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pH of aqueous very dilute hydrochloric acid

Calculate the $\mathrm{pH}$ of pure water in which $\ce{HCl}$ was added. The final concentration of $\ce{HCl}$ is $\pu{10^{-6}M}$.

So what I did was to add the concentration of $\ce{H3O+}$ in pure water ($1.0\times10^{-7}$) to the concentration of $\ce{HCl}$ and calculating the $\mathrm{pH}$. But I got to $6.0$, which I don't feel is the correct answer. According to me, it should be lower right. I need help in this.

You are correct. The $\mathrm{pH}$ should be around six. The actual value can be found by calculating $-\log(1 \times 10^{-7} + 1\times 10^{-6})$, which gives me a final $\mathrm{pH}$ of 5.96. While you are correct that the $\mathrm{pH}$ of a strong acid solution tends to be significantly below 7, the concentration of $\ce{HCl}$ in your solution is so small that it does not have a very large effect on the overall $\mathrm{pH}$. By convention, you consider the pH of a solution to be determined by the largest source of hydrogen ions, meaning that the $\mathrm{pH}$ of this solution should be exactly six. Despite this, $\mathrm{pH}$ is actually a measure of the total concentration of $\ce{H+}$, so in this case, since the concentration provided by the acid is so small, both the hydrogen ions from the dissociation of water and those from the acid should be considered.

The question says the final concentration is $10^{-6}$. That will dominate the equilibrium in any case. Neglecting the deviation of the activity from the concentration, which is quite feasible with this low concentration, the pH becomes by definition 6.


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A quickly evaporating degreaser that does not ruin plastics?
I was wondering if there is a quickly evaporating solution that effectivly removes fat from glasses, but does not ruin plastic or metal. That is any eyeglass frame material..

Diethyl ether, found in gasoline engine starter sprays, is a very effective solvent for fats, and it does not attack most plastics too rapidly. Try a bit on an out-of-sight location to be sure it does not damage the frame or lenses.
Note that some engine starter fluids contain an oily lubricant, which would defeat your purpose. Also note that ether is extremely flammable. A gently pressed spray can releases only small amounts at a time, but it will still catch fire if there is ny source of ignition nearby.

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Why do alkali metals have an exothermic electron affinity?
Alkali metals should have positive electron gain enthalpies as they are electropositive elements and also their atomic sizes are larger in their periods so they should be reluctant to take electrons but they show negative electron gain enthalpies. Why?

Why is the ionization energy of sodium positive? It is highly electropositive, it should be happy to give electrons. Why should you give some energy to force the electron to leave sodium's sphere of influence?
There is no general answer to this question. Electron Affinities are determined experimentally and there is no way one can predict electron affinity theoretically.
The sodium atom attracts the electron towards itself. A shock? The nucleus is positive and the electron is negative, they attract each other. There are however electrons which repel the incoming electron.

I think you mean to ask about the electron affinity of the alkali metals. That is, the enthalpy gain or lost when adding an electron to an atom.
While the electron affinity for alkali metals are much smaller than the halogens, they are slightly positive. That is, quoting from Wikipedia:

Electron capture for almost all non-noble gas atoms involves the release of energy and thus are exothermic.

The catch, is that a positive electron affinity (Eea) refers to an exo-thermic process:

Confusion arises in mistaking Eea for a change in energy, ΔE, in which case the positive values listed in tables would be for an endo- not exo-thermic process.

You refer to the alkali metals as "electropositive" -- in this case, you're referring to their electronegativity relative to, say, halogens. Yes, a sodium atom is likely to become a sodium cation and give up an electron. But the electron configuration is $3s^1$ and consequently, can also gain an electron and become the more stable $3s^2$ state.
It's more likely that a sodium atom will lose an electron, and become a cation, but the electron affinity reflects the modest gain in stability as $3s^2$ after attaching an extra electron.

After taking one electron ns2 orbital is fulfilled and become stable. So Alkali metals has negative electron gain enthalpy.

The first electron gain enthalpy is positive only for :

Inert gas elements

Be, Mg and N

It is so because of the dominating energy
due to stable configuration and small size factor .

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Tridecane isomer name
What is the name of this $\ce{C13H28}$ isomer:
$\hspace{5.5cm}$
I think it must be "4-ethyl-6,7-methylnonane," but I saw somewhere that its name is "6-ethyl-3,4-methylnonane". Why? The ethyl number must be smaller, isn't it? (because E is before M). Am I wrong?

Alphabetic order dictates which order the substituents are written in, but not the numbering of their positions.  Numbering is determined by what minimizes the numbers.  E.g. 3-ethyl-2-methylhexane is correct, but 4-ethyl-5-methylhexane is not.


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Fake smoke for a fire drill on a boat?
Is there any way to generate fake smoke that could be used in a fire drill on boat? Basically the idea is that the captain would set the "fire" and it would start smoking. The crew then has to find it and put it out.
Ideally it needs to smell, if not like real smoke, at least like something.
Also, it needs to billow out and obscure vision the way real smoke does.
I seem to remember a reaction that billows out black smoke but I cannot remember what it was.

Is a colored smoke bomb or classic smoke bomb sufficient for your needs?

Having some level of experience in fire training in various scenarios over a number of years, I highly recommend that you take a commercial approach to generating your smoke, rather than a homemade recipe that compounds your liability. Any smoke you generate from a homemade smoke bomb is likely to be considerably toxic, and should be avoided. Fire fighters would not enter a smoke-filled boat to attack an active fire without the appropriate safety gear which would include correct breathing apparatus.
You can purchase smoke cartridges of various sizes, colours and burn times, and these are designed for this very purpose. Examples can be found using a quick search for fire training smoke cartridges: here, here and here are examples from companies that I in no way endorse. They are pretty cheap for the effects they produce. I use the 20m3 burn capsules for fire fighting training with vehicle fires - these have a burn time of about 5 minutes, and they are very authentic, apart from limited smell. The smell from a fire in a boat is likely to be an acrid toxic affair, quite unlike your typical woodfire smell, so you would be best to avoid this. For better effects, burn a couple together, and you can get burning cups with electronic ignition that allows you to remotely activate it whenever you are ready.
Another option for generating limited visibility that is frequently used is a smoke/fog machine that is used for theatrical smoke and at nightclubs. Search for fog machine or smoke machine. Here is an example - again, I have no affiliation with this company. I have used ones that have come off ebay, and they are very cheap, but do the job nicely. I have used these a lot simulating structural fires and vehicle fires, and they can produce great amounts of 'smoke'. You need to take care that before any water is applied that the smoke machine has been turned off and removed well aware from the area. These also work very well for using with ram fans to visualise wind flow patterns through buildings/structures to understand convection flows and the path of travel for fires.
Be sure to read and understand the SDS of all materials before you use them in an enclosed space such as a boat, and ensure you have a plan to sufficiently ventilate any space between fire drills and afterwards.

You need to ask yourself the questions "is there no fire without smoke" and "is there no smoke without fire". I can think of some types of fires which will not form visible smoke. The amwericium containing smoke detector (ionization type) are very good at detecting very fine particles of smoke which are invisible which are emitted in the early stages of a fire.
One problem I see in a ship is that a small but very horrible fire could break out in a hidden location such as a duct, if this is damaging a key system then the crew will need to locate it quickly and take action against it.
By the way the soviets used to use plutonium in their version of the smoke detector. The UK and USA smoke detectors normally used americium except for the very early ones which used radium-226.
I have to agree that the use of homemade smoke bombs is bad as it will open a vast can of worms in terms of legal issues.

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Chemical desalination
Is there a way to remove salt from water, even just partially, using chemical or filtration methods prior to a reverse osmosis?
The problems is that salt is very hard on a reverse osmosis membrane which is expensive. Is there some way to attract some of the salt out of the water before using the membrane?
One possibility I have heard of is decanoic acid which can be used to sequester salt.

You could distill the water, but that pretty much ruins the point of trying to use reverse osmosis!

The method you mentioned is known as directional solvent extraction and seems quite straight forward and efficient. It is a relatively new method of desalination. Soybean oil, hexanoic acid, decanoic acid, and octanoic acid can be used (the last two are most effective) [ source: https://dspace.mit.edu/handle/1721.1/78539 ]. This video shows the setup used.


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Do all adiabatic processes involve a change in temperature?
I learnt that adiabatic processes involve no heat exchange between the system and the environment. Does this mean that all adiabatic processes involve some changes in temperature then?
Thanks a lot!

Yes... free expansion is an exception (its actually not right to call free expansion as an adiabatic process)

Adiabatic expansion does involve certain change in temperature; also, what's wrong about calling it adiabatic? Essentially, if we fix two variables in the equation of state, we are left without any degrees of freedom. Thus the answer is "yes", unless we have multiple components, in which case it turns to "no". Mixing of two gases at the same temperature can be an example of adiabatic processes without change in temperature.

I believe its physically impossible to have an adiabatic process by mixing two gases.

What's physically impossible about having two gases at the same temperature, and then gently mixing them?

How are you going to gently mix the gases?

Let's say I'll just open the valve between the two tanks. We don't really have to be gentle: it will be an adiabatic process anyway.

The first law of thermodynamics says,
$$\Delta U = Q - W$$
By definition, no heat is exchanged between the system and the surroundings in an adiabatic process. Therefore, the 1st law reduces to,
$$\Delta U = -W$$
In an adiabatic process, $PV^\gamma =const. $ has to be followed. If you change pressure, the above equation implies that volume changes. If the volume changes, there will be some work done which in turn will change the internal energy of the system. Therefore, the temperature changes.
You can also come to the same conclusion using ideal gas equation.
We know the ideal gas equation,
$$PV = nRT$$
If there should be no change in temperature, $PV$ should remain constant.
But in case of an adiabatic process, we know that
$$PV^\gamma =constant$$
So if you change one of those parameters, the product $PV$ will obviously change (so as to keep $PV^\gamma$ constant). Hence, temperature always changes in an adiabatic process.
*I have ignored ridiculously insane cases where matter can exit/enter the system through an adiabatic wall (which isn't possible). If you could discover a wall that allows matter to move across and does not allow heat to conduct through it, you deserve a nobel prize.
*Free expansion and mixing gases are two processes where the process is adiabatic and there is no temperature change.

Why cant only P decrease when V increases , so as to keep the equation constant?

If you change $P$ and $V$ such that $PV^\gamma$ is constant, then $PV$ cannot remain a constant unless $\gamma$ is unity.


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organic chemistry and magnet
Is it possible to use magnet to influence the reaction in order to favor one outcome over the other? ie addition reaction that favor the minor.

In short, it is not. Some reactions can be influenced by the magnetic field, but the effect is minuscule anyway.

Magnetic fields have been used to effect short-lived radical ions pairs generated via photoinduced eletron transfer reaction. But this was in spectroscopical studies only. I'm not aware of any use in preparative organic chemistry.

http://www.sciencedirect.com/science/article/pii/S0301462204003965


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Relation of Boiling Point and Melting Point with the reactivity of an element
When we look at the trends of periodic table, there is an interesting correlation to be noticed. Along a period, when we look at the trends in reactivity of an element( Both due to electropositivity AND electronegativity) the trend is that from the right to the left, first there is a decrease and then an increase; which is definitely sensible. The question that arises is due to this observation that same happens with the boiling point and the melting point of that particular element. It first decreases and then increases. So this leads to the intuition that the reactivity (ie combining capacity ) also affects the physical characteristics of that particular element. If such is the case, the WHY EXACTLY does this happen- what is the explanation behind this phenomenon. If this intuition is false, then correction would be highly appreciated.
Thanks and yes, you are awesome!

Yes there are trends in the periodic table. There are also many questions on Chemistry SE wondering why something seems out of place. Don't rely on trends - they can mislead. But, for a high-altitude view, what does a low melting point indicate about the strength of bonds in the solid?


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Ions and bonding
I am currently learning about ions and I am a bit confused about the bonding part.
I understand that if an atom has an 'unstable' amount of electrons (like 2 electrons, then 8, then 4) that it can 'borrow' some electrons from some other elements. An example of this would be sodium and chlorine.
But again, I am just confused on the bonding part. So when an atom loses or gains electrons it obviously changes to a + or - but does that particular element (atom) have to join with the other atom, or do they just 'give' each other electrons and 'be happy'?

If you rub a balloon and hold above your head, you will notice that your hairs are lifted. What is happening is that your hair and the balloon is charged. One is slightly positive and one is slightly negative. This is what causes them to attract. And this attraction is what lifts your hair up. The hair and balloon situation is somewhat similar to ionic bonds. The ions become charged as a result of the transfer of electrons. Since they are charged, they become attracted to each other (similar as to the hair and balloon).

This is why breaking ionic bonds are somewhat easier to break apart than covalent bonds. All you have to do to break ionic bonds is to neutralize the charge of the ions. This is what happens when you dissolve salt in water.

Covalent bonds, however, share electrons. It is harder to make a compound or chemical to share electrons for reasons that are complex (I don't think I understand all of it to be honest).

@Quantum you don't netralize the ionic charges wen you dissolve $\ce{NaCl}$ in water.  You insulate them from each other instead, with solvation and the dielectric constant of the water.

@OscarLanzi http://www4.ncsu.edu/~hubbe/Defnitns/Neutrlzn.htm

I am not sure whether you are referring to something more specific, but charge neutralization doesn't have to occur only via electron transfer.

Ok so this is very simple concept
There are 3 ways known in which atoms bond ie join to each other. Every atom needs 8 electrons in their outermost shell(valence shell) to be stable. Stable means it does not react with other atoms and form more bonds. So if it is able to get 8 electrons in their outermost shell by combining with other atoms, then it becomes stable. Now once this part is clear, we can proceed further. Now an atom can have various no of electrons in their outermost shell(valence shell) ranging from 1 to 8. For example Sodium has 1, Magnesium has 2, Aluminium has 3, Chlorine has 7 and Neon has 8. Neon is stable( as u mention happy) because it has 8 electrons in the outermost shell.But what about the others (Sodium , Magnesium etc etc?) In their case, they need to seek help of other atoms by either taking electrons, giving electrons or sharing electrons.(also donating electrons but that is much more advanced. Right now we can skip that).
Now these atoms are pretty smart. If they have electrons less than or equal to 3, they will give away their electrons to some other atom which needs some. So instead of trying to find 5-7 more electrons, they simply give away their 1-3 electrons and depend on the prior shell which already has 8 electrons.On the other hand, atoms with 5 or more electrons will search for atoms which are ready to give away such electrons and take from them instead of losing 5-7 electrons which is a large no. This, is called electrovalent bond. A trick to remember
less thn 4= give electron
more than 4= take electron.
Whichever combination forms 8, match made :) for eg Sodium gives one electron, and Chlorine takes one which has 7 electrons. Or it could also be this way- 1 Magnesium has 2 electrons to give, while chlorine required only one, so they play smart. Instead of one chlorine, 2 chlorines come into picture as 2 chlorines require 2 electroms. They form MgCl2 (Magnesium Chloride).
Now the part where you may most likely be confused is covalent bonds, which are formed by "sharing" of electrons. In the previous case, it was not sharing, it was give and take. Sharing means both the participating atoms give, and both take. In previous case, it was 1 give one take. In this this case, it is both give and both take. How? Lets take example of O2 in this case. An oxygen atom has 6 valence electrons. So it requires 2 more.. how can it do so with another oxygen? Well, its quite simple. Oxygen takes two of its electrons and shares it with another oxygen, which in return also shares it with 2 electrons. They do not exactly LOSE their electrons to the other one, the 4 electrons 2 from atom a and two from atom B are shared, are in between, kind of like a tug of war with equal force on both sides. Thats covalent bond.
Phew, enough said, its more about how you visualize it. If you can visualize this, then congrats, you mastered the basics.

It would be useful to note that you can never have 100% ionic bonds. The ionic bond is regarded as a rather extreme case of covalent bonding and will therefore always have some degree of covalent character. When the electronegativity difference between two elements is less than 0.5 the bond is said to be a non-polar covalent bond with more or less equal sharing of the electrons between the atoms. When the difference is between 0.5 and 1.7 then the bond is said to be a polar covalent bond with the electrons spending more time around the element with higher electronegativity. Above 1.7 difference, the bond is considered to be ionic but as I said, that really is a very very polar covalent bond. The 0.5 and 1.7 cutoff numbers are just rough approximation.  For example, the C-Li bond has electronegativity difference of about 1.5 and can either be regarded as ionic or covalent. It is more appropriate to think of bonds belonging to a spectrum which transitions from non-polar covalent, through polar covalent, towards ionic bonds. There are formulas you can look into which allows you to calculate the percentage ionic character of bonds which tells you how close to 100% ionic the bond really is.


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Why isn't 4s1 electron of Chromium the last electron?
I know that to aid the symmetry of singly occupied orbitals one electron jumps from 4s orbital to 3d orbital thus giving Chromium the configuration 3d5 4s1.
But my question is, when we're evaluating the quantum numbers of the last electron, shouldn't we definitively use the farthest electron from the nucleus? Irrespective of which electron is assorted last. Because it's not like a Chromium atom is created by sequentially inserting electrons one after the other voluntarily.
In a perfectly stable Chromium atom, which has the configuration Ar(18) 3d5 4s1, the last or ultimate electron in terms of energy level, orbital and radius, should be the 4s1 electron. Why is it that most references I found tell me to consider the 3d5 electron?

3d is the highest energy, 0.04 Hartree higher than 4s according to NIST's Atomic Reference Data for Electronic Structure Calculations, Chromium

Oh so we calculate the electron with the highest energy? I reckon then the 4s1 electron comes all the way after the five 3d electrons. Right?

@AniruddhaGS I don't understand what you mean by "comes all the way after"?

I mean when we calculate quantum numbers for the last electrons. 4s1 doesn't even come into consideration until we've discussed all five 3d electrons.

Also I should say that the reason why I'm asking this is that in a discussion we were asked to take into consideration the quantum numbers of the last electron of Chromium. I considered 4s1 as the last electron and so did a few others. But most considered 3d5 as the last electron. Nevertheless, which one is the correct consideration?

@AniruddhaGS "last electron" doesn't mean anything to me.  See this question about which electrons are written last in a configuration: http://chemistry.stackexchange.com/questions/27538/abbreviations-in-electronic-configurations/27553#27553


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Is the final temperature of a solution affected by the order of mixing components?
For example, I have two beakers each containing water. One is at 90 $^{\circ}$C and one is at 10 ${^\circ}$C.
Will the final temperature of the solution I make from mixing the two components be different if I slowly poured the 90 ${^\circ}$C water into the 10 ${^\circ}$C water and vice versa?

No - the temperature of the final solution is not dependent on the order in which the components are mixed together.

If that's the case I would appreciate it as an answer. Also, not sure why I have a downvote.

If the components may chemically react together, the answer is surely yes. I'm sure any reasonably competent chemist could even come up with examples where mixing the components together in one order is perfectly harmless, while mixing them in another order will produce a violent explosion.

Temperature is a so-called state function:

In thermodynamics, a state function or function of state is a function defined for a system relating several state variables or state quantities that depends only on the current equilibrium state of the system. State functions do not depend on the path by which the system arrived at its present state. A state function describes the equilibrium state of a system.

The final temperature of your system does not depend on how the system got there (order of mixing, for example).

Whilst the order of mixing does not matter in theory (see e.g. Todd's answer) it can play a role in praxis: Consider an imperfectly isolated system (e.g. a coffe cup) in air. The loss of heat is among other things proportional to the temperature gradient (basically local difference in temperature). An intuitive example would be that it matters whether you let your tea cool for a while and then add milk, or add milk and then let it cool. The former will be colder after the same time.
Another thing to consider is that it gets more complicated once you have changes in phase. E.g. a molten salt will likely have much more energy than a solution at the same temperature. If you heat the components separately and later mix you will have a higher temperature than when you first  make the solution and then heat them. (The example with salt might be a bit contrived as they generally have high melting points.)


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Can sodium acetate anhydrous be used as a desiccant?
Sodium acetate trihydrate can be converted to sodium acetate anhydrous easily with an oven. I've been told to quickly store sodium acetate anhydrous in a sealed container because it takes water from the air and turns back into the trihydrate form. It's also non-toxic, and is even sometimes used as a food additive. For these reasons, would sodium acetate anhydrous be viable as a desiccant? Perhaps as a less-toxic replacement for silica gel?

Many salts can be (and indeed are sometimes) used as dessicants, but what's wrong about silica gel, really? That it is not edible? Why would it be? Your computer is not edible either, so what?

@IvanNeretin For lab use, silica gel and other common desiccants would certainly be better, but I figured that something non-toxic (edible, even) would be better for use in, for example, beef jerky packages.

Now that makes sense. Well, $\ce{CaCl2}$ is another lab dessicant which is more or less edible, but it might be too strong to use on food... Wait, why would beef jerky need a dessicant in the first place? Isn't it usually vacuum-packed?

@IvanNeretin I'd imagine it would be for when you open and reseal the package. You'd let moisture from the air inside the bag.

No, not in most applications. While it may act as a dessicant, it is fairly volatile, so you'll get acetic acid evolving from it in the presence of water. Generally, you don't want to expose what ever you're keeping dry to acid. The two obvious issues when it comes to food uses (ie toxicity concerns) is smell (and taste) and acceleration of spoilage (although in the cases of acidic foods, it might not be an issue). You don't generally want your dessicant mixed into the product so keeping them separate is optimal but with the volatile acetic acid it could be a problem. If you wanted to use its acidity to enhance stability or taste (mixed into the food), then you'd have problems with changes in consistency as water was absorbed.

A quick look at Wikipedia says that the specific surface area of silica gel is  800 m$^{2}$ /g as it has a nanoporous structure.  I don't know the figure for sodium acetate but I would think that it does not have such a nanoporous structure and therefore would not have as much capacity to absorb water (note that technically silica gel adsorbs water onto the surfaces inside its pores, rather than acting like a sponge).
Also, silica gel itself is not toxic per se, rather it is the coloured indicators that are added that are not good to eat (cobalt (II) chloride is used, of which the anhydrous form is blue and the hexahydrate purple).


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Agar gel vs aqueous solution
Can someone tell me why - in experiments to observe corrosion of iron nails - we often use a preparation of agar gel (with $\ce{NaCl}$, phenolphthalein, etc.) instead of an aqueous solution?

If you are trying to observe corrosion in a specific area, agar is a convenient way to prevent convection currents from obscuring what caused the corrosion and for localizing the indicator.


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Efficient oxidation of KCl to KClO3
I am trying to convert $\ce{KCl}$ to $\ce{KClO3}$ through an efficient process.  I realize that I could electrolyze a solution of $\ce{KCl}$ to obtain some $\ce{KClO3}$ as a precipitate, however this seems inefficient as some chlorine will inevitably escape as a gas. Could an oxidizer such as hydrogen peroxide or sulfuric acid achieve this?  I realize that $\ce{Cl-}$ would be oxidized to $\ce{Cl}$ but I thought that there may be a way around that.  What is the best method?


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Cross-linking in Latex
My understanding was that cross-linking would generally make a material harder and less flexible, but if latex does have cross-linking, how would this affect its shape and hence its properties, especially as a material with elasticity?
In addition, is there a method of testing if a substance has crosslinks?
I have googled to no avail.

"Raw" latex consists of long hydrocarbon chains. The chains themselves have covalent bonds and are therefore very strong; imagine them like hemp threads. A volume of latex is basically a huge mess ("ball of wool") of those chains, which are weakly held toghether by van der waals forces.
This raw latex is therefore a thermoplast: It is soft and sticky when hot and rather brittle when cold. This must have been rather annoying when only this form was known. It is also not very elastic but malleable,  (is that the word?)
But now it was discovered that sulfur likes to form cross-connections between said chains. This basically makes the latex a single, huge molecule (hence a duroplast). But there are not too many of those cross connections, and the material looks rather like a loose net. If this vulcanized latex is streched, it can regain its form rather well; it is therefore called an elastomer. The more sulfur-bridges there are, the harder is the material; i.e. a hair band has less bridges than an ice hockey puck.
I think this view should explain the behaviour of latex quite well and on an intuitive level.

Testing for crosslinks seems rather difficult in general; but in the case of latex you can test for sulfur however. If you only have an emulsion this is very easy, if it smells repugnant there is sulfur in it. If you have solid material you might as well just test for elasticity.
In general it seems a save bet that elastic materials have crosslinks.

Ah thank you! So what you mean that in the process of vulcanisation, the crosslinking is actually responsible for the properties that allow an elastomer to stretch and bounce back into shape without breaking or deforming?
And that it would make sense to conclude that then they must have some cross-linking, but not to the extensive degree that would make them a thermosetting polymer?
I think malleable is an appropriate descriptor, though I think I've only heard it used in the context of metals in chemistry.


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How do sugars react with cobalt(II) and ammonia?
I did an experiment regarding the identification of a range of monosaccharides and disaccharides and wanted to know what the exact purpose of this experiment was and what is was testing.
The following steps were carried out:

Add 10 mL of 5% solution of fructose, glucose and sucrose to separate test tubes.
Add 4mL of cobalt (II) chloride to each test tube.
Heat each test tube to near boiling, then cool.
Add 8 drops of ammonia to each test tube.
Record any colour changes.

Fructose was the only one with an observable colour change.
Why / how does the fructose behave differently from other saccharides? Is it because fructose breaks down in a different way from the other sugars?


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Sublimation of Iodine from Iodine Povidone
Iodine povidone is a complex of iodine and the polymer povidone. I am wondering if one were to heat it enough, the iodine would sublimate and could be collected. Normally iodine boils at 184.3 °C, according to wikipedia.
A video of how the procedure might work, except iodine crystals would replaced with boiled down iodine povidone: Sublimation of Iodine Time lapse
This can be used to purify iodine, so from my understanding, you would be able to vaporize the iodine and leave the polymer behind.
Does the complex with povidone change how this would work?
I apologize if my question is ignorant, as I do not have a great understanding of how complexes work.

Yes you can because it is in a complex with the povidone. You just have to heat it a little bit extra. This is because a complex is "tangled up" molecules. In this case it is Povidone and Iodine. Nile Red used a different process (liquid-liquid extraction), but here is the link. https://www.youtube.com/watch?v=FNf8PSda7iI

Yes, I've seen Nile Red's video. Since he removed most of the polymer through his process, I was not sure if you could merely sublimate the iodine out. Any idea if what kind of heat you might need to do so?

Would about ~200 C be enough?

Should be. Experiment around

150-180 C is when Povidone Melts

As long as your povidone doesn't evaporate/boil your experiments should go fine

Keep Note that only 1% is iodine

another povidone question: How to align povidone-iodine molecules in order to look for optical polarization

its not sublimation, its evaporation

A procedure is available on Erowid for this process.

Make or buy a coldfinger (see below), boil your fluid to absolute dryness, without burning the dry restant, then hang your coldfinger in an appropriate Erlenmeyer, and heat gently the now dry restant. Make your coldfinger fat-free with acetone first (on the outside where the I2 will condense). The I2 will sublimate from the solid form into the gas form, without getting a fluid first (that's sublimation), and the gaseous I2 will resublime from the gaseous form into the solid form against the cold wall from the coldfinger. A little bit will of course also resublime against the upper part of the Erlenmeyer, but that you can also scrape off later, when all your 10% I2 hangs on the cold glass walls.


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Is biphenyl considered to be aromatic?

Biphenyl has 12 π-electrons and according to Hückel’s rule, it should not be aromatic because $12 \neq 4n+2$, but it nevertheless is aromatic. Why is that so?

@henry It isn't a single aromatic system but two independent ones (there is mostly free rotation around the c-c bond that joins the two rings).

Huckel's rule only applies to monocyclic systems without side groups. In fact, this rule is so frequently misused I'd rather have people forget about it altogether.

@IvanNeretin According to Geoff Hutchison's answer here, Pyrene too doesn't follow Hückel's Rule, but still it is aromatic. He has provided a modification to Hückel's Rule, where we count only $\pi$ electrons on the border. Can any such modification be also applied here?

I see no point in such a modification. How would you know when to apply it and when not to? Will it work with corannulene, for example? (It will not.) No, stick to the Huckel's rule when it is applicable, and when it is not, resort to the more complicated math (finding the eigenvalues, etc). Alternatively, you can point out that in the real biphenyl molecule the rings are twisted (non-coplanar), and hence not conjugated, so that each $\pi$ system is pretty much on its own.

@IvanNeretin How can we use math?! (If math starts to apply in chemistry, I will top the examinations in chem :D)

@Henry Pyrene is a fused-ring system which is why you need more sophisticated rules about aromaticity. That isn't needed here as the two rings are not fused and can be treated independently due to the relatively free rotation about the linking bond (strictly there is a small steric barrier to rotation which results in a preferred confirmation where the rings are ~40 degrees from being planar, which also implies they are two separate aromatic systems and not one single one.)

Math does apply in chemistry, but... Well, see this: https://en.wikipedia.org/wiki/H%C3%BCckel_method

For a discussion of measures of aromaticity in polycyclic systems, see e.g. https://www.nrcresearchpress.com/doi/pdfplus/10.1139/v75-133

The two rings in biphenyl are inclined at an angle of 44° to each other because of the large steric repulsion between the ortho hydrogens on each ring.

For Hückel's rule to be applied, the molecule must be able to be planar and it must be monocyclic.
Due to the molecule not being planar, the two rings are not strongly conjugated to each other and you cannot apply Hückel's rule to the whole system. Additionally, the whole system is not monocyclic. Hückel's rule can be applied to each ring individually and they are both aromatic.

I think I'd disagree that there isn't conjugation between the rings in biphenyl. Yes, there is more delocalization if you can force the system into planarity, but there's definitely electronic communication even in twisted systems.

When you say that both the rings are inclined to each other at an angle of 44°, it is the most stable conformer you are talking of, isn't it? Just being sure


text:
Is the cyclopropenyl anion antiaromatic?
According to my book, a compound is antiaromatic if it is cyclic, planar, and possesses a fully conjugated system of p-orbitals with $4n$ π-electrons.
However, I have also been told that the cyclopropenyl anion is neither aromatic nor antiaromatic.

How is this so? It has four π-electrons and therefore should satisfy the criteria listed above.

Related: Is cyclobutadiene anti-aromatic?; Why can't antiaromatic compounds just escape planarity and become non-aromatic to avoid destabilization?

When you have a ring that has $4n$ π electrons it's anti-aromatic if you force it into a fully conjugated cycle.
Often such rings find a way to break the unfavorable (or at least, less favored) anti-aromatic conjugation.  Large ones like cyclooctatetraene are likely to bend so all ring atoms are no longer in the same plane. Small ones like the cyclopropenyl anion shown here can't bend the ring out of planarity easily (or do it at all with only three atoms), but they can still break the conjugation by using unequal bond lengths or moving a ligand out of the plane giving that ring atom a pyramidal bonding geometry.
The pyramidal bonding geometry seems to be what cyclopropenyl anion actually does.  A computational result is shown by Kass [1]; the top carbon atom has its bonds directed towards the observer indicating a pyramidal geometry (see below, from  https://comporgchem.com/blog/archives/2987 and [1]):

Reference
1.
Kass, S. R. "Cyclopropenyl Anion: An Energetically Nonaromatic Ion," J. Org. Chem. 2013, 78, 7370-7372, https://doi.org/10.1021/jo401350m.

I would not say that unequal bond lengths break the conjugation. See, the p-orbitals are still there, and still able to overlap and interact. Or look at butadiene: isn't it one conjugated structure?

Cyclobutadiene (just butadiene" is a difgerent, non-cyclic molecule) has unequal carbon-carbon bond lengths and its ring is not considered fully conjugated.

Yes, I intentionally asked about butadiene, which is different and non-cyclic. Though it applies to cyclobutadiene as well. Now you say they are not fully conjugated; this implies they are still conjugated a little, aren't they? Anyway, I see your point: the pattern of orbitals is no longer the same as predicted by the Frost circle, the degeneracy is broken, so cyclobutadiene is not a biradical, but just a diene, so it is not anti-aromatic, but just not aromatic. This makes sense.

@OscarLanzi I couldn't get your point. What does cyclopropenyl do to break conjugation?

I checked Wikipedia, they say cyclobutadiene forms an equilibrium between square and rectangular structures, as if the square structure were a transition state between two petpendiculatly oriented rectangles.  We get teo stereoisomers of 1,2-dideutetocyclobutadiene depending on whether the deuterium atoms are attached ot a long of short side of the rectangle.  Whether any conjugation remains in the rectangular forms is ultimately immaterial:  the molecule is destabiluzed by stetic strain, and only aromatic conjugation could overcome that.

Henry please read my answer.  Cyclopropenyl anion breaks the planar structure by pyramidalizing a bond to a ligand atom.

@OscarLanzi "Cyclopropenyl anion breaks the planar structure by pyramidalizing a bond to a ligand atom." Can you please provide a diagram for reference? Thanks.

http://comporgchem.com/blog/?p=2987 has a good computational description and a 3D rendering which might help @Henry. Although mathematically, three points always form a plane, remember that the charged carbon has the option of adopting either an sp2 or sp3 hybridization. it can only be aromatic if it adopts sp2 so that the remaining p-orbital is planar and can interact effectively with the other p-orbitals. However, the bond angle puts a strain that is quite significant whereas adopting sp3 will reduce this significantly. But if that carbon is sp3, then there can be no continuous p-orbitals

@IvanNeretin Altering bond lengths does not necessarily break the conjugation, you're right about that. It does lower the symmetry of the molecule, which means it splits degeneracies of orbitals. The main reason for the destabilisation is that you have either too many or to few electrons to fill those degenerate orbitals completely or avoid it. In this sense, it breaks the ideal conjugation.

Wouldn't the carbanion keep its lone pair in a hybrid orbital instead of a p orbital to avoid anti-aromaticity, hence making it non-aromatic due to absence of complete conjugation?

That is what pyramidalization effectively does.

I don't understand why cyclopropenyl anion is planar(necessary condition for anti-aromaticity). I learnt that a carbanion is always sp3 hybridised.

Putting the lone pair in an sp3 hybrid orbital would break the conjugation and result in a more stable, non-aromatic rather than very unstable anti-aromatic.

Well, that looks kida like TL:DR for Oscar's answer - rather not enough to stand on its own.


text:
Electronegativity and anion stability
If two similar sized atoms have different electronegativity values, the atom with higher electronegativity holds the electrons more tightly, and thus is more stable. Example: hydroxide vs azanide (NH2-).
If two atoms have different sizes (down the periodic table), the larger atom has electrons more spread out and thus is more stable.
Example: fluoride vs iodide anion
To me, these two statements are contradictory. One says tight electron cloud means not reacting while the other says spread electron cloud means stability (less repulsion). Is one of the statements wrong? I'm sure I am missing something here. Thank you in advance for your help.

The word "stable" has several quite different (and indeed sometimes contradictory) meanings. How is fluoride anion less stable than iodide? The latter is much easier to oxidize. But it is "more stable" in a different sense: it is harder to protonate ($\ce{HI}$ is a stronger acid than $\ce{HF}$).

When I mean stability, I mean how well are the electrons kept to the atoms themselves and not shared with their neighbors? I was thinking about repulsion between negative charge of electrons. For specific case, I was thinking about nucleophilicity in protic solvents. With electrons less likely to stay with the atom, the electrons are more likely to attack (electronegativity is small). With larger atom, electrons are more spread out, and there's less repulsion, but larger volume for interaction with other molecules. Somehow larger atoms react less with hydrogen of solvent.

I may be confusing between nucleophilicity and basicity as well, because it seems hydrogen bonding is more likely to occur with smaller atoms such as fluoride ions. My thought is that if fluoride ions attract more hydrogens, they would also attract more electrophiles. Of course, somehow basicity and nucleophilicity has inverse correlation in protic solvent for some unknown reason.

@IvanNeretin Stability you talk about is achieved in aqueous solution. In anionic gaseous form the stability is different and I- is more stable than F-.

@IvanNeretin I meant stable as in electron's tendency to stay there. It seemed that I was simply confused on the concepts of electronegativity and electron affinity.

I believe that you are mixing up the concepts of electronegativity and of electron affinity.
The first statement you present is related to Electronegativity.

If two similar sized atoms have different electronegativity values, the atom with higher electronegativity holds the electrons more tightly, and thus is more stable.

Yes. This is generally true.
Electronegativity establishes how well an element can pull electron density towards itself.
The second statement is about electron affinity.

If two atoms have different sizes (down the periodic table), the larger atom has electrons more spread out and thus is more stable. Example: fluoride vs iodide anion

Generally true as well. Flouride anion is less stable than iodide anion due to the repulsion of electrons in the outermost octet of fluoride. But this rule is not absolute as chloride has the highest stability (in the gaseous state) though it is above bromine and iodine.
Edit:
Considering some confusion in the comments I shall put this here.
The electronegativity is generally not compared based on inter-electron repulsion because the shared pair doesn't completely enter the orbits of the electronegative atom but is shared and so electron density is between the two atoms and not concentrated on one atom for repulsion to take much effect.

The point of my question is that increase in electron density and increase in electron distance both decreases electronegativity and electron affinity. I don't think this is an issue of electronegativity vs electron affinity since the two statements are true for both concepts.

Okay, after hours of frustration, I think I understand more of what you mean. Electronegativity does not take into account electron repulsion but electron affinity does. This is because electronegativity is about tendency of electrons to move to one side in a bond (thus somewhat distant from the atom) while electron affinity is about an electron completely becoming part of the atom and thus getting the full effect of electron repulsion.

@KennyKim These facts are all predicted by quantum mechanics and confirmed by experiment. It's hard to say just by qualitatively comparing ionic size and charge density.

@KennyKim You seem to have cleared up confusion. Yes, the electronegativity is generally not compared based on inter-electron repulsion because the shared pair doesn't completely enter the orbits of the electronegative atom but is shared and so electron density is between the two atoms and not concentrated on one atom for repulsion to take much effect.

@shre_sudh_97 Doesn't greater electronegativity pulls electrons towards smaller volume therefore increase charge density?

The electronegativity stabilizes due to the lower coulombic potential between opposite charges. However, as we go down the periodic table, we are adding electron shells to the atom which dramatically increases its size. This means the negative charge is distributed over a wider space, reducing repulsion among all electrons.
You might say its a competition between attraction to the nucleus and repulsion among electrons, and reducing repulsion among all electrons takes precedence over increasing attraction because of how quickly the volume expands. The key point is that repulsion is mostly mitigated by adding shells, which does not take place in your example of NH2 vs OH.


text:
Gilman Reagent with vinyl chloride
What are the products formed when vinyl chloride reacts with dimethyl copper?
I thought the product formed would be propene. But answer given is PolyVinylChloride. How does polymerization take place?

I thought Gilman Reagent is lithium dialkyl copper. I would expect propene to be formed too.

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

@orthocresol so the reaction will start with a nucleophilic attack by a methyl carbanion on vinyl chloride. Will this nucleophile attck the anti-bonding pi orbital?

Yes. You can't make propene - that would involve a SN2 (which organometallics are notoriously bad at) at a sp2 centre (which is extremely rare to begin with)


text:
Hybridization of Cu2+ in tetraamminecopper(II) sulfate complex
$\ce{Cu^2+} $ has 9 electrons and the $\mathrm d$-orbital is almost completely filled (except for 1 electron vacancy), eventually forming $\mathrm{sp^3}$ hybrid orbitals, where each $\ce{NH3}$ donates a lone pair of electrons.
But it has been found to have a square planar geometry, hence completely contradicting the above reasoning. So, please explain what is the hybridization of the $\ce{Cu^2+}$ in the given complex?

Hybridisation is determined by geometry. It is purely a mathematical trick to create nicer orbital representations. You cannot assign it without knowing the geometry in advance.

@bon Given that it is sqaure planar,whats the hybridization

https://en.wikipedia.org/wiki/Tetraamminecopper%28II%29_sulfate - it isn't square planar

@Mithoron then what is it?

Do not use hybridisation to describe coordination compounds; it is bad.

I feel a bit like a broken record. Do not use hybridisation to describe coordination compounds; it is not helpful.
The cation you called tetraamminecopper(II) is more accurately described as tetraamminediaquacopper(II) $\ce{[Cu(NH3)4(H2O)2]^2+}$, because two aqua ligands are there, just further away. The four ammine ligands donate two electrons each, as do both the aqua ligands, totalling 21 electrons overall. Due to the strongly Jahn-Teller distorted geometry and other reasons, the bonding interactions are overall stronger than antibonding ones making the complex favourable.
The interaction between the metal and ligand involves $\mathrm{e_g, a_{1g}}$ and $\mathrm{t_{2u}}$ orbitals (when viewed as an undistorted octahedral complex). That would include two d-orbitals, three p and one s. If your life depended on it and you are the last person on Earth, you may attempt to extract a hybridisation from that. But don’t do it.


text:
Hydrolysis of Aspartame with a warmed alkali
Aspartame has this structure below

Shouldn't the peptide group be oxidised along with the ester group?
Then why is the answer to the question below B?
EDIT : I'm sorry I misunderstood a friend's marking, the answer is A. I deeply apologise for my stupid mistake.

Thank you very much!

Hi I've put in a new image

No oxidation is taking place. It is just a hydrolysis reaction. I do agree with you though that both groups should be hydrolysed.

Amides are much slower to hydrolyze than esters, so you could obtain the bond 2 product under mild conditions or if you limited the reaction time. However, unless the specific conditions and hydrolysis rates for this reaction were covered in the readings or lecture notes, you should get credit for answer A.

But why would the answer be B?

This looks like it's from an exam so there's a very high possibility that the person setting the question and answer intended the ester group on the right to be something else. I am largely of the opinion that the answer should be A.
However, there is one way I can imagine that you can try justify "B" being the answer (although it is a long shot):

Under alkali conditions, we can expect the acid group from aspartic
  acid to be deprotonated to carboxylate which can act as an
  intramolecular nucleophile.
We take note that intramolecular reactions are generally faster than
  intermolecular reactions. The carboxylate can attack either the ester
  group or the amide group. An attack on the amide results in a
  5-membered ring transition state, and an attack on the ester results
  in a 9-membered ring transition state. Since 5-member rings are more
  stable, the carboxylate attacks and breaks bond 1, leaving bond 2
  alone. Aqueous alkali then destroys the acid anhydride that is formed.

That said, this is highly unlikely given that esters are less stable to base-catalyzed hydrolysis than amides. So if Bond 1 Breaks, Bond 2 must have also broken.


text:
Can I calculate the percentage of alcohol if I only know that 100ml wheight 90,135 g
Imagine I have a bottle of 100ml which weight 90,135gr. I know that there is only water and alcohol in it. I know that the weight of water is 0,998 gr/cm3 and alcohol is 0,79 gr/cm3. And when I try to figure out how it is divided I'm doubting whether it is at all possible or it is by trying out around 42% alcohol. But is that true and how can you easily calculate this?

If this is a homework question the answer may be fairly simple. If it isn't you might want to make into account that water and ethanol are not ideal liquids and the volume of the mixture will not equal the volume of the two components added up.

It is not home work

100 $cm^3 =90.135 g$
density is $0.90135 g / cm^3$
$0.90135 g / cm^3$ = $0.998x grams + 0.79y grams/ 100 cm^3$
$x+y=100$
$y=100-x$
redefine it as
$0.90135 g / cm^3 = (0.998x grams + 0.79(100-x) grams)/100$
$90.135=(0.998x grams + 0.79(100-x) grams)$
$90.135=0.998x grams + 79 - 0.79x grams$
$90.135=79 - 0.208x grams$
$11.135=0.208x grams$
$11.135/0.208=x$
$53.5336538=x$
$53.5336538 mL$ of water.

For homework questions we have a policy where you do not simply post the answer without explanation but rather that you give hints that lead the asker towards their own solution. We are not a homework completion service. Also, this is horrible to read. Please format it using LaTeX.

Sorry I messed up on formatting and I removed the answer to the question.

I would contest that the answer is incorrect as volumes of solvent mixtures are not simply additive,  nor are they simply linear relationships


text:
Does aromaticity require a co-planar pi system?
It's well established that angular twisting can decrease the degree of delocalization across a $\pi$ system (e.g. biphenyl).
Moreover, there's considerable debate about aromaticity in spherical species like $\ce{C_{60}}$.
So let's ask a naive question for discussion. Does aromatic stabilization require a planar delocalized $\pi$ system?

Good prompt points would include buckybowls and Mobius aromatic systems.

Then helicenes would also fit I guess.

The $\pi$ system is not obliged to be planar at all; the only limitation is to keep the angular twist between the adjacent atoms low enough so they still can be conjugated. (Biphenyl violates that, so it is out of luck.) The whole structure can be pretty much anything. It can be slightly out-of-plane, like corannulene or circulene; it may have cylindrical shape (see the example below), in which case the Hückel's rule still applies, or it may be all twisted to the topology of a Möbius strip, in which case the said rule is reversed.
Fullerenes are not aromatic, unless you stretch the definition of aromaticity to utter pointlessness. Some of their derivatives are, though. $\ce{C60Cl30}$ comes to mind: its 30 remaining $sp^2$ carbons are grouped in two 6-membered benzene-like rings and one 18-membered belt, which (judging by its equalized bond lengths) is an example of a cylindrical aromatic system.

Maybe I'll expand on spherical aromaticity (or rather lack thereof).

Done: http://chemistry.stackexchange.com/questions/53400/why-are-spherical-compounds-with-a-half-filled-homo-concidered-aromatic/53464#53464


text:
Why does 1,2-dichloro-4-nitrobenzene undergo SNAr para and not meta to the nitro group?
In the reaction of 1,2-dichloro-4-nitrobenzene with sodium ethoxide, why does ethoxide end up substituting the chlorine para to the nitro group rather than the chlorine meta to the nitro group?

I think that C–2 should be the most electron-poor carbon (and hence prone to nucleophilic attack) since it experiences a combined electron withdrawal from the nitro and chloro groups flanking it. Shouldn't the nucleophilic ethoxide anion attack there instead?

Related: Nucleophilic substitution mechanisms in aromatic compounds

You can obtain an answer by looking at the electronic structure.  $\ce{NO2}$ is a meta director for electrophilic aromatic substitution (EAS).  The question asks about nucleophilic aromatic substitution (NAS). Aromatic substitutions involve a free pair of electrons resonating around a structure. Electrophiles may add at any resonance site where the pair of electrons are. Nucleophiles may inversely add to any site where electrons are not present shown in the picture below. Therefore if a functional group is meta directing for EAS it will be ortho/para directing for NAS and vice versa.

The nitro group is strongly meta-directing ... for electrophilic substitution.  But look a little more closely.  Is this really electrophilic substitution?
Once you figure that out, draw the mechanism as suggested in @jerepierre's comment.  Now you should see what really happens.


text:
How can value of acid dissociation constant is vary with concentration of acid?
First of all I have learned that $K_\mathrm{a} = \frac{[\ce{H+}] [\ce{A-}]} {[\ce{HA}]}$ . Also I have been informed that $K_\mathrm{a}$ value is independent of concentration.
So let us take an example. Suppose 1 mole of acid $\ce{HA}$ dissociates to form 0.9 moles of $\ce{H+}$ ions and 0.9 mole of $\ce{A-}$ ions and 0.1 mole of undissociated $\ce{HA}$. So, the $K_\mathrm{a}$ value will be $K_\mathrm{a} = \frac{[0.9] [0.9]} {[0.1]}=8.1$.
But if we take 2 moles of same acid then, 1.8 moles of H+ ions and 1.8 moles of A- ions will be formed leaving 0.2 moles of undissociated HA. This time $K_\mathrm{a} = \frac{[1.8] [1.8]} {[0.2]}=16.2$.
So we can see that $K_\mathrm{a}$ has increased with increase in concentration. Can you tell me where my assumption is wrong?

Before we begin, note that $K_a$ is defined in terms of concentration and not amount. Furthermore, $K_a$ is a constant so it must be the same provided the conditions, other than concentration, are the same.
The nature of equilibrium is such that it does not always give the same absolute reaction (i.e. a fixed percentage of reaction, or the reaction does not proceed to the same "position" each time). This is unlike a fixed forward reaction which always ends at 90% completion. In other words, although the 1 molar solution of HA gave 0.9M of $H^+$ and $A^-$, the 2 molar solution would only give out...
$$ K_a = 8.1 = x^2/(2-x) $$
where $x$ is the number of moles of $H^+$ produced. When you solve this, you would get $x=1.66M$

But why Does this happen?Why is 90% dissociation not seen in 2 moles? Can you explain please?

@Jajati you could get 90% dissociation in 2 moles, if the concentration was still 1M (i.e. You had 2 dm3 instead of just 1 dm3). But if you're increasing the concentration by dumping dry HA into the system that is already in equilibrium, you could think of it as the system trying to keep the number of moles down in accordance to ale Chatelier's Principle.

OK.But if there is a solution of 2M . In that case will the % of dissociated HA molecules be less than the % in 1 mole?Does that mean that in high concentration,strength of an acid decreases?Because we see less degree of dissociation?

So yes, % dissociated decreases but that doesn't mean that the acid gets "weaker". So if you look at the 2M vs 1M solution, the 2M solution still has more dissociated protons than the 1M solution.

ok thank you very much for your help. Then last question. If the % dissociation of 2M solution is less than 1M then does that mean that if we obtain 0.5M solution then we will see more % dissociation . If yes then why so?And if not then can you please explain the relation of %dissociation and molar concentration .

Can I get the answer sooner please. It is urgent.

Hi @Jajati, it's all in the equation already, and I would recommend against generalizing because this really depends on the conditions of the system (i.e. is there another proton source, or is there already an excess of counter-ion, etc.). If you apply Le Chatelier's Principle it should become clear to you why dissociation would increase/decrease as you reduce/increase the concentration.


text:
Temperature in the Gibbs Free Energy equation
My professor gave us 2 different ways to calculate this problem. I did it using the Clausis-Clapeyron equation.

However, I don't quite get the second method because I thought that if the change in enthalpy and entropy of formation are in standard states, the temperature (T) in the Gibbs Free Energy equation meant to be 298K? And then making the temperature in the second equation 1100K, to find the equilibrium constant?

http://chemistry.stackexchange.com/questions/41862/what-is-the-real-difference-between-%E2%88%86g-and-%E2%88%86g/41864#41864

In fact, Gibbs free energy is not defined at exact temperature. It is defined for standard states of all compounds in equation. So you really can use this solution of problem. However, it is a very simplified solution, because you assume that standard entropy and enthalpy do not change with temperature. This is not entirely correct assumption but if you took change of enthapy and entropy with temperature into account, it would be much more complicated.
You can find information about Gibbs free energy here
http://www.chem1.com/acad/webtext/thermeq/TE4.html#3A

To add to what you said, if the OP used the Clausius Clapeyron equation as his first method, he must also have assumed that the enthalpy change did not vary with temperature.  The two methods are mathematically equivalent, both if the temperature dependence (of the entropy change and enthaply change) is neglected and if the temperature dependence is included.


text:
Which isomers (cis or trans) are more water soluble?
As a general sense I know that cis isomers are less symmetrical and more polar in nature whereas trans isomers are less polar. Now given those information, would it be appropriate to assume that cis isomers are more water soluble owing to their polarity?
Also there was a discussion in our class where my teacher told us that between the cis and trans Butenedioic acids, the cis acid is not water soluble on account of the fact that it undergoes dehydration and the carbonyl groups are bonded by an oxygen atom in between eliminating hydrogen bonds (??).

I would like to know if there is any constant character when it comes to solubility of cis and trans isomers.

That's not quite correct and can be quite easily to be shown as wrong by looking up data tables and/or MSDS. Fumaric acid (the common name of the trans-isomer) is less soluble, with a solubility in water of 4.9 g/L whereas maleic acid (the cis-isomer) has a solubility of 478.8 g/L.
Maleic anhydride would also react in water to form maleic acid.

And regarding my last question? Is there any common rule governing it? Are either of cis or trans isomers more water soluble than the other?

@user30287 As with most things in chemistry, there is no hard and fast rule. I'm sure given a sufficiently complex molecule you can find an exception.

Actually in the cis isomer there is a repulsive force acting between the molecules but when we see the trans structure there is a compact structure, there is less repulsive force... so due to this repulsive force the molecules in cis isomer will be farther from each other and hence will be more soluble.

What exactly do you mean by repulsive force. I'd say there's simply much less H-bonding to overcome during dissolution.


text:
How to calculate peak ratios in mass spectometry for a polyatomic species from given isotope ratios?
Element Z has 2 isotopes, 55Z and 57Z with the abundance ratio of 2:1. It exists in nature as Z3. The ratio of the peaks at m/e ratios 165:167:169:171 is 8:x:y:1. What are the values of x and y?
I tried to sketch the graph..shouldn't the graph have 100% and 50% abundance of element Z? Ive been doing this for 3 hours now...need help! thanks!

This is a permutation and combination problem.
If the element existed as Z, and not $Z_3$, then yes you are right the peaks should be 2:1, but more correctly abundance would be 66%:33% rather than 100%:50%. Keep this in mind, it will be helpful when you do the calculation.
But now we are told that it exists as $Z_3$, which means we have to have 3 atoms of Z.

There is only one "way" to get 171, and that is to have all three atoms as 57Z. So from probability, this would look like $$P(M=171)=(1/3)^3 = 1/27$$
Similarly, we consider the case of getting 165, which is all three atoms have to be 55Z. Mathematically, this would then be $$P(M=165)=(2/3)^3 = 8/27$$
In the case of M=167, we need two 55Z's and one 57Z. So after doing the math we find that it is $4/27$.

Since this is homework, I'm sure you can do the case of M=169 by yourself now.


text:
What molecule is mostly responsible for the smell of petrol/gasoline when I refill my car?
Is it benzene or is there another aromatic molecule in gasoline with "heavier scents"? Does anybody know of a scientific investigation to provide evidence for an answer?

As you state, aromatic compounds such as benzene, naphthalene, xylene and toluene have noticeable odor (ergo "aromatic"). In addition, odorous oganosulfur compounds may remain, though in Europe and North America effort is made to remove them (kerosene and heating oils may have higher amounts of oganosulfur contaminants).
In the US, Methyl tert-butyl ether (MTBE) was a common strong-smelling additive that raised octane number and reduced CO emission, but it has been largely phased-out because of it propensity to leak into and damage aquifers.
A Safety Data Sheet list some other chemicals that may have a noticable odor, such as ethanol (added for the same purpose as MBTE).
A trained nose can detect and identify any or all of these compounds, so it really depends on the particular petrol you buy. As a rough guide, the aromatics may have a "sweet' or fruity smell to some people, though naphthalene has the smell of moth-flakes, MBTE a "burning" odor, and sulfur compounds may smell like rotten garlic.


text:
Entropy Change of Resistor

A very large swimming pool filled with water of temperature equal to $20\ \mathrm{^\circ C}$ is heated
  by a resistor with a heating power of $500\ \mathrm{W}$ for $20$ minutes. Assuming the water in the
  pool is not in any contact with anything besides the resistor:
Is the change of entropy of the resistor positive, negative, or zero?

My Attempt
We know that entropy is given by the formula:
$$\mathrm dS = \frac{q_\text{rev}}{T}$$
My first thought was that since the resistor is giving off heat energy to the water, $q$ must be negative and therefore the entropy change must be negative. However, does the fact that there is a continuous current of electrons passing through the resistor complicate this simple calculation? I am thinking that the current of electrons increase the entropy of the resistor and offsets the decrease in entropy due to heat loss. Am I correct?

The answer in the book is that the entropy change is $0$.

Resistor is  heating itself too, but as amount of water is huge there's almost stationary state here.

@Mithoron What do you mean by 'stationary state'?

Passing current produces 500W of heat which is passed on to water completely so it's temp doesn't change - process is stationery

@Mithoron Oh, okay. So q of the resistor is $0$ then, right?

Here entropy change will be zero because resistor convert the electrical energy into heating so it take as work not heat so entropy change will be zero


text:
What is at 298 K at standard entropy?

The standard entropy of a substance is the entropy of $1$ mol at $298$K and $1$ bar pressure.

What exactly is at $298$K? The system or the surrounding?
Let's take vapour as an example where $ \ \ce{H2O(g)} \ \ \ \ S^{\Phi}_{298}\approx188.8  \ JK^{-1}mol^{-1}$. Is the $298$K referring to the temperature of the gaseous $\ce{H2O}$? I'm confused.

298K refers to the temperature of the substance and its surroundings. 298K is about 25 degrees Celsius. In other words, the statement can be simplified as:

The standard entropy of a substance is the entropy of 1 mol at room temperature and pressure.

the part that's confusing me is if the substance $H_{2}O$ is also at $25ºC$ then how can we say it is a gas  and find its entropy?

As you mention, vapor, is water in the gas phase at temperatures lower than 100 oC. That's like humidity.

298 K and 1 atm is a hypothetical state for pure water vapor, based on treating the water vapor as an ideal gas.  You take the entropy of water vapor at 100 C and 1 atm., and then use the ideal gas heat capacity at constant pressure to then get the entropy at the hypothetical state of 298 and 1 atm using $dS=C_pdT/T$.
If you know the entropy of formation of liquid water at 25 C and 1 atm, another way of getting the entropy of formation of water vapor in the hypothetical ideal gas state would be to determine the entropy change for the following path:  drop the pressure on the liquid water to the equilibrium vapor pressure at 25C, vaporize the liquid water to water vapor at 25 C, compress the water vapor using the ideal gas relationship for compression of an ideal gas, to 1 atm.


text:
Semi-sandwich complex IUPAC name

I've tried looking here, but I can't figure it out.
As a start, what is just the cobalt and cyclopentadienyl anion compound called.
How does that name get added to the trimethylphosphine and carbonyl groups.
I want to ask further questions about this compound on chemistry stack exchange, and I think it would be a good idea to be able to use the name in those questions.

The name of this compound is carbonyl(η5-cyclopentadienyl)(trimethylphosphane)cobalt. Note that, generally, the prefixes denoting the organic groups and any other ligands are placed in alphabetical order before the name of the metal.
The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) presents the rules for formulating and naming coordination compounds (Chapter 9) and organometallic compounds (Chapter 10). An abridged version is included in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.
In particular, the ‘η5-cyclopentadienyl’ term used here is explained in Section IR-10.2.5.1. as follows:

The special nature of the bonding of unsaturated hydrocarbons to metals via their π-electrons has led to the development of the ‘hapto’ nomenclature to designate unambiguously the unique bonding modes of the compounds so formed. (…) The Greek symbol η (eta) provides a topological description by indicating the connectivity between the ligand and the central atom. The number of contiguous atoms in the ligand coordinated to the metal is indicated by a right superscript numeral, e.g. η3 (‘eta three’ or ‘trihapto’), η4 (‘eta four’ or ‘tetrahapto’), η5 (‘eta five’ or ‘pentahapto’), etc. The symbol η is added as a prefix to the ligand name, or to that portion of the ligand name most appropriate to indicate the connectivity, (…)
The ligand name η5-cyclopentadienyl, although strictly speaking ambiguous, is acceptable as a short form of η5-cyclopenta-2,4-dien-1-yl, due to common usage.
These ligand names are enclosed in parentheses in the full name of a complex. Note the importance of making rigorous use of enclosing marks, etc. to distinguish the above bonding modes from the other four cases below. (…)


text:
Water has H Bonds?
So I'm studying for  a chemistry test and one of the practice problems is "What is the intermolecular force present in water?" We made handy flow charts for this and so I determined that it was polar. The next step asks me to use a Lewis dot diagram to find the shape; symmetrical means London Forces and Asymmetrical means either Dipole-Diploe or H bonds. Easy enough; it's bent, which would make it symmetrical, right? The answer key says it has H-bonds and that makes sense, but I concluded that it had London Forces because bent is symmetrical? What am I missing?

Even if it's the kind of bent where there are two extra electron clouds (which it is) that doesn't change the fact that it is symmetrical...

Whether it has symmetry or not is completely irrelevant. Everything has London dispersion forces but in some molecules they are masked by stronger interactions such as hydrogen bonds.

Strictly speaking, all: Lewis, dipoles, and H-bonding are present in water. I suspect that there's a item writing problem here (read=poor wording). If the question is about just IMF's, then all of them, if it asks for the $strongest$, then H-bonding.

Any answer to your question depends on speculations about what your "flow chart" specifically is. You seem to confuse molecular shape symmetry with H-bonding. There is no simple relation between them. HOH using Lewis octet rules, gives the O atom two lone pair and so a bent bond angle is predicted. This in turn would suggest that water has a large dipole moment and so is a polar molecule. This says nothing about whether the H atoms or the central O atom can hydrogen bond. Water has two planes of symmetry: all three atoms (by definition 3 points define a plane) are in a plane so are symmetrical about that plane, and the two H atoms are mirror images in a plane through the central O atom and perpendicular to the first plane I mentioned. But so what? None of this speaks to whether either the H or the O atom form H bonds. The ability of an H to form an H bond depends on the Z-H bond polarity, NOT the polarity of the entire molecule. (where Z is the atom the H is (covalently) bonded to). Some Z-H bonds, for instance in methane, the C-H bonds make the H atoms very unlikely to H bond. Consider the chlorite anion ClO2(-) we can correctly predict structure using Lewis dot rules, we can predict that it will be polar (not to mention being negatively charged), but what can we predict about each atom's H bonding propensity? Nothing as far as I can see. Will any of the 3 atoms H bond? Well, since it is negatively charged, I'd expect it will. So, which of the atoms will do so? Lewis structure and symmetry don't help, as far as I can see. What would your flow chart predict (and what is the correct answer) ? IDK.


text:
Stereochemistry of cyclic hemiketal formation
$\hspace{5cm}$
I know that the acid protonates the carbonyl oxygen and this opens a pathway to an intramolecular cyclization reaction because the $\ce{-OH}$ group will attack the carbonyl carbon and a six-membered ring is formed.
I know that the carbonyl part is flat, and that the $\ce{-OH}$ can either attack from either the re or the si face. I'm not concerned about that. What happens to the methyl group on the wedge? Does that change depending on which side the $\ce{-OH}$ attacks the carbonyl? I am having a hard time visualizing what might happen to it.

No, the stereochemistry of the methyl group is not dependent on which face the $\ce{-OH}$ attacks from. This is apparent in the products of the cyclization of aldopyranoses; only two isomers (called anomers) are formed, which are classified as either $\alpha$ or $\beta$:

Depending on which position we are considering the stereochemistry of, a wedge in the straight chain could appear as either a wedge or a dash in the cyclic form. For the molecule you are considering:
$\hspace{.7cm}$
To visualize this, convert the skeletal structure into its Fischer projection:
$\hspace{4.5cm}$
$\hspace{2cm}$

I would expect the hemiketal with two equatorial methyl groups to be a tad more preferred in the reaction.


text:
Why does nitration of aromatic amines give meta directed nitro compounds?
Amines are activating groups and donate electrons to the resonance structure using the lone pair of electrons on Nitrogen.
But when aromatic amines are nitrated, for example when benzyl amine is nitrated, it gives a meta nitro benzyl amine.
Given the activating nature of amine groups, shouldn't the compound be an ortho-para directing variant?
I tried figuring it out. One explanation I've come up with is that when it's nitrated, the proton from the nitrous acid forms a coordinate covalent bond with the amine group closing the lone pair electron to the bond. As a result it becomes an $\ce{NH3+}$ group. That would mean that it has a deactivating effect or electron withdrawing effect. That's why the incoming nitro groups are attached to the meta positions. But then why doesn't the same happen for nitration of hydroxide groups, for instance phenol?

Have a look at this question http://chemistry.stackexchange.com/questions/7/why-does-nitration-yield-a-meta-product-in-this-reaction

Phenol isn't very basic, but aniline is pretty basic.

Oh thank you. I searched before making the question. But I guess I didn't use the right keywords for my search. Should have found that question.

A couple things here.  First, it's nitric acid, not nitrous acid, that is the nitrating reagent.  Otherwise, your own explanation is correct.  Orthocresol points out that phenols are sufficiently inert to protons, to avoid this situation.
Second, you can make ortho/para directed nitrated aromatic amines, if you are a little more clever and if you have a tertiary amine (http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch22/ch22-3-6.html).  Use nitrous (not nitric) acid, which is weaker and will react only with the unprotonated amine.  That gives you a nitroso $\ce {-NO}$ group which you can then oxidize to the nitro group.


text:
Number of stereoisomers of 3-ethylpent-1-ene-1,4-diol

I am trying to figure out how many stereoisomers this compound has. I know that there are two chiral carbon atoms, which allow $2\times2 = 4$ possible isomers. There is also a double bond which, as far as I know, doubles the number of stereoisomers, since there are now four possibilities for the cis-isomer, and four for the trans-isomer. So I thought the total number of stereoisomers was eight, but my teacher told me the answer was six. I have thought about it for a while, but I can't figure out why it should be six. What did I do wrong?

Ask your instructor for the six stereoisomers. The answer is 8 to me.

question is misleading. Only 4 are possible. We should also take into account of tautomerization.

The answer lies in the double bond.  The alcohol on the double bond undergoes keto-enol tautomerism. The greatly more stable aldehyde effectively makes the vinylic alcohol non-existent. This removes a factor for stereoisomers.  six is an odd number of stereo isomers to have since 6 is factored to 2 and 3.  This implies a stereoisomer with 3 forms.  I think your teacher may be thinking of the trans, cis and aldehyde state of the alcohol to come up with six, but this would neglect a carbon stereo center. in this case 12 would be the correct answer. However you should ignore the tautomers and just assume aldehyde which means the answer should be 4.

I agree that the aldehyde would be favored over the enol, but the question is asking for the number of stereoisomers for this particular compound.


text:
Simmons-Smith Cyclopropanation and Sterics

I know this is a Simmons-Smith reaction. Wikipedia tells me this:

The Simmons–Smith reaction is generally subject to steric effects, and thus cyclopropanation usually takes place on the less hindered face.

Therefore, will the cyclopropane be added to the opposite side of the methyl group; i.e. will the cyclopropane group be drawn with dashes rather than wedges?
I understand that if the methyl group is not present then we would get equal amounts of two products, which are enantiomers (as shown by the second image)?

Yes, the cyclopropane ring will be on the opposite side of the ring of the methyl group. If the methyl group of (R)-3-methylcyclohex-1-ene was not there, it would be just be cyclohexene, and there would only be one product (no stereoisomers). In the reaction with 1-methylcyclohex-1-ene, each side of the ring is equally hindered, and which is why the enantiomers are formed in equal amounts.


text:
Electrolysis and reduction potentials
Why does the hydrogen gets reduced at the cathode and not $\ce{Na}$ when aqueous solution of brine is subjected to electrolysis?
My book says because discharge potential of hydrogen is lower than that of sodium but in the reduction potential table, hydrogen is assigned a value = 0 and sodium a value of -2.7. So isn't the negative value of sodium lesser than that of hydrogen? Or is it being talked just magnitude wise?
Please explain me the whole concept of what factors decide the reduction and oxidation of some elements over others.

The whole concept of what factors decide how a redox happens is a bit too broad.

A positive value for the reduction potential means that it has a higher propensity for reduction, whereas a lower value for the reduction potential means that it is less likely to be reduced (i.e. it will resist reduction).
Since $\ce{Na+}$ has a lower reduction potential than $\ce{H+}$, we can expect that $\ce{H+}$ will be preferentially reduced.
For other elements (especially those with more than one oxidation state), it's a little more complicated. You would need to know the condition under which the electrolysis is done. Familiarity with Latimer diagrams will also help you figure out which is the most likely oxidation state.

Okay so why does my book say "since the discharge potential of H+ ions is lower than that of Na+ ions, therefore H+ ions are discharged in preferance to Na+ ions" what does it mean? And could you please explain me the latimer diagrams with an example if possible?

If "discharge potential" refers to the discharge of positive ions, then the more negative values on the Reduction Potential Tables means higher discharge potential, and the more negative values mean lower discharge potential.
Also, I think for simple systems you won't need the Latimer diagram and all you have to do is to compare the relative potentials according to the principle I've described in the answer.

I think there is some mistake in the comment you made


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Why are Alkali Metal Ions Soluble?
I think the question needs no elaboration. Why are salts containing Group 1A cations extremely soluble in water?
I speculate that the reason involves the low charge density of those ions in a given period. Thus, they make better ion-dipole bonds with water molecules than other cations. But the sizes of the Group 1A metals vary, so an argument based on charge density fails as a generalization.

Ions must be solvated. I think you mean why are salts of these ions generally soluble (in water).

@Dissenter  Sure, any salts that contain those ions.

related http://chemistry.stackexchange.com/questions/7869/why-are-caoh%e2%82%82-and-aloh%e2%82%83-less-soluble-than-naoh-in-water

@YunfeiMa There are plenty of poorly soluble alkali metal salts, even inorganic. E.g. $\ce{Li2CO3}$, $\ce{NaSb(OH)6}$, $\ce{NaMg(UO2)3(CH3COO)9}$, $\ce{NaAlSi3O8}$, $\ce{Na3AlF6}$, $\ce{KClO4}$, $\ce{K2PtCl6}$, $\ce{K2SiF6}$, $\ce{KAlSi3O8}$, $\ce{KC4H5O6}$ (potassium bitartrate), $\ce{RbClO4}$, $\ce{Rb2PtCl6}$, $\ce{Rb2SnCl6}$, $\ce{CsClO4}$, $\ce{Cs2PtCl6}$, $\ce{Cs2SnCl6}$.


text:
Is phosphorus a stereogenic centre in this phosphate ester?

My textbook says that in the above diagram that there is only 2 stereoisomers. I am guessing this is due to the stereogenic carbon atom and not due to the phosphorus atom which apparently isn't a stereoisomer.

However, it states that the above molecule has 4 stereoisomers. This must indicate that in this case the phosphorus atom, it is a stereogenic and gives rises to 2 stereoisomers.
Why is the phosphorus atom in A not a stereocenter while the phosphorus atom in B is a stereocenter?

In compound A, the negative and double bonded oxygens bound to the phosphorus are equivalent:
$\hspace{5.1cm}$
In compound B, they are not equivalent:
$\hspace{7.5cm}$

An even better answer would point out that there isn’t really a $\ce{P=O}$ double bond in that molecule and that $\ce{\overset{-}{O}-\overset{+}{P}R2-\overset{-}{O}}$ would be the better depiction ;)


text:
Finding the amount of nitrous oxide produced in the reaction between lead sulfide and nitric acid

Lead(II) sulfide dissolves in excess nitric acid according to the equation below. Calculate the volume of $\ce{NO{(g)}}$ at $\pu{27 ^\circ C}$ and $\pu{1.10 atm}$ produced from $\pu{4.7 g}$ of $\ce{PbS(s)}$.
$$\ce{3PbS{(s)} + 2NO3–{(aq)} + 8H+{(aq)} -> 3Pb^2+{(aq)} + 3S{(s)} + 2NO{(g)} + 4H2O{(l)}}$$
a) $\pu{0.29 L}$
b) $\pu{0.44 L}$
c) $\pu{0.66 L}$
d) $\pu{30 L}$
e) $\pu{45 L}$

I used the ideal gas formula and converted the mass of $\ce{PbS}$ into number of moles of $\ce{PbS}$ however got the answer b) $\pu{0.44 L}$ which is incorrect. The correct answer is a) however I am unsure about. Why this is?

You skipped a step. You got the number of moles of $\ce{PbS}$, that's good. Now you need to find the number of moles of $\ce{NO}$ which result from the reaction, and only then apply the ideal gas law.

Nitrous oxide is not nitric oxide, i.e., NO!

The first step is to calculate the number of moles of $\ce{PbS}$:
$$n(\ce{PbS}) = \frac{\pu{4.7 g}}{\pu{239 g mol-1}} = \pu{0.0197 mol}$$
The next step is to calculate how many moles of $\ce{NO}$ will be produced based on the stoichiometry of the reaction. As stated in a comment, it looks like this is the step that was left out of your calculations. Because only $\pu{2 mol}$ of $\ce{NO}$ are produced for every $\pu{3 mol}$ of $\ce{PbS}$, only $2/3$ of $\pu{0.0197 mol}$, or $\pu{0.0131 mol}$ $\ce{NO}$ will be formed.
At this point we know all of the variables required to invoke the ideal gas law:
$$
\begin{align}
V(\ce{NO}) &= \frac{nRT}{p}\\
 &= \frac{\pu{0.0131 mol} \times \pu{0.0821 L atm mol-1 K-1} \times \pu{300K}}{\pu{1.10 atm}}\\
 &= \pu{0.29 L}
\end{align}
$$
And this gives us the correct answer of a. Again, as your answer was off by a factor of $2/3$, it is pretty clear that the second step above regarding the stoichiometry of the reaction is the only step that you left out of your calculations.


text:
Failed esterification of 1-butanol and ethanoic acid
I did an esterification experiment of 1-butanol and ethanoic acid.
30 mL of 1-butanol was added followed by 30 mL of ethanoic acid, and then 1.5 mL of concentrated sulphuric acid and some boiling chips.
The mixture was colorless at first. However, it turned brown during the reflux of esterification immediately after heating the solution, and the solution had a burnt fruity smell.
I suspect this is because I didn't swirl the mixture before heating. Also, before heating, the bottom of the round bottom-flask felt warm, and I suspect this could be the reaction between the concentrated sulphuric acid and ethanoic acid.
One thing that I can't seem to understand is, how did the color of the solution turn brown? Could it be that the product of the reaction between sulphuric acid and ethanoic acid reacted with the 1-butanol to produce a colored solution?
P.s: I couldn't obtain any pure ester after the distillation of the ester as the temperature didn't rise above 100 °C.

The reaction should have remained clear the entire time, so it sounds like the sulfuric acid reacted with an organic contaminant in your reaction vessel. Such a reaction can be seen when dripping it on paper.

Would heat be required for the sulfuric acid to react with the possible organic contaminant? Because the color change only occurred when heat was supplied.

I do not think that the brown color necessarily must come from the presence of organic contaminants. There might be decomposition/condensation byproducts that are brown. However, I have not done this experiment before, so I'd be happy to let me put right. To get good yields of the ester you probably should do an azeotropic distillation to continuously remove the water formed during the reaction. https://en.wikipedia.org/wiki/Dean-Stark_apparatus

@Jerome Concerning the last remark in the question, note that butyl acetate and the remaining butan-1-ol form an azeotrope, which makes the purification by distillation difficult. Therefore, you should use an excess of acetic acid in order to convert virtually all of the butan-1-ol.

I don't think the brown color necessarily means that the reaction failed. It could be due to some contaminant in your reagents but in a small quantity that wouldn't cause the reaction to fail. The fruity smell could very well be due to your product see: n-butyl acetate has a sweet banana smell. I would take the distillate you collected (or skip the distillation altogether) and try an aqueous work-up: Add very carefully aqueous $\ce{Na2CO3}$ until basic $\ce{pH}$ and then extract this mixture with ether. This way any ester product will be in the the organic layer and sulfuric/acetic acid will stay in the aqueous together with most of the alcohol.
To finish the work-up dry the ether layers over $\ce{Na2SO4}$, filter and remove the ether (unless you have access to a Buchi evaporator, the safest way is to leave it in an open flask/beaker overnight). You can further purify the final product by distillation.
You might improve the yield if you cool the reaction mixture in an ice bath before adding sulfuric acid


text:
Understanding chemical kinetics

For a chemical reaction
$$\ce{2 A + 2 B -> C + D}$$
the order of reaction is one with respect to $\ce{A}$ and one with respect to $\ce{B}$. The initial rate of the reaction is $\pu{4e-2 mol L-1 s-1}.$ When $50\%$ of the reactants are converted into products, the rate of the reaction would become
(A) $\pu{2e-2 mol L-1 s-1}$
(B) $\pu{1e-2 mol L-1 s-1}$
(C) $\pu{4e-2 mol L-1 s-1}$
(D) $\pu{2e-1 mol L-1 s-1}$

This is the first time I am attempting kinetics problems. Is the question asking the concentration when the system has reached its half-life?
Could I use the half-life formula here?
My attempt at a solution:
I used the rate law which I equated to the definition of rate.
I have serious conceptual issues which I hope to clear up. So is this this right?
$$\frac{1}{c(t)} - \frac{1}{c_0}  = kt$$
Would this formula solve the above problem?

Is this the right formula to use to solve the above problem

Yes, this is it, if the initial concentrations of the reagents are the same (which is not mentioned explicitly in the problem statement, but probably is implied, because otherwise the problem would be unsolvable.)
Indeed, as A and B are spent in equal amounts, their concentrations are bound to be the same not only initially, but at all times: $[\ce{A}]=[\ce{B}]=c(t)$. Therefore $\dot c=-k\cdot[\ce{A}]\cdot[\ce{B}]=-k\cdot c^2$, which you seem to already know how to solve.
Update. Come to think of it, you don't actually need to solve for $c(t)$. You know that the rate is proportional to the first power of $[\ce{A}]$, and also to the first power of $[\ce{B}]$, and that both $[\ce{A}]$ and $[\ce{B}]$ have decreased by 50%; what would happen to the rate?
The concept of half life does not apply here, since the decay is not exponential.
Also, welcome to Chem.SE, and pay attention to the formatting.

Is the answer    $ 1 * 10^{-2}$

Looks like so. The concentrations decrease by 50%, and the rate depends on $c^2$, so it must have become 4 times lower.


text:
Why is cyclopentadiene more stable than its aromatic counterpart cyclopentadienyl anion?

What will be the order of stability?

The answer key says that the stability order is: cyclopentadiene > cyclopentadienyl anion > cyclopentadiene  cation
According the me it should be: cyclopentadienyl anion > cyclopentadiene > cyclopentadienyl cation
because the order of stability is aromatic > non-aromatic > anti-aromatic.

This is a nice question because it forces you to step back and think rather than fall into the obvious trap.
You are entirely correct that aromatic compounds are more stable than non aromatic compounds which are more stable than anti aromatic compounds assuming that all other things are equal. In this case, all other things are not equal because one of the species is neutral while the other two are charged. Having a positive or negative charge on carbon is not very favourable, even if the negative charge does make the species aromatic. The $\text{p}K_\text{a}$ of cyclopentadiene in DMSO is 18[1] which means the equilibrium constant for dissociation of cyclopentadiene into the cyclopentadienyl anion and a hydrogen ion (which will actually be attached to a DMSO molecule) is $10^{-18}$! In other words, cyclopentadiene is much more stable than the cyclopentadienyl anion.

I am not sure about this. This link say otherwise.


text:
How to work out equations with period 3 oxides?
I know the general equation of reacting an acid with a base to be the salt and water, but I am not able to apply this to the following equations. So in what way can I derive the end product if I have never seen the reactants before?
For example, I can see the water is liberated out in the reaction between silicon dioxide and sodium hydroxide, producing this equation:
$$\ce{SiO2 + 2NaOH -> Na2SiO3 + H2O}.$$
However if the reactants were aluminium oxide and sodium hydroxide, how could I work out the equation:
$$\ce{Al2O3 + 2NaOH + 3H2O-> 2NaAl(OH)4}$$
I would have thought the product formed would be: $$\ce{Al2O3 + 3NaOH ->2Al(OH)3 + 2Na+}$$

It is actually possible to precipitate $\ce{Al(OH)3}$ from $\ce{Na[Al(OH)]4}$ solutions. This is a key step of the Bayer process.

The first thing to say is that you final equation doesn't balance. In fact it is impossible to balance your equation using hydroxide because you can never conserve charge with only neutral products.
You can write a reaction that forms $\ce{Al(OH)3}$: $$\ce{Al2O3 + 3H2O -> 2Al(OH)3}$$
However, beyond about pH 6 the $\ce{Al(OH)3}$ will start to dissolve to form $\ce{[Al(OH)4]-}$. Once you get to around pH 7.5 all of the solid has essentially dissolved and you are left with solution of $\ce{Na+}$ and $\ce{[Al(OH)4]-}$ (see footnote). So unless you are very sparing with your addition on $\ce{NaOH}$ you will get the tetrahydroxo product in solution.
$$\ce{Al(OH)3 + OH- -> [Al(OH)4]-}$$
Footnote:
The aluminium species present in aqueous solution are strongly dependent on the pH, particularly around the neutral point. There is a nice graph here showing the different species. The notation is $\ce{Al}$ corresponds to $\ce{[Al(H2O)6]^{3+}}$ and then $\ce{AlH_{-n}}$ denotes the loss of n protons from the hexaaqua complex. This was originally found by @Jan for this question.

At first it'll form an Al(OH)3, but we should remember that Al(OH)3 is a Lewis acid since it'll need a pair of electron to complete the 'octet rule', and when it's formed in the solution (aqueous), the water (H2O)'s Oxygen will solvate the Al(OH)3 and it'll form an Al(OH)4-.
Adding excess base will also resulting in the making of Al(OH)4-


text:
Is there powdered blood in TANG-pineapple powdered juice mix?
My TANG pineapple juice drink mix is a yellow powder. When I add a little water to it, it dissolves and red dots are briefly observed before mixing with the yellow dots is complete, after which time the red dots disappear.
What happened? I'm scared: are the red dots from powdered blood?

What?? Why would you suspect blood?

I'm voting to close this question as off-topic because it's not really about chemistry.

I think this question might be slightly off-topic...  but in case you are really concerned, I think the answer is no.
What you might be seeing is an effect of the Tang powder getting wet before it dissolves fully to form your Tang drink.  See this post from the physics Stack Exchange or this post from 'Ask the Van' at the University of Illinois
Things appear to be darker when they are wet, for reasons explained in the posts I linked to (briefly, dry objects reflect more light than wet ones, as the water allows more light to pass through a wet object).  So your yellow powder might be appearing as a darker shade of orange, the 'red spots' you see.
Also, if Tang were really putting blood in its drinks, various food standards regulatory bodies would probably notice and stop them from doing it, and there would have to be court cases etc.  So keep drinking the Kool aid (or whichever brand you prefer)!


text:
How to determine the order of a reaction from given concentrations?

In the presence of an acid, the initial concentration of cane sugar was reduced from 0.20 to 0.10 molar in 5 hours and from 0.2 to 0.05 molar in 10 hours.
What is the order of the reaction?

My attempt
So i found the first 2 rate
R1 = 0.015
R2 = 0.010
And i tried comparing this with the final concentrations of the 2 processes
I am getting the answer as 3
How do i solve the problem

You can solve this by determining the half life of the reaction, so from the first one we have 5 hours to half the concentration, and from the second data we get the same half life.
So in conclusion, the rate order is 1 because the half life is constant

I have a question   What if the concentration is not half after a period of time what do i do ?


text:
Why would a reaction take place in this instance?

This makes little sense to me, both of the electron configurations are d6, and would presumably be low spin - therefore both reactants should be inert, and yet my notes suggest that an electron transfer would occur to form the products Co(II) and Ru(III). Surely this is a less stable arrangement?

You have false assumptions: just because something is d6 LS it doesn't mean inert.

Perhaps not, but surely the products, both with an unpaired electron are still less stable than the reactants. Perhaps the driving force is entropy?

Cobalt(II) is most certainly high-spin in this environment. Cobalt(III) is likely to be. However, I am at a loss for the answer …