. Why is this favorable over having less but heavier particles?

Heavier molecules would fly slower so as to exert exactly the same pressure as lighter ones.

The equipartition principle applies to energy rather than momentum. The heavier molecules have lower values of velocity-squared (rather than simply having lower velocities if it were momentum to which that equi-partitioning were occurring.) Since most chemical reactions have both forward and backward rates that are going to be "driven" by energy as well, I would guess that other concerns will be dominant. The smaller volume taken up by bonded atoms appears to be that "other issue".

The Maxwell–Boltzmann distribution is based on the principle that the distribution of kinetic energies of gas molecules is the same. So in a mixture of $\ce{N2}$, $\ce{N2}$, and $\ce{He}$, the molecules of each gas would all have the same KE distribution. However since He is much lighter, its molecules would have higher velocities.
Normally we'd think of the gas mixture as being homogeneous in a liter bottle in the laboratory. However if we consider the atmosphere then gravity has an effect. So helium will drift upwards in the atmosphere since it is less dense. In fact at the edges of the outer atmosphere a helium atom can get enough velocity to escape earth's gravity. So all helium that leaks into earth's atmosphere is ultimately lost into space.

Root mean velocity and most probable velocity, both vary inversely with M^1/2, therefore heavier particles will move slower, keeping momentum constant.


text:
Titration curve of a diprotic acid with a strong base titrant
I have wonders regarding the titration curve of the titration of sulphuric acid, or any diprotic acid, with a strong base. I am aware of the fact that a diprotic acid is protolysed in two steps but, how does this affect the titration curve?

The pKa values for sulfuric acid from Wikipedia are −3, and 1.99. So for all practical purposes the first ionization happens completely and would be unobservable with a pH electrode in aqueous solutions. So if you had a 0.1  molar solution of sulfuric acid it would almost look like you were titrating a 0.2N of $\ce{H^+X^-}$ where that acid had a pKa of 1.69 (pH at half NaOH used). This would be a strong acid too and the inflection point would be that for water at pH 7.

Note that the pKa 1.69 is assuming $\ce{H^+X^-}$ and is taken at 1/2 of the $\ce{Na^+OH^-}$ used. Knowing that $\ce{H2SO4}$ has two protons, the second pKa should be taken at 150% on the plot which would give the correct pKa value of 1.99.
If you had an acid with two lower pKa values then you'd see two pH transitions in the titrations. Look at the figure below. With two widely separated pKa values maleic acid shows two nice transitions. With two pkA values closer together the two transitions for malonic acid aren't as sharp. For succinic acid the knee at a volume of about 5 ml is a hint that there are two dissociations but you don't see two nice complete transitions like you do for maleic acid.

To go back to the sulfuric acid tritration, we'd really expect to see a knee at a low pH something like the titration curve we got for succinic acid. Such a knee isn't seen because a pH electrode just doesn't work at such high acid levels (pH range from -3 to 0).

PS - Thanks to the commenters for help to vastly improve the answer.

It's not true! Strong first dissotiation makes it even more visible that it's diprotic! Please don't add obviously wrong answers.

I didn't do the math, but if I have 0.1 N $\ce{H2SO4}$ I'd expect about 0.1 moles $\ce{H^+}$ and $\ce{HSO^{-}_4}$ and very little $\ce{H2SO4}$ or $\ce{SO^{2-}_4}$. Right?

Geez, if you titrate even stronger HCL curve is completely normal! Base is reacting especially with dissociated protons. @samueltober you shouldn't accept wrong answers.

For HCl the transition from acidic to basic solution would happen at pH 7 which is indicative that the HCl was completely dissociated.  If you titrate acetic acid you just get one break, not one for acetic acid and another for water.

Sorry but i did not know that this answer was wrong

I don't know that the answer is wrong. I still don't understand the objections. The one break for acetic acid happens at a pH lower than 7 indicating the dissociation constant of the acetic acid.

http://www.titrations.info/acid-base-titration-sulfuric-acid seems you're partially right, I misunderstood you; but it's one transition because of that the second proton is almost completely dissociated, if it was weakly dissociated there would be two breaks - edit it and I'll undo downvote

That being said, I doubt your last sentence is right : indeed, the two pKas 4.5 and 2.5 are so close that we'd probably again observe only one jump. See for instance the titration curves of succinic acid. @MaxW

Ok, I did some major edits on the answer. Thanks for the help improving the clarity.

paH isn't here equal pKa, your value was ok

@Mithoron - I assume you mean that the titration shows pKa of 1.69 whereas Wikipedia has 1.99 for pKa2?

Hmm, there's 0,1 M H2SO4 titrated (not N) so pH  at half should be higher than 2 but maybe because it's activity not concentration, it's lower... dunno, but at 1 M (not 0,1) pKa should be equal pH, if I recollect right

KNOWING that there are two H+ the second pKa would actually be at 150% on the graph shown, not 100%.


text:
Why does water condense on grass and iron grills in cold weather?
In my book it is said that in cold weather fog appears. This fog is nothing but water droplets. After a hot day if the night is cold then the moister in air starts to condense. What I do not understand is why does water condenses heavily on specific places like grass and iron objects only? I first thought that since iron is good conductor of heat so it becomes cold at night early which then captures nearby moisture more effectively than usual objects. But then grass is a poor conductor of heat. Another thought is that most of the objects absorb water. Those which do not absorb are metals and grass.
Please explain.

Gravity takes colder air down. In the yard the "highest" thing to come in contact with the cold air is the grass blade. Also the grass cools faster than the dirt.

@MaxW But then with the same logic colder air should come down on the concrete roofs of houses too -- but this doesn't happen.

A concrete roof has a lot of thermal mass. It doesn't cool off easily.

In early morning good conductors will eventually lose heat and cool things like wood concrete will cause the condensation  only because they are still cool. Hence the poor conductors like cement road should capture water droplets.

Well, in the bantering here I realize that there is a factor that you didn't consider. "Dew" does not necessarily come from settling "fog." The air can (usually?) becomes supersaturated with water, but lacks a sufficient nucleation site. The grass, iron grill or whatever provide such a site - if they are cold enough.

While grass IS a poor conductor of heat, it is super thin and will cool off really quickly because of of that alone. The increased surface area also means a greater area is exposed to the night air during the time when the moist air drops to the ground and will thus collect more dew than maybe a stone or a tree trunk.

Temperature is directly related to the speed of particles. Colder temperature means that particles, on average, will be moving slower.
Slower means that any interaction of the particles (e.g. gaseous water molecules) can be prolonged in time and thus (in the case of water), hydrogen bonding is induced between molecules to form droplets. If it gets cold enough, the water molecules will be more intensely bonded (i.e. ice, solid phase water). Hydrogen bonding looks like this between water molecules:

The same interaction principle applies for interactions of water molecules with other stuff, like the molecules on grass blades and the metallic lattice on a grill. The electrostatic interactions aren't "stronger", there's just more probability of prolonged interaction and thus adhesion/cohesion because in colder environments the gas just moves slower. But -- the water has to get up in the air in the first place, which is why what you observe typically happens after a hot day (moving particles faster in the air) cools down substantially.
This is a principle described by Boltzmann called "Boltzmann's distibution of velocities":

Here you can consider a=1 cold, a=2 warmer, and a=5 hot. The x-axis is for velocities and the y-axis is for probabilities of respective velocities.
Metallic surfaces are favorable because they have high conductance -- i.e. susceptibility to electrostatic interaction (compared to, e.g. the paint on the walls outside). Grass blades "attract" the water mostly because, simply, many water molecules cohesively bind to make droplets, which have enough mass to be effected by earth's gravitational pull, and fall/collect to whatever is on the ground. There are also biochemical mechanisms that help certain grasses pull in water from the air -- I think.

As an aside -- this is why the speed of sound is less in cold weather -- the gas molecules are what "carry" the sound waves, and if they move slightly slower, so does sound. No matter, no sound.

What kind of electrostatic interactions happen on metallic surface? Is it something like Hydrogen spillover where hydrogen atoms make electrostatic bonds with the metal surface? 2. Why doesn't water droplets, which are already cohesively bound and are pulled by gravity, fall over cement surface like road?

Coulombic interactions are the primary cause of atomic attraction -- the negative electronic charge and positive nuclear charges attract, as a principle. Spillover involves bond breaking/formation -- the interactions we're talking about are generally weaker.
Why grass is "favored" over a roadway can be answered practically -- cars drive on it and move the molecules before they can collect/settle -- and chemically -- the interactions of grass/water are simply stronger than cement/water because of chemical properties.

"cars drive on it and move the molecules before they can collect/settle" -- I don't think that's the reason. Because in early morning there is no dew on my house's roof(cement). Secondly what kind of chemical properties of a substance decide its surface's water holding capability? Does cement absorb water and grass doesn't?

Porosity of cement can be a partial explanation -- the bottom line here is that there are many cooperative and dynamic forces at work, the fundamental, quantitative answer to your question lies in understanding of quantum mechanics; i.e. the physical movements and distribution of particles at the atomic level.
The most simplistic answer is that not all atoms/molecules interact in the same way -- some interactions are stronger than others.
You seem curious -- put a few grams of salt on the ground outside and look at it the next day -- NaCl interacts with water very differently than cement.

The fundamental properties you're effectively looking for are: mass, charge, and arrangement. We can discover a lot about macroscopic properties from these things.

I think there must be a qualitative answer without going into much details of atomic structure. Because 1. this question is from a 6th grade book. 2. It has something to do with biology(as you already mentioned) -- most of the plants, there leaves and flowers included, I've observed collect dew in early morning. May be their soil absorb too much dew which is then released from stomata?


text:
Quasi-steady state hypothesis application
Consider the reaction scheme:
$$\ce{S + E ->[$k_1$] C1} \qquad \ce{C1 ->[$k_2$] E + P} \qquad \ce{S + C1 <=>[$k_3$][$k_4$]C2}$$
where $\ce{S}$ is the substrate, $\ce{E}$ is the enzyme, $\ce{P}$ is the product, $\ce{C1}$ and $\ce{C2}$ are enzyme substrate complexes. Let $[\ce{S}] = s$, $[\ce{E}] = e$, $[\ce{C1}] = c_1$, $[\ce{C2}] = c_2$ and $[\ce{P}] = p$ be the concentrations of each respective chemical. I have simplified this system down to
\begin{align}
\frac{\mathrm ds}{\mathrm dt} &= -k_1se_0 + (k_1-k_3)sc_1 + (k_1s+k_4)c_2 \\
\frac{\mathrm dc_1}{\mathrm dt} &= k_1se_0 - (k_1s+k_2+k_3s)c_1+(k_4-k_1s)c_2 \\
\frac{\mathrm dc_2}{\mathrm dt} &= k_3sc_1-k_4c_2 \\
\end{align}
using the conservation equation $e=e_0-c_1-c_2$. I have found that $p(t) = k_2\int c_1(t) \,\mathrm dt$. Now I need to use the quasi-steady state hypothesis to show that $$\frac{\mathrm ds}{\mathrm dt}= - f(s),\qquad f(s) = \frac{k_1e_0s}{1+\frac{k_1}{k_2}s + \frac{k_1k_3}{k_2k_4}s^2}.$$
Now I'm not really given much information on this hypothesis. I have been told it means

We assume that the initial stage of complex formation is very fast. After which it is essentially at equilibrium.

So how do I apply the hypothesis to transform $\mathrm ds/\mathrm dt$?

You understand that the ratio of k3/k4 gives you the equilibrium constant for the third reaction? S, c1 and c2 thus follow equilibrium.

So if $[C_1]$ is a constant (quasi-steady state hypothesis) then dp/dt is $k2[C_1] $

I'm not really sure what you mean by that, I come from a mathematical background so don't have much chemistry knowledge

Initially there is no [P]. The assumption is that [P] is so slow to form in the quasi-steady state that the mass balance is S (at t=0 ) = $S + C_1 + 2C_2$

Sorry to be annoying. I understand that assumption, but I don't understand how that leads to your second point and what your second point means.

Given that k1 is fast, then the steady state can only be reached by most of the S going to C2.

If k2 was about k3 then P would be forming as fast as C2 and there would be no quasi-steady state.

If you go back to your original question, my very first comment on it describes the assumptions that you need to make for the steady-state approximation. I did not elaborate any further or write an answer because at that point in time it seemed that the SSA was not what you were looking for, but it might now be worth going back and looking at it. More info here  http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html

What you do is assume that the time derivatives of c1 and c2 are close to zero (quasi steady state), so you set the right hand sides of these equations equal to zero.  You then have two linear algebraic equations in c1 and c2 that you solve, and substitute these solutions into the equation for ds/dt.  You should then end up with f(s).

I'll just point out that this is an extension of an earlier discussion on the problem.

Thanks for the interesting question.  Before we get into the details, let me note a few things:

Your system is a model of substrate inhibition, a phenomenon where higher-than-optimal concentrations of a substrate for an enzyme decrease its own rate of conversion to product.
It sounds like you have a good handle on the enzyme mass balance, i.e. that $e_0 = e + c_1 + c_2$ is a conserved quantity whose time derivative is everywhere zero (no approximations or hypotheses required) in this kinetic scheme.  This makes sense because enzymes are catalysts that accelerate reactions without any net participation in them.

OK, onward to your questions.

The assumption of fast equilibrium for $c_2$ is a different assumption than the PSSH hypothesis.  Assuming a fast equilibrium for $c_2$ means that reactions 3 and 4 happen much more quickly than reactions 1 or 2.  Thus the concentration of $c_2$ is determined solely by the thermodynamics of reactions 3 and 4: $$k_4 c_2 = k_3 s c_1 \;\;\; \Rightarrow  \;\;\; c_2 = \frac{k_3}{k_4} s c_1$$
The pseudo-steady-state hypothesis could be applied to $c_1$.  $$\frac{
dc_1}{
dt} = k_1se_0 - (k_1s+k_2+k_3s)c_1+(k_4-k_1s)c_2 = 0$$  Substituting in the result of the fast equilibrium assumption (point #3) gives $$k_1se_0 - (k_1s+k_2+k_3s)c_1+(k_4-k_1s)\frac{k_3}{k_4}s c_1 = 0$$  You can solve this equation for $c_1$ and plug the result into your $\frac{
ds}{
dt}$ equation.
If you are interested in understanding more about the mathematical validity of the PSSH, you should know that it is related to singular perturbation.  For a treatment of a "simple" Michaelis-Menten system by singular perturbation, see this book excerpt.  The PSSH solution is an "outer" singular perturbation solution.

This is the same exact method I outlined in the 9th comment to the original question.

Thanks for letting me know.  I don't always read comments and didn't in this case.  I prefer to write answers instead.

I also corrected a couple of errors from the equations in the original post that propagated into your answer.

Thank you!  I just copied the equations from the original post so I'm glad someone is out there checking this!

I realized that there were errors in the formulation only when I tried to apply our solution method, and the answer didn't work out.


text:
LogP of halogen derivatives of aromatics and aliphatics explanation
Benzene logP = 2.13 (logP=log([benzene in octan-1-ol]/[benzene in water]))
Fluorobenzene logP = 2.27 (slight increase, slightly more hydrophobic)
Chlorobenzene logP = 2.84
Bromobenzene logP = 2.99
So as we go down group 7, logP or hydrophobicity increases. What can explain these trends? Furthermore on aliphatic compounds, we observe this trend which confuses matters more.
-Fluorine DECREASES logP slightly (by -0.17) on an aliphatic compound. The book doesn't give a reference as to which alpiphatic compound this is and which hydrogen is substituted, but oh well.
-Chlorine increases logP (by 0.39) on the same aliphatic compound.
-Bromine increases logP even more (by 0.60) on the same aliphatic compound.
So a few questions:
1. Why does a fluorine substitution decrease logP slightly on the aliphatic but increase slightly on the aromatic?
2. Why do chlorine and bromine substitutions increase logP in aliphatics and slightly more in aromatics.
Answers to these trends would be greatly appreciated, thanks.

Here is a chart of the solubility of the species in water and the log of the partition coefficient you noted.
                           solubility       solubility
                  MW       g/100ml water    moles/liter   log(P)
  Benzene         78.11    0.179 (1)        0.0229         2.13
  Flororbenzene   96.103   0.155 (2)        0.0161         2.27
  Cholorbenzene  112.56    0.050 (2)        0.0044         2.84
  Bromobenzene   157.01    0.041 (2)        0.0026         2.99

(1) Wikipedia
(2) Cation Complexes of Compounds Containing Carbon-Carbon Double Bonds. VI. The Argentation of Substituted Benzenes
L. J. Andrews, R. M. Keefer
J. Am. Chem. Soc., 1950, 72 (7), pp 3113–3116
DOI: 10.1021/ja01163a085
weblink http://pubs.acs.org/doi/abs/10.1021/ja01163a085
There is a Wikipedia chart in article on Ionic radii.
 Ion     radii (picometers)
  F-      119
  Cl-     167
  Br-     182

The halide on a benzene ring is going to be relatively covalently bound. It definitely won't dissociate to form ions and only very moderately participate in hydrogen bonding.
So as the size of the halide increases the "volume" of the molecule increases. Thus the molecule would preferentially like to be in the non-polar phase.
Here is a scaled image of fluorobenzene molecule:

and a scaled image of the bromobenzene molecule.

Haloalkanes
The "anomalous" fluoroalkane is explained in a different manner. To get an n-alkane that has a boiling point above room temperature we need to go to n-pentane.
                   BP                 solubility      solubility
                   (deg C)   MW       g/100ml water   millimoles/liter
 n-Pentane         36        72.15    0.040 (1)         0.55
 1-Fluoropentane   63        90.14    1.270 (2)         14.1
 1-Chloropentane  108       106.6     0.197 (3)         1.85
 1-Bromopentane   130       151.05    0.127 (4)         0.838

(1) Wikipedia
(2) NIH
(3) Wikipedia
(4) Handbook of Aqueous Solubility Data, Second Edition
 By Samuel H. Yalkowsky, Yan He, Parijat Jain
There are two competing effects.
(1) The size of atom bound to the alkane.

Br > Cl > F > H
Scaled model of Fluoroethane

(2) The second effect is the electronegativity of the X group in the $\text{Alkane}\text{X}$ bond.
 F    >   Cl   >   Br   >   H
 4.0      3.2      3.0      2.2

The fluorine atom has a slightly larger radius in a fluoroalkane than the corresponding C-H group. But the halide groups allow more dipole forces. So by "volume" the overall effect is:
 fluoroalkane
 alkane
 choloroalkane
 bromoalkane

Why does increased volume lead to the molecule preferring the non polar phase? I don't understand how increased dipole forces reduce the volume of the fluoroalkane. Furthermore why wouldn't the volume of chloro and bromoalkanes be reduced in this case? Also, why is it that fluoroalkanes actually prefer water MORE than alkanes?

See this problem for an answer to the volume effect. http://chemistry.stackexchange.com/questions/40108/why-is-logp-quite-a-bit-lower-in-cyclohexane-than-in-hexane/40119#40119


text:
Directing effects of substituent groups on aromatics
Normally, directing effects of substituent groups on aromatics are explained by drawing resonance structures of the Kekule representation of the compound. My question is that since the Kekule representation is in fact incorrect but rather there is equal electron density in all parts of the pi cloud in benzene, I am not completely satisfied with the explanation which relies on the fact that the bonds are localised.
Is there any other way to explain directing effects of substituents on benzene and other aromatic systems?
Obviously stuff like NMR analysis of phenol proves that the curly arrow predictions of the Kekule structure are true (ortho and para hydrogens have a shift of 6.7 and 6.8 ppm respectively - more shielded than normal benzene hydrogen of 7.2ppm so higher electron density in ortho and para carbons). But that could just be due to chance that the prediction is right. Also how do the curly arrows show that the ortho carbon is very slightly higher in electron density than para?

The explanation you are talking about explicitly says that the bonds are not localised by invoking resonance structures so your argument is somewhat invalid. Molecular orbital theory is the other main way of describing bonding and is probably what you are after but they both produce exactly the same result.


text:
Can hemi-acetal rings form under basic conditions?
I recently encountered a problem on a test.  It was a standard synth problem, "Provide reagents to produce y from x".  The starting material was DHA and the ending molecule was a branched pentose. (I included an image of the solution I came up with).  In my solution I use acetaldehyde under basic conditions to form the pentose.
My question is would pentose form under basic conditions or do I need to acidifying at the end?  Would acidifying just encourage the formation of side products?  Is there some other way to get this pentose as a primary product that I am not thinking of?

If your reaction was carried out in basic conditions the product is an hemicetal.
If your reaction was carried out in acid conditions the product is an acetal.
Why? This is catalysis question.
When you have the hemicetal to form an acetal you need the protonation of the -OH group, the oxonium ion formation and a new atack. But this can't occur in basic conditions.
If you need more explation you can read this:
Jonathan Clayden: Organic Chemistry
Chapter 14: Nucleophilic substitution at C=O with loss of carbonyl oxygen

Using base means that the reaction will stop at the hemiacetal and acetal formation would not take place. Acidifying at the end might encourage acetal formation.

Using base means that the reaction will stop at the hemiacetal and acetal formation would take place. Aren't these two statements contradictory? Also please can you explain your answer in more detail. At the moment this might be deleted for being too short.

I was hoping for some kind of explanation.  I have a very fuzzy understanding of that hemi-acetal mid step.  Are you saying that neither acidic or basic conditions would work in this reaction?

Oops that's a typo; I'll try to write up an explanation when I get home in an hour

authors: عبدالعزيز عبداللطيف


text:
How to calculate the rate constant of a zero order reaction?

The reactant concentration in a zero-order reaction was $7.00×10^{−2}~\mathrm{M}$ after $160~\mathrm{s}$ and $2.50×10^{−2}~\mathrm{M}$ after $345~\mathrm{s}$. What is the rate constant for this reaction?
What was the initial reactant concentration for the reaction described in part one?
The reactant concentration in a first-order reaction was $7.70×10^{−2}~\mathrm{M}$ after $50.0~\mathrm{s}$ and $1.60×10^{−3}~\mathrm{M}$ after $85.0~\mathrm{s}$. What is the rate constant for this reaction?

Background info: The integrated rate laws for zero and first order reactions may be arranged such that they resemble the equation for a straight line,
$y=mx+b$.
\begin{array}{lccc}
\text{Order}& \text{Integrated Rate Law}& \text{Graph}& \text{Slope} \\\hline
  0 &  [A]=−kt+[A]_0      & [A]\text{ vs. }t    & −k\\
  1 & \ln[A]=−kt+\ln[A]_0 & \ln[A]\text{ vs. }t & −k\\\hline
\end{array}

Hint - For (a) how many points does it take to define a straight line? That should get you started.

I'm still lost :(

Draw X/Y axis. Label X as seconds and Y as concentration. For (a) calculate the slope of line.

According to your table, for a zero order reaction, you make a graph of the measured concentrations vs the corresponding times.  It will be a straight line, and the slope and intercept will give you the rate constant and the concentration at time zero.
For a first order reaction, you make a graph of the natural log of the concentrations vs the corresponding times.  It will be a straight line, and the slope and intercept will give you the rate constant and the concentration at time zero.

Question 1
First, note the integrated rate law for zero-order reaction: $$[A] = -kt + [A]_\circ$$ As you have already realised, this is in the form of a linear equations: $$y = mx + b$$ where $y$ corresponds to [A],
$m$ which is the gradient correpsonds to $-k$
$b$ which is y-intercept corresponds to $[A]_\circ$
The question is asking for the rate constant, so all we have to do is find the gradient of this line and that will be equal to $-k$. $$m = \frac{[A]_2 - [A]_1}{t_2 - t_1} = \frac{7.00\times 10^{-2}\ \mathrm{M} - 2.50\times 10^{-2}\ \mathrm{M}}{160\ \mathrm{s} - 345\ \mathrm{s}} = -2.43\times 10^{-4}\ \mathrm{M\ s^{-1}}$$ Therefore $k = 2.43\times 10^{-4}\ \mathrm{M\ s^{-1}}$
Question 2
Now that we know what $k$ is, we have a equation where there is 4 variables and we know 3 of them. Therefore, all we have to do is substitute the values that we know and then solve for the unknown variable, $[A]_\circ$. $$[A] = -kt + [A]_\circ$$ $$[A]_\circ = [A] + kt$$ $$[A]_\circ = 2.50\times 10^{-2}\ \mathrm{M} + 2.43\times 10^{-4}\ \mathrm{M\ s^{-1}}\times 345\ \mathrm{s} = 1.09\times 10^{-1}\ \mathrm{M}$$ Therefore the initial concentration of the reactant is: $1.09\times 10^{-1}\ \mathrm{M}$
Question 3
This question is exactly the same as question 1, except the only difference is that the reaction is first order, not zero order. To solve this question, just follow the same procedure that we used to do question 1. The only difference is going be is that the integrated rate law is going to be slightly different.
I will leave this question for you do. If you have any problems feel free to ask.

$$m = \frac{[A]_2 - [A]_1}{t_2 - t_1} = \frac{2.50\times 10^{-2}\ \mathrm{M} - 7.00\times 10^{-2}\ \mathrm{M}}{345\ \mathrm{s} - 160\ \mathrm{s}} = -2.43\times 10^{-4}\ \mathrm{M\ s^{-1}}$$


text:
Can nucleophilic attack be faster than rearrangement?
Let's say in a reaction, unimolecular (SN1), we have $\ce{CH3-CH(CH3)-CH2+}$ carbocation, and a nucleophile $\ce{Nu-}$. Generally the carbocation is first rearranged and then there is attack of $\ce{Nu-}$, but can't it be that a very strong nucleophile, strongly attracted towards the cation, attacks the cation befor it rearrange to give kinetically controlled product (KCP) (at least at lower temperatures) ?

You are asking what would happen if Something like 1-bromo-2-methylpropane were to undergo an $\mathrm{S_N1}$ reaction. The simple answer is: Your premise won’t happen.
The first step in an $\mathrm{S_N1}$ reaction is the formation of a carbocation by the leaving group leaving. A general equation for this would be:
$$\ce{R3C-Br <=> R3C+ + Br-}$$
Where this reaction has its equilibrium and thus what concentration of the carbocation can be expected, depends on the relative stabilities of reactant and carbocation (the bromide is always the same so it doesn’t matter). Now primary carbocations are so unstable (in a thermodynamic sense) that thermodynamics predicts an equilibrium far to the left for primary haloalkanes:
$$\ce{RCH2-Br <<=> RCH2+ + Br-}$$
(This reaction arrow doesn’t do it justice. It should be a humungous left-pointing arrow and a minute right-pointing one.)
For tertiary or otherwise stabilised carbocations, we can expect a significant concentration of the right-hand side of the equation at equilibrium. Thus, we have a potential carbocation concentration that can potentially react with another nucleophile in an $\mathrm{S_N1}$ reaction. But primary haloalkanes will not even allow the carbocation to form, or in a different view, will pull back the bromide to regenerate the haloalkane before any nucleophile has a chance of attacking the carbocation.
That’s why primary haloalkanes predominantly react via the $\mathrm{S_N2}$ mechanism: There simply isn’t any significant amount of carbocation present for $\mathrm{S_N1}$ or any type of Wagner-Meerwein rearrangement.

I think you're missing the point.
Consider the SN1 mechanism shown in Wikipedia. The overall reaction is:
$\ce{(CH3)_3C-Br + 2H2O -> (CH3)_3C-OH + H3O^+ + Br^-}$
There is no way that 3 molecules are going to simultaneous collide and make this reaction work. The "SN1 reaction" breaks this particular type of reaction into its intermediate steps.
If the nucleophile did some other sort of reaction, then it would be some other reaction mechanism, not a SN1 reaction. For instance there is a SN2 mechanism as well.
So chemistry isn't just about doing overall mass balanced equations. It is trying to show you how you got from point A to point B.
Multiple Reactions
It is possible that multiple reactions can happen. Chemists can change the reaction conditions, or use a catalyst, to change the kinetics (relative reaction rates) to favor a particular reaction.

Alright you seem to have gained a friend here since I came here from the VLQ review. Note to self: Another instance of unjustified flagging

@MaxW what I am asking is that in SN1 after the leaving group separate out, and the nucleophile is about to attack the cation, with a possibility of rearrangement (of cation), can the nucleophile  attack the cation befor it rearranges ?

Look at this reaction. https://en.wikipedia.org/wiki/SN1_reaction#Stereochemistry Assuming 4 different groups, is what you want an "optically active" product?

@MaxW i am not talking about optical activity. I am asking that can CH3−CH(CH3)−CH2-Nu be a product of SN1 reaction. (Nu = nucleophile )

Let's assume the carbon reacting has four different groups so that it is stereoactive. The "rearrangement" of the carbocation results in a "planar" carbocation which can be attacked from either side. This would make racemic mixture. // If the carbocation kept its pyramid form then the product would be stereoactive too.  Isn't that what you are getting at? This would be a special case of a SN1 reaction. // If this were a SN2 reaction then the product would be the opposite of the original optically active reactant.

@MaxW I am not asking about stereo activity either. I am asking is, can nucleophilic attack be faster than rearrangement of carbocation? and not about the rearrangement & optical activity of carbocation

By "rearrangement of carbocation" you mean going from a "pyramid structure" to a "planar structure"?

The formation of a distinct carbocation followed by migration of a neighboring group could follow your scenario. Thinking about it tho the nearest Lewis base to the incipient C+ ion is the adjacent group and it seems likely a concerted shift ionization would be a lower energy transition state and your scenario is best described as a $\pu{S_N}2$ reaction and would show bimolecular kinetics.  Kinetics should show a distinct difference between secondary halide solvolysis without and with rearrangement.


text:
Why doesn't 1-cyclopropylethyne give a positive reaction in the bromine-water unsaturation test?
1-cyclopropylethyne has a triple bond:

Hence it seems like it should yield a positive unsaturation test with bromine water. But, my book states that the opposite is true.
I guess this may be due to steric crowding on the C atom bearing the cyclopropyl ring, but I'm not sure.
What is the exact reason that 1-cyclopropylethyne does not yield a positive bromine water unsaturation test?

I'd guess that it is too insoluble in water to react with an aqueous water solution.

@MaxW I would assume that wet THF would be a fine solvent for this type of reaction...


text:
Stereochemistry of hydrogenation with Raney nickel
If I react the following with Raney nickel and $\ce{H2}$,
$\hspace{65 mm}$
then I am getting a meso compound, as the $\ce{H-}$ anion will have anti addition, right? One $\ce{H}$ will attack on right $\ce{C}$ from bottom and the other on the left $\ce{C}$ from the top. That makes it a meso. But in the book it says, there is racemization. How?

check edit @bon

The hydrogenation of alkenes using heterogenous catalysts such as Raney nickel is syn-stereospecific; i.e. both hydrogen atoms are added to the same face of the double bond.
Therefore, the hydrogenation of (2E)-(2,3-2H2)but-2-ene yields
(2R,3R)-(2,3-2H2)butane and
(2S,3S)-(2,3-2H2)butane.

Good old paper drawing has blindsided you. Think of the $\ce{H2}$ as adding on top of the molecule as you have drawn it on paper. Not all the molecules know that they are supposed to land "right-side" up.

Some will land up side down.

You get stereoisomers when there are four different groups bonded to a single carbon atom. Such a mixture will effect polarized light. One configuration will twist the polarized light positive and the other negative. An equal mixture of such stereoisomers is a racemic mixture and is optically inactive.
Right-side up you get (H)(D)(CH3) going clockwise looking down the C-C axis that was the C=C double bond. Upside-down you get (H)(CH3)(D) going clockwise.

sorry, didnt understand what you mean by (H)(CH3)(D) going clockwise

so, is my book correct ?

Yes. Edited my answer to make that point more clearly.


text:
How mass spec formate and acetate adducts masses are calculated?
If I look at any tables or calculators, I find that negative ion formate adduct has a monoisotopic mass of +44.9982 units. If I try to calculate myself, I get slightly different result:
H + C + O + O =
1.007825 + 12.0107 + 15.9994 + 15.9994 =
45.017325

The difference is 0.019, what is already not indifferent, as usually I calculate with a tolerance of +- 0.02 units, which means this can cause me miss many hits. I could not find out what may cause this difference.
Later had similar experiences with the acetate, as in tables +59.013851 can be found, but if I calculate:
3 x H + 2 x C + 2 X O =
3 x 1.007825 + 2 x 12.0107 + 2 x 15.9994 =
59.043748

Which means even more, 0.0298 difference.
My question is what causes the difference, and what is the correct wat to calculate these masses?

In your calculation, you are using the average molecular mass of the structure, where atomic masses are based on the natural abundance of all isotopes of the element.
H:  1.00794 u
C: 12.01070 u
O: 15.99940 u
O: 15.99940 u
─────────────
   45.01744 u

However, for the exact molecular mass (or monoisotopic mass) of the structure, the atomic masses of each atom are based on the most common isotope for the element.
H-1:   1.00783 u
C-12: 12.00000 u
O-16: 15.99491 u
O-16: 15.99491 u
────────────────
      44.99765 u

Furthermore, you are missing the mass of the extra electron of the formate ion. The electron mass is $m_\mathrm e=0.000548579909070(16)\ \mathrm u$. Thus,
H-1:   1.00783 u
C-12: 12.00000 u
O-16: 15.99491 u
O-16: 15.99491 u
e−:    0.00055 u
────────────────
      44.99820 u

Thank you! I was believing I have already the monoisotopic masses in my table... so I need to look for another table. And special thanks for pointing out the extra electrons should be added.

And if anybody else needs the monisotopic masses, here is a nice table:  http://www.ciaaw.org/atomic-masses.htm


text:
What type of bond(s) are present in the borohydride ion?
In the $\ce{BH4-}$ ion, what bonds are present between boron and hydrogen?
Are there three covalent bonds between boron and three hydrogen atoms (using the boron 2s, 2px, and 2py orbitals), with the fourth hydrogen being a hydride ion $\ce{H-}$ which forms a coordinate or dative bond to the boron 2pz orbital?

No, this is simply a bond between four hydrogen atoms ($\ce{H}$) and a boron anion ($\ce{B-}$). These bonds are identical to and indistinguishable from the other $\ce{B-H}$ bonds. They are all covalent bonds of length $1.21\:\mathrm{Å}$.
Your idea of $\ce{H-}$ completing $\ce{BH3}$'s $2\mathrm{p}_z$ orbital isn't quite right either. Atom's individually have atomic orbitals, but when they form bonds these atomic orbitals combine to form molecular orbitals. This is picture more accurately demonstrates the bonding between $\ce{B-}$ and the $\ce{H}$ atoms.

[Photo Source]

(source: mcat-review.org)
You are quite correct, the boron forms covalent bonding with three Hydrogens, and then a coordinate bond with H- ion (1s2)
Extra info:
This ion is fairly water soluble and acts as a ligand, using bridging hydrogens as three-centre two-electron donor atoms, forming complexes like Al(BH4)3 and Be(BH4)2
Reference: Principles of Descriptive Inorganic Chemistry
 By Gary Wulfsberg

what is opinion about ringo's answer?

I looked up what ringo answered, and it appears that his answer is more accurate. I am basing my answer on basic middle school chemistry, whereas his is a little more advanced. I don't know a lot, as of now, I am only in high school after all :p


text:
pH of a mixture of base and an acid
I have the following question: Assume that you have two solutions. One of  $50\ \mathrm{cm^3}$ $\ce{HCl}$ $\mathrm{pH} = 1.3$ and one of $150\ \mathrm{cm^3}$ $\ce{Ba(OH)2}$ $\mathrm{pH} = 12.3$. If we mix these two solutions what will the $\mathrm{pH}$ of the mixture be?
I have done the following:
$\mathrm{pH} = 1.3 \Rightarrow [\ce{H+}] = 10^{-1.3} = 0.05\ldots\ \mathrm M  \Rightarrow n = 0.05\ldots \times 0.05 = 0.0025\ldots$
$\mathrm{pH} = 12.3 \Rightarrow \mathrm{pOH} = 1.7 \Rightarrow [\ce{OH-}] = 10^{-1.7} = 0.019\ldots \Rightarrow n = 0.019\ldots \times 0.15 = 0.0029\ldots$
My problem is I do not know how to proceed

Please typeset your post. That will help others help you get the information you request.

The OH- and the H+ will react, neutralize each other to form H2O, H+ + OH- = H2O. Hence the ions remaining is: 0,0029... - 0,0025... = 0,00048... which will be OH- ions. I believe you can proceed on your own from here. Just calculate the concentration and then the pH.


text:
How to change color of water based on its acidity?
I recently saw a video on YouTube in which water, when mixed with a solution (I believe iodine was involved,) changed color based on its acidity. I am attempting to replicate this experiment, but cannot concoct the solution. Any help would be much appreciated.
EDIT:
To clarify, if I had water with a ph of 10 it would turn red, but if the water had a ph of 4 it would turn green.
EDIT 2:
I found the video in question
http://youtu.be/_yi83kOB8ug

Can you give us more details on the experiment? A link to the youtube video would be helpful for a start.

Say I have highly acidic water. When I combine it with the solution, the water should turn red. If however, it had low acidity, it would turn blue.

I'm assuming that you're doing this at home. A popular (and SAFE!!) experiment at home is to use red cabbage, vinegar, and baking soda.
The indicator comes from the cabbage.
Vinegar turns the solution acidic.
Baking soda turns the solution more alkaline.

What would the operating procedure be? How would you use this to, as an example, make highly acidic water a certain color?

If you are trying to reproduce the exact experiment that you saw on youtube, then I'll need a link to the video. // In general an "indicator" is used to signal pH changes (or the pH) in a solution. The indicator undergoes a acid-base transition too, but is low concentration and deemed a "spectator" rather than a participant in the reaction.  // There are a zillion pH indicators that can be purchased. https://en.wikipedia.org/wiki/PH_indicator I didn't see one like you described, but indicators are sometimes used in combination too.

Alright, I added the link. Thanks for the information regarding the indicator. Let me know what you think.

Yuck. I think they are trying to sell their water purifier. The music is god-awful. I saw a couple of pH changes. I'm too lazy to watch the whole thing and track all the changes. // Any bottled water or tap water would be safe to drink. Fancy "sparking water" has CO2 and will be acidic.  // It is a mixed indicator for sure. I'd guess that a store which sells pool supplies would have some sort of such pH testing kit. Either paper strips or dropper solution.

Agreed. In any case, it seems as though the best way to run an accurate version of this would be to use an indicator like you mentioned before. Thank you for helping me understand.

For a water purifier I'd be more worried about the bacteria that will grow in the filter than I would be about the chemicals being removed.  Really hard water, or water with $\ce{H2S}$ would need some sort of treatment to not have an off taste.


text:
If the reactivity of group 1 elements increases down the group, why is this not the case for halogens?
I was wondering, as you go down a group, the atoms get bigger... This means that the electrons in the outer shell are further away from the nucleus, and so the attraction decreases and it is easier for the atom to lose electron, thus becoming more reactive, as is the case for group 1 (the alkali metals).
However, as you go down group 7 (the halogens), reactivity actually decreases, and the reason given is that it is harder for it to gain electrons, which contradicts the previous statement? Any ideas?
I'm guessing that because halogens only need 1 more electron to complete their shell; they need to be small (to have a strong attraction between the nucleus and the electrons) and vise versa for the alkali metals.
P.S. Please keep answers brief and generalise as much as possible. The answer need only be correct to a GCSE standard, thanks.

" The answer need only be correct to a GCSE standard, thanks." -- Is that another way of saying "the answer doesn't need to be correct"? :P

Lol maybe, it's all a matter of perspective.... I just put that in there because whenever I ask question, I get this degree standard explanations that are simply unecessary.

In one case you gain an electron, in the other you lose one, so you don't have a contradiction.

Yes, you are correct in your approach.
Larger alkali metals means that there is less Z-effective or effective nuclear charge between the central nuclei and the valence electrons due to a screening/shielding effect. This allows them to be more reactive as they can lose electrons easily.
In halogens, to react more, they will need to be able to accept more electrons. This means that if the atom is smaller, the Z-effective is large, the nuclear attraction is high and it is more reactive. For a large atom, it is tougher to accept electrons due to reduced nuclear charge because of increased screening/ shielding effect.

Ahh I see, thanks. So what you're saying is that the effect of the size of the atom (the nuclear attraction) changes depending on whether you need more or less electrons.

Kind of, but not exactly. If the nuclear charge increases, it becomes harder for the atom to lose the valence electrons. Reactivity depends on how easy or difficult it is for an atom to lose the valence electrons.

The answer to your question is quite simple.The alkali metals of group 1 grow larger in size downwards and since the effective nuclear charge remains the same as we go down group 1 the number of shells also increases and so it is easy to lose an electron and therefore as we move downwards the reactivity increases.
However in group 17 the halogens have to gain an electron. So when the size increases and the number of shells increases the attraction of electrons decreases as we move away from the nucleus and so it is difficult to gain an electron as we move downwards.