text:
Is triclosan harmful?
I keep hearing about this product known as triclosan, and that it is used in sanitizing soap. And I also heard that it is harmful, and causes some sort of muscle problems.
However some people aren't affected by it. Why is this, and should I
 be worried about washing my hands?

Triclosan (2,4,4′-trichloro-2′-hydroxydiphenyl ether) is a nonionic, broad spectrum, antimicrobial agent that, because of its favorable safety profile, has been incorporated into a variety of many personal care products, including deodorant soaps, underarm deodorants, shower gels, and health care personnel handwashes. Triclosan exhibits a moderate degree of substantivity to the skin, and, in many products, it imparts a remnant antimicrobial effect.Here is the link.
U.S. Food and Drug Administration (FDA) states that "Triclosan is not currently known to be hazardous to humans."
Despite having significant uses, it also posses little amount of adverse effects on prolong use.It impairs muscle function and skeletal muscle contractile property.This research has only been conducted in mice and results were not good; when mice were exposed to triclosan, heart muscle function was reduced by 25 percent and grip strength was reduced by 18 %.
So result has concluded that it might cause skeletal and heart muscle failure in humans as well.Here is the link.
Exposures to triclosan can also cause dermatitis, or skin irritation, and an increase in allergic reactions, especially in children.Due to its link with dioxin, it has been suspected to cause other health problems like weakening of the immune system, decreased fertility, miscarriage, birth defects, and cancer.Here is the link
But to eradicate these effects,FDA has been engaged in an ongoing scientific and regulatory review of this ingredient. FDA does not have sufficient safety evidence to recommend changing consumer use of products that contain triclosan at this time.
So we can conclude that, it cannot be considered entirely harmful for human body.

-1, this is an extremely selective reading of the evidence. Triclosan has low enough toxicity to be used as a toothpaste additive with good results (http://onlinelibrary.wiley.com/doi/10.1002/14651858.CD010514.pub2/abstract;jsessionid=27AD5AFB113E1833634F174D7C55FFFA.f01t01). It is not a dioxin like molecule, and the trace dioxins which might be produced by its degredation are not considered harmful. The FDA are correct in thinking it not to be currently hazardous, because beyond the rare occurence of allergic reactions there is no evidence that its use is problematic.

Hey i respect your comment. All the things that i posted as answer is based upon the websites which i have put on as proof. The article that you mentioned has said that "The studies investigated local
adverse effects such as tooth staining, altered taste and included
clinical examination of oral and perioral soft and hard tissues.".

Please check the first link. It has mentioned that "Tricolsan impairs muscle function and skeletal muscle contractility, researchers report in a new study done at the University of California Davis. Although the study was done in mice, researchers said the effects of the chemical on cardiac function were "really dramatic.""

Yes, but that's just one study. If tricolsan had such a large impact on human health, then it would be clear given over forty years of use. There is plenty of evidence that it is non-toxic; see http://www.ajicjournal.org/article/S0196-6553(96)90017-6/abstract, http://informahealthcare.com/doi/abs/10.3109/10408444.2014.910754 for some sources. I'm not saying that its use is entirely risk free or that there are not more appropriate antibacterial agents but there is no need for hysteria about it.

Yeah. i was just pointing out that "there are some adverse effects on using products containing triclosan" being little skeptical . and these might cause on long exposure. Just recommending "not to use them if you think it is best for you " to asker. Well every thing has its own positive and negative sides .

Do you want me to edit the answer then??

I think that would be nice if you balanced your answer a little, made it NPOV.

Here you go. I have edited my answer favoring both of us. Thank u for suggesting .

However as we all know, if there are cases of something being harmful, then we should probably not use it. Just to be safe. Then again people who favor this product might also try to find evidence to prove this wrong. So at the end of the day we, not the FDA, make the decision to use this certain substance. Just trying to stay neutral.


text:
le Chatelier's Principle
I am really confused about how reducing partial pressure of a gas by a certain percentage would impact the behavior of an equilibrium system.
e.g $\ce{POCl3_{(g)} <=> POCl_{(g)} + Cl2_{(g)}}$ is at equilibrium.  How would this system behave if the partial pressure of $\ce{Cl2}$ is reduced by 75%?
I understand the partial pressure is related to moles but having a hard time applying it here.  My guess is that the reactant would be consumed to produce more of the $\ce{Cl2}$ since its pressure and thus moles are decreasing.  However, my instructor said the system would react by consuming $\ce{POCl}$, the reason for which was not explained too well.
If I could get an explanation, it would be much appreciated.

The partial pressure can be converted to moles and then you could use Kc. Or you could simply use Kp. However, I would think that you are correct: given no other changes, if products decrease, the system would act to create more products, until equilibrium

There are several good questions that already address this.  A change in partial pressure of a component means a change in concentration.  This perturbs the system away from equilibrium which then adjusts to reach a new equilibrium.  I agree with @Andy on the result: some POCl3 will dissociate to produce more products.  http://chemistry.stackexchange.com/questions/4130/, http://chemistry.stackexchange.com/questions/4262/, http://chemistry.stackexchange.com/questions/9118/

I think what your instructor meant was that the reaction would involve the consumption of $\ce{POCl3}$ and that in turn leads to the formation of $\ce{POCl}$ & $\ce{Cl2}$.  Thus resetting equilibrium.

Partial pressure can be imagined to be the pressure of a gas in the absence of other gases.
Le Chatelier's principle states:

When a system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

Hence when the pressure of $\ce{Cl2}$ is reduced by 75% in the $\ce{POCl3_{(g)} <=> POCl_{(g)} + Cl2_{(g)}}$ system, the system naturally acts to counteract the change: by increasing the pressure of $\ce{Cl2}$.
Using the ideal gas law, $pv = nRT$ so pressure is directly proportionate to the number of moles of $\ce{Cl2}$ present. The position of equilibrium therefore shifts to the right, favouring the forward reaction that produces more $\ce{Cl2}$ (though not as much as to increase partial pressure of $\ce{Cl2}$ to its former value). This would mean it also consumes more $\ce{POCl3}$.
As for more $\ce{POCl}$ being consumed, if all other factors remain constant I'm not sure why this would happen.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
How to calculate the temperature at which a reaction becomes spontaneous?

At $25~^\circ\mathrm{C}$, $298~\mathrm{K}$, the reduction of copper(I) oxide, $ \Delta H = 58.1~\mathrm{kJ}$, $\Delta S = 165~\mathrm{J/K}$, is nonspontaneous, $ \Delta G = 8.9~\mathrm{kJ}$.
  Calculate the temperature at which the reaction becomes spontaneous.

So for this question I'm thinking I use the equation
$$ \Delta G = \Delta H - T\Delta S$$
Then plug in the values and solve for $T$? $$8.9~\mathrm{kJ} = 58.1~\mathrm{kJ} -T(0.165~\mathrm{kJ}) $$

the formula you want is : T = delta H/delta S the one that said to set G to zero was correct but they didn't show how that formula looks when you do that, so here it is.

You have the right equation.  But the problem already gives you the temperature at which $\Delta G$ is 8.9 kJ.
Assume that $\Delta H$ and $\Delta S$ are constant and do not vary with temperature.  $\Delta G$ still will though because of the $T$ in the $\Delta G = \Delta H - T\Delta S$ equation.
So plug in the values for $\Delta H$ and $\Delta S$, and also the $\Delta G$ value that you think represents the transition from spontaneous to non-spontaneous.  You should be able to take it from there.

Actually he had the wrong equation.  The correct equation to use was $$\frac{d\left(\frac{\Delta G}{T}\right)}{dT}=-\frac{\Delta H}{T^2}$$
Sign error corrected on 10/16/15.

What you really want to do here is use $ΔG = ΔH - TΔS$, use the given data for enthalpy and entropy. Set $ΔG$ to $0$, and solve for temperature, as that is the criteria for equilibrium and the point for $ΔG$ below which the reaction will be spontaneous

Any value above 352K is the answer. Sum of temperature times delta S must be over 58.1(total of delta H) so that gibbs free energy is 0 or less . Thus divide 58.2 by 0.165 to find energy needed for delta G =0 any temp above this will result in a negative (spontaneous reaction).

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

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Predict Entropy based on the Balanced Equation
You cannot use values but must determine from looking
The equation I have trouble with is as follows:

$$\ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2}$$

According to my teacher the entropy is positive. I also got the entropy to be positive, but now that I think of it, the moles is decreasing and one is becoming aqueous. So does turning into a gas trump everything?
What if NaOH was solid? Would it mean still that the entropy would be positive?
Also finally according to the textbook the entropy values is as follows:

$$\ce{s<l<<g}$$

But where would the (aq) be?

Aqueous salts will always have less order than solid salts. There is much less order when ions are floating around in solution than when they are in their rigid lattice structures. This is what I think you are failing to consider. The reaction is not so much:
$$\ce{2Na_{(s)} + 2H2O_{(l)} -> 2NaOH_{(aq)} + H2_{(g)}}$$
But really:
$$\ce{2Na_{(s)} + 2H2O_{(l)} -> 2Na^{+}_{(aq)} +2OH^{-}_{(aq)} + H2_{(g)}}$$
What results from the reaction of 2 atoms of a very structured metal and 2 molecules of liquid water, is 4 aqueous ions that are free to move within the solution, and one molecule of gas. Not only are there now more species present, but they are also in a relatively more free states than before.

Yeah thanks, I was thinking about that dissociation.


text:
Number of gauche interactions in an alkane

The textbook states that (a) has 2 gauche interactions, (b) has 3 gauche interactions and (c) has 3 interactions.
But how do we determine this? I thought a gauche interaction would occur when two methyl groups are side by side, however I'm unsure where these answers are coming from.

When we sight down a carbon-carbon single bond and draw a Newman projection of a particular conformation, any two groups on the adjacent (vicinal) carbons that have a dihedral angle of 60° between them are said to have a gauche relationship.
In the following figure, the $\ce{X}$ and $\ce{Y}$ have a gauche relationship.

$\ce{X}$ and $\ce{Y}$ can be the same (they can both be methyl groups, other alkyl groups, other atoms such as halogen, etc.) or they can be different from each other. In most organic classes, the gauche interaction between methyl groups or alkyl groups is frequently discussed.
The answers you included (2,3,3) are correct.  Looking at the middle molecule (b) in your figure, there are 3 gauche interactions.  They are between

the front methyl and the back ethyl
the same front methyl and the back methyl
the other front methyl and the back methyl


text:
Why does the cathode have to be made out Cu in a Zn-Cu voltaic cell?
The classic voltaic cell has a $\ce{Zn}$ anode and $\ce{Cu}$ cathode. The reduction half reaction occurring at the cathode combines aqueous copper ions and electrons supplied by the anode to form solid $\ce{Cu}$. So what is the purpose of the $\ce{Cu}$ cathode? $\ce{Cu}$ is produced anyways from the electrons supplied by the anode. Why can't we have instead of the $\ce{Cu}$ cathode, simply the wire connecting to the anode which will supply electrons to fuel the reduction half reaction?

You could certain have a zinc-copper cell in which copper was not the cathode material.  Graphite or silver or whatever could be the electrode material instead.   But once you close the circuit:

Current flows
Cu is deposited as a metal solid on the electrode surface, whatever it is made out of.
Copper metal is, as it always is, conductive.
As a result of steps 2 and 3, there is now a copper metal cathode instead of a cathode made of another material.

So even if you start without a copper cathode you wind up with one.

You could have a cathode made from something other than $\ce{Cu}$, however, this reaction will only produce $\ce{Cu{(s)}}$ as a result if the metal used for the cathode is a weaker reducing agent than $\ce{Zn}$, otherwise the $\ce{Zn}$ electrode would become plated in $\ce{Zn{(s)}}$.
Using a $\ce{Mg}$ electrode in the $\ce{Cu^{2+}}$ solution, for example, would result in the $\ce{Zn}$ electrode to become plated with $\ce{Zn{(s)}}$.

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


text:
Paint changes color when applied to Stucco. Chemical rxn assumed, but what?
PROBLEM:
When I apply paint to my home’s exterior stucco (currently white), the paint changes color. The grays change to purple. The whites change to yellow. Greenish-beige becomes orange. I have purchased many samples of paint, and a color change is apparent for all but the darkest colors. The change in color starts happening immediately when I apply the paint, and it becomes more intense over subsequent days. After a few days, there is a radical difference between the paint’s intended color and the new color.
Apparently a chemical reaction is occurring between the paint and a substance in/on the wall, but I don't know what it is, or how to remedy it.
SURFACE FACTORS:

The house is stucco, so perhaps the surface is very alkaline. The house was built in 1971, and several coats of paint have been applied by the previous owners since then, so I would think the alkalinity would no longer be a factor, but perhaps it is. (Would it even be possible for improperly cured stucco to remain highly alkaline for 44 years?)
The previous owner told me that, around fifteen years ago, he painted the house with "pool paint." (Unfortunately, I don’t know what type of “pool paint” was applied, or why it was applied.) My guess is that this was in response to the color-changing problem rather than the cause of the problem, but I don’t know for certain.  (It is interesting that the pool paint has remained white all this time, while I’m unable to get a true white.)

I have painted the wood next to the stucco, and the paint on the wood appears exactly as it should and it remains constant(aside from becoming just slightly lighter as it dries). In other words, the paint only changes color on the stucco. Since the wood was never painted with pool paint, and since the wood is obviously not stucco, I still don’t know which factor (pool paint, stucco, or something else) is causing the color change. However, this does show me the paint color is truly changing on the stucco.
I have some chemistry background, but I'm really stumped by this problem.
Any help in figuring out what’s going on would be very much appreciated.

A agree that a house cannot stay alkaline for decades given that we have rain to wash off most of the base and CO2 in the air to neutralize the rest. We don't know composition of  "pool paint" or current paint (those are trade secrets). Besides this field is quite small, so I doubt  we can answer it. Your best route is either to ask manufacturers of current paint or just double/triple paint the house.

Did you apply a sealer and primer prior to painting your topcoat?

You haven't told us the color of the stucco you are painting, nor the type of paint.  Chemical reaction is very unlikely, paint that changes color would make for bad paint.  Pigments are chosen to be stable.  Old stucco would be only a little alkaline, but it is porous.  This sounds more like color leaching out of the stucco and into the new paint.  You need paint with better coverage (covers contrasting colors) or better sealing ability.  Seriously, don't ask us, talk to your paint supplier at a paint store.  Maybe get them to refer you to their in-house experts.  You probably will have to spend more for the paint than you are planning on.

Shouldn't be happening!
Neither of the stated special conditions have any apparent reason to cause this problem.
Stucco applied in 1971 will very probably be a mortar of Portland cement and sand. As you note, it is quite alkaline but this is perfectly normal and doesn't normally affect the colour of paint. The stucco is often applied over a layer of wire mesh to help it adhere to the wall -- the alkalinity actually inhibits corrosion in this mesh. If the mesh does corrode, this could certainly affect colour, but it is a gross defect showing obvious patterning. When it is well-cured, damp stucco can support a range of fungal and algal growth but again, these are fairly obvious in appearance.
Pool paint can be any of a wide range of paints, at least including acrylics, epoxies and chlorinated rubber. What they have in common is that they are designed to adhere well to Portland cement, and to provide a waterproof layer. Ironically this is likely to raise moisture levels inside your stucco as it will prevent damp from evaporating easily; but again there is no obvious reason this might affect colour.
Old paint exposed to the sun "weathers", and it can be hard work to strip this off a rough surface. However it tends to affect adhesion, rather than colour.
Approach systematically
I would recommend that you do some controlled tests to isolate the problem. Get some small sample pots of your preferred paint, and a couple of other types. Ideally find a patch where the existing paint is peeling. Apply small colour strips to: exposed stucco; peeled-off paint fragments, as collected; peeled-off paint fragments that have been carefully washed; to nearby Portland cement that is not part of the stucco; and to a "control" surface that is expected to show the colour well. Carefully record which strip is which. The test strips can be very small, so long as you can clearly identify the colour change.
This inexpensive test should help you to identify if the problem is associated to:

the stucco
subsequent paint layers
a contaminant on the exterior of the painted stucco
your paint batch or brand
the local environment generally

It may or may not help to identify the root cause, but could at least help to find a work-around.


text:
Do hard drive platters contain toxic chemicals?
So, I'm considering an Artsy Fartsy project with hard drive platters. Being mildly aware of the world around me, I've come to the conclusion that I probably shouldn't grind up modern (or any) electronics and drink it for tea. To that end, is there anything in hard drive platters that is significantly toxic, or are they fine lying around the house? I don't intend in grinding them or anything, I'll be using them whole, so dust inhalation presumably isn't a big issue.

I think that you're safe. The core is aluminium, on which several layers of metallic material are sputtered. Are you planning to spin the platters at high speed (microcontroller + ESC)? The worst thing that can happen is a platter that breaks loose and starts flying around ;)

Actually, I've got a handful from a dead server. I'm making a set of coasters. Thanks for the answers. :) @KlausWarzecha , can I ask why your comment isn't an answer? I'd go ahead and accept it.

@Sidney My first comment simply was a bit too short for an answer. It might be more fitting now.

I never opened a newer 2.5" hard disk, but I've read that these may contain glass platters.
However, I've seen and touched stacks of platters from old 3.5" hard disks - a friend of mine makes clocks from hard disk cases.
These platters have an aluminium core, on which another metallic layer, that serves as a kind of primer, and the magnetic material are applied by sputtering. The surface is covered by a protective polymer layer.
Platters have been drilled or cut in a lot of electronics projects, e.g. to make hdd clocks based on the persistence of vision effect. Apart from the usual safety rules for metal works (protect your eyes, don't touch the hot drill, etc.) one only has to make sure that the rotating platter is properly mounted and doesn't start flying around!
When the platters aren't mechanically altered and only painted, there is even less danger. The platters are not toxic and do not even have sharp corners - it is safe to keep them around in the living room.
On a side note, the magnets from these hard disks are more dangerous. These magnets are pretty strong and when playing with them, you can easily jam your fingers between them ;)

I can verify about getting my finger caught between some of the magnets...ouch!

@santiago Yeah... I did that the other day! XD

I have opened my old HDD with many bad blocks. After opening it, I learnt about its workings, and then I took a pair of tongs, and then I tried to bend the platters... It just broke off like a glass (with very sharp edges, but so smooth thet you can't cut your fingers with it). The glass was so thin but hard that I thought it was just a strong metal. You can't break it by simply dropping it from 6 feet height. So I would say it's not always aluminium plates...

Discs from 3.5" drives are perfectly safe. They are made of a majority of aluminum, with hints of platinum, palladium, and other precious metals (not a lot though — the Pt coating is only a few atoms thick).
Stay away from 2.5" drives though. Their platters are made of some sort of glass, and if you put too much pressure on them, they will shatter with razor-sharp chunks of glass flying everywhere.

If you're thinking "This post is a few years old, and these discs from my 2.5" drive really look like metal" you are wrong - they are glass. Do not test them, they will shatter and cut you.

I think you're OK  if they're just lying around.
Some drives, maybe only older ones, do contain nasty stuff such as beryllium, so I'd be wary of opening them up or crushing them in any way.
https://www.osha.gov/SLTC/beryllium/healtheffects.html
https://www.seagate.com/au/en/global-citizenship/product-stewardship/


text:
If losing electrons means getting oxidized, then does it mean that ALL metals are reducing agents, since they themselves get oxidized?
I was just going through my last year's school textbooks and while I was reading this chapter on Redox, this above question crossed my mind.
Shouldn't it be true that since ALL metals lose electrons to get stable, they should also get oxidized. And since they getting oxidized themselves, they should act as reducing agents... Always?
If the answer is no, please provide an example of the exception.

Your reasoning is sound. All (pure) metals can act as reducing agents, but they do not have to all the time. There are a few examples of metals in negative oxidation states, where they would act as oxidising agent.
Probably the most prominent examples are Zintl phases, where alkaline or earth alkaline metals reduce post transition metals (or metalloids like antimony). One of the first reactions observed was the sodium lead reactions in liquid ammonia in 1907. Back in the day they proposed the following reaction equation:
$$\ce{Na + excess~Pb ~->[\ce{NH3}][\mathrm{-50^\circ C}]~ Na+ + $“$~Pb2^{-}~$”$}$$
It was later found that the actual composition is $\ce{Na4Pb9}$. With crown ethers and/or cryptants it is possible to isolate discrete anions, too.
There are more examples like this, and a nice overview is given in John E. Ellis, Inorg. Chem., 2006, 45 (8), 3167–3186.

No, not always.
For example, in the classic Galvanic cell:
In the above image, you have zinc metal acting as a reducing agent at the anode, true to your line of thinking. However, it is also the case that the copper ions at the cathode are being reduced.
This can happen quite frequently in electrochemistry as the reduction reaction that occurs is largely governed by the reduction potential of that reaction. In the Galvanic cell example, the copper reduction reaction has a higher standard reduction potential (+0.34V) compared to that of the zinc reduction (-0.76V), meaning that the copper ions are more likely to get reduced, and so the copper reduction will proceed over that of the zinc.
Image: http://chemwiki.ucdavis.edu/@api/deki/files/119/ZnCu_detail.gif?revision=1

Elemental Cu is not being reduced. Cu ions are being reduced to elemental Cu. Cu 2+ ions are an oxidising agent as they can be reduced to Cu metal. Elemental Cu is a reducing agent as it can be oxidised to Cu ions. Under normal circumstances, elemental Cu cannot act as an oxidising agent. There are some cases where a very strong reducing agent can cause elemental metals to act as oxidising agents, forming anionic species. http://en.wikipedia.org/wiki/Zintl_phase gives for example 4(Na)+ . (Si4)4- If you don't consider Si a metal, there are similar compounds with Sn and Pb (and In and Tl)

You are right. I was being overly broad by not distinguishing between the two forms of copper. I'll change my answer to show that.


text:
How does radiation facilitate the formation of tetravalent iron?
In reading the article abstract What Oxidation State of Iron Determines the Amethyst Colour?, the author states that in regards to the gemstone amethyst, from clear quartz:

The crystal was transformed into violet amethyst by gamma-irradiation. The change in colour was accompanied by changes in the Mössbauer spectrum that can be interpreted as the conversion of trivalent iron into the tetravalent state: $\ce{Fe^{3+}→Fe^{4+}}$

How does gamma radiation facilitate the conversion of oxidation state of iron, from $\ce{Fe^{3+}}$ to $\ce{Fe^{4+}}$?

The ionization energy of $\ce{Fe^3+}$ to $\ce{Fe^4+}$ is 54.8 eV.
The energy of typical gamma radiation is much higher (in the order of roughly some keV to a few MeV). Depending on the energy, the interaction of average gamma radiation with the affected electron is mainly due to Compton scattering.


text:
Naproxene syntheses: electrophilic aromatic substitution on activated naphthalene
I am studying Syntex's naproxene syntheses and I came across a couple of electrophilic substitutions whose mechanism I can't understand.
The first one is a Friedel-Crafts acylation:
the products are the ones indicated in the picture and they are separated before proceeding. No other byproducts are shown.

Why did the acetyl go in pos. 6? As far as I know, When an electrophilic substitution is made on an activated substituted naphthalene, the electrophile will go in the substituted ring, in the pos. 1 (kinetic control) or in pos. 2 (thermodynamic control), which is in ortho or para to the activating substituent. So I would expect these products, in theory:

I assume the contributing structures that represent the electron-donating effect are the following:

Therefore, although I would expect a 6-acetyl derivative as a byproduct (at least in theory, since (4) might allow it), I don't understand why it is the main product. No clues at all about the 5-acetyl byproduct, since the 5-acetyl intermediate is not stabilized by the electron-donating effect: could it possibly be an error?
The second reaction is a bromination on 2-naphthol:

Likewise, I can think that bromine would first go in pos. 1, and then, since it is only slightly deactivating, another bromine can be added. But again, why in pos. 6? Why not in pos. 3, in ortho to the activating substituent?
And in the end, are these reactions done under kinetic or thermodynamic control?

No other byproducts are shown

I haven't checked the original Syntex patent (if you've ever written a patent you know that sometimes important details are omitted), but it seems odd to me that those two isomers are claimed to be the most significant isomers formed. As you point out, the 2-methoxy-5-acetoxy isomer is not expected to be formed in significant amounts based on electrophilic substitution patterns in substituted naphthalenes (Note: here is a nice concise reference that shows how to apply the same concepts used to predict which isomers will  predominate in electrophilic substitution with benzene to substituted naphthalenes).
In this abstract the Friedel-Crafts acetylation of 2-methoxynaphthalene is studied.  By changing solvents they can shift the major product from 2-methoxy-3-acetoxy to 2-methoxy-6-acetoxynaphthalene.  These two isomers are among those expected and the experiment shows that these two isomers are close enough in energy that small factors can significantly impact the product distribution.  Two other factors to keep in mind

the 6-acetoxy isomer corresponds to "para" substitution
When running a Friedel-Crafts with oxygenated (hydroxyl, methoxy) aromatics, the aluminum chloride will (depending to some degree upon solvent) complex with the oxygen.  In the case at hand this could make the methoxy a bulky substituent disfavoring reaction at the ortho positions.

But again, why in pos. 6? Why not in pos. 3, in ortho to the
  activating substituent?

Careful low temperature mono-bromination of phenol shows that para-substitution is strongly favored (>95%, reference).  If additional bromine is available a second bromine will be substituted at the ortho position. It seems that the same thing is happening here in your 2-naphthol example.
Like you, I might not have predicted these outcomes, but that's why we run experiments.

are these reactions done under kinetic or thermodynamic control?

Sorry, without more information (like product composition as a function of time, temperature, etc) I can't answer that.

Thank you very much for your extremely helpful explanation. My source is not the original patent but course notes, so sometimes I wonder if I puzzle over actual errors, which this may be the case as it seems! My question about the reaction control was made with the assumption that it could be indicated just by the products themselves. I don't have a lot of practical experience yet, so I happen to forget to think beyond the textbooks sometimes :) Thank you again.

The first link in your answer seems to be broken :(


text:
How to calculate the amount of gas in a compressed cylinder
I am trying to calculate a cost per hour use of a gas we buy in a cylinder. The details I have been working with are:
Gas used CP-grade N2 material number: 110628-L can

Gas: N, density = 1.251 g/L @ 0.101 MPa M (molar mas) = 14.007
Gas cylinder: Pressure = 200 bar or 20 MPa Cylinder volume 9.45 m3 or 9450 L
Operating pressure of gas from cylinder = 4.5 bar or 0.45 MPa and flow rate = 60 cL/min

I have been applying the ideal gas law to PV=nRT to figure out how many total litres of N are in the can at 4.5 bar pressure to then get a rough estimate of how many hours of flowing gas I get. However I seem to be getting unreasonable numbers, potentially from me using the equation or reasoning incorrectly:
When I apply the gas law under isothermal V2=V1/(P2/P1) conditions to see how much the internal gas volume expands from the 20 MPa pressure, as a sanity check I wanted to see how many L at 1 atm came out. The result was 1.89 ML which makes little sense as the conversion of 1.89 ML N2 to mass = 2.364 Mg.
Similarly it equates to 420 kL @ 0.42 MPa.
Where am I going wrong?
Maybe the specifications mean that its 9.45m3 or 9450 L of compressed gas to 200 bar?

9450 L is about 10 cubic meters, so unless your cylinder is really big -- say like 1 meter in diameter and 12 meters high, it isn't 9450 L big.  So you are probably right that what you are receiving is 9450 L of gas at STP.  A further note: nitrogen gas is $\textrm{N}_2$, a diatomic molecule, so its molecular weight is 28 g/mol, not 14.

I worked it out in the end. The specifications given of 9450 L are indeed the volume of the uncompressed gas, which compresses to 20 MPa results in about 47 L which fits the dimensions of the L gas cylinder. I simply understood the information backwards. Noted for the 28 g/mol. Thank you.

Have a look at this video. This Should you help you out. https://www.youtube.com/watch?v=Yo9IZ2RJCCU

Turns out my assumptions were wrong about the stated $9\,450\ \mathrm L$ of $\ce{N2}$ gas. The $9\,450\ \mathrm L$ is the total volume of standard pressure gas compressed into the cylinder to $20\ \mathrm{MPa}$.
Using $V_2=V_1/(p_2/p_1)$ having $p_1=0.101\ \mathrm{MPa}$, $p_2=20\ \mathrm{MPa}$, $V_1=9\,450\ \mathrm L$. The total volume of gas within the cylinder is about $47\ \mathrm L$ a reasonably fitting size for the cylinder dimensions.
Hence to answer my own question, at the pressure of usage ($0.45\ \mathrm{MPa}$) the volume of gas (using same equation as above) is $2\,121\ \mathrm L$ which at $60\ \mathrm{cL/min}$ equates to $59\ \mathrm h$ of flow roughly.


text:
Alkylation of alkenes by nucleophilic substitution
A double bond/alkene reacts as a nucleophile. I have never seen a reaction where it takes part in an $S_N1$ or $S_N2$ reaction with a halogenoalkane for example. This would presumably be a very useful reaction yet I have never seen it. I assume this is because it doesn't work. Is this the case? If so why?

A double bond/alkene reacts as a nucleophile. I have never seen a reaction where it takes part in an $S_N1$ or $S_N2$ reaction : If you've never seen a reaction where a double bond participates in a nucleophilic reaction, how can you say that it reacts as a nucleophile?

...with a halogenoalkane (as it says in the question)

The pi electrons in an alkene or alkyne can act as a nucleophile.  This class of reactions is often taught a bit later than the standard $\ce{S_{N}1}$ and $\ce{S_{N}2}$ reactions and comes up when neighboring group participation or anchimeric assistance is covered.
The pi electrons in a double or triple bond are not strongly nucleophilic but when you arrange things in their favor, pi electrons can act as a nucleophile.  The reactions involving double bonds that I am familiar with involve an $\ce{S_{N}1}$ mechanism.  Here are a few examples.

(image source)
In example (1) notice how the compound with the double bond solvolyzes $10^{11}$ times faster than the saturated model compound!  Note too that only the product with acetate group anti to the double bond is formed.
Triple bonds and benzene rings can participate as well.  Often in cases involving double or triple bonds new carbon-carbon bonds are formed between the unsaturated linkage and the carbon with the leaving group [examples (2) and (3) above]. In cases involving aromatic rings aromaticity is restored so no new carbon-carbon bonds are formed between the aromatic ring and the carbon with the leaving group.

Just in terms of a non intramolecular reaction. Like, for example, the reaction of propene with methyl iodide, is this possible? I suspect not but I can see no clear reason why.

Pi electrons are a weak nucleophile.  They need that entropic help that comes with intramolecular reactions.  Still, like you say, I don't see why it can't happen intermolecularly, but I can't think of an example.  Perhaps someone else reading this can come up with such an example.  I'll think about it some more.


text:
Charged nucleophile and SN1/E1/SN2/E2 reactions
How can we say that if a charged nucleophile/base is present we can safely rule out the possibility that the reaction can take place via SN1/E1 mechanism? I have been trying to figure this out since hours. Help would be appreciated. Thank you. Source

Could you please take a little more time to expand your question and add more details? It's pretty cryptic right now (at least for me).

I have added all I could. To see from where my problem arose-check the link.

It says so in rule 3: you cannot form a carbocation for obvious reasons.

Strong nucleophiles tend to undergo SN1/E1 reactions whereas weak nucleophiles tend to undergo SN1/E1 reactions.
An increase in electron density will make a nucleophile stronger. This effect can be clearly seen when conjugate bases of common nucleophiles are stronger than the nuclephiles themselves. For example: $\ce{NH_2^- > NH_3} $.
In SN1/E1 reactions, the weak nucleophile can only attack the carbocation once the leaving group has left. This is precisely because it is a weak nucleophile.
IN SN2/E2 reactions, the strong nucleophile attacks the carbon before it can even form a carbocation. This forces the leaving group to leave. This is precisely because it is a strong nucleophile.


text:
How to synthesise N,N,3‐trimethylbutanamide from 1‐chloro‐2‐methylpropane?

Suggest a synthetic route to amide C from the alkyl chloride B (reaction mechanisms are not required). Standard reagents and solvents may also be used.

In this question immediate reaction is to make a grignard reagent with B by reacting it with magnesium and have it attack $\ce{COX(N(Me)2)}$ to form C.
I'm not sure how it would be possible to make $\ce{COX(N(Me)2)}$ from standard reagents.
Are there any obvious synthesis ideas I'm missing in making $\ce{COX(N(Me)2)}$ as surely if I started from $\ce{COXY}$ where $\ce{X}$ and $\ce{Y}$ are Leaving groups, I would end up with $\ce{CO(N(Me)2)2}$ on reaction with $\ce{NH(Me)2}$ which would give an alcohol as the end product, as the amine isn't a good leaving group?
Also would $\ce{COXY}$ i.e. $\ce{COCl2}$ even count as a standard reagent? I have no idea what that really entails.

COCl2 (phosgene) is a standard reagent, but you can't use it here because you won't be able to stop at a single addition of Grignard to phosgene. Making a Grignard is a good idea, but try adding it to a different one carbon electrophile that gives a carboxylic acid (or carboxylate).

I don't think it's necessary to start with a Grignard; it is not stipulated in the question? Correct me if I'm wrong.

You can make dimethylcarbamic chloride (Me)2NC(O)Cl from DMF with thionyl chloride or just buy it, it is commercially available for quite cheap. Addition of your Grignard to it should be simple as you suggested.

Phosgene, $\ce{O=CCl2}$, is a potent toxin and in no way would I consider it a safe or even standard laboratory reagent (at least on an undergraduate level).
Allow me to lead you down the garden path of organic synthesis.
We have a haloalkane. What do we know about such compounds; they have a polarised $\ce{C-X}$ bond ($\ce{x}$ is a halogen) and are prone to nucleophilic attack/substitution. In this case, we have a nice primary, haloalkane, hence, any nucleophilic substitution is likely to be SN2.
Now, the target is an amide, but what do we note is different about the target versus the starting material (SM)? There's an extra carbon present (and yes there's no halogen).
Warning bells should be going off at this time. What nucleophilic, reagent do we know of that could extend a carbon chain?
I can think of one: $\ce{CN-}$ (cyanide ion).
So, first react with $\ce{KCN}$, second react with conc. $\ce{H2SO4}$ to give a carboxylic acid, third react with $\ce{NH(Me)2}$ to give the product.
Now, once you've isolated the carboxylic acid, you can react it 'neat' with $\ce{NH(Me)2}$ and molecular sieves. The molecular sieves simply ensure a dry reaction and very high temperatures are needed (approx $\pu{160 ^\circ C}$). You could convert the carboxylic acid to an acyl chloride and react the amine to ensure a very high yield and 'fast' albeit more 'vigorous' reaction; in this case any ordinary organic solvent should suffice.

react with conc. h2so4  under aqueous conditions would hydrolyse it to give a carboxylic acid

@goods This strategy is fine, but you definitely want to derivatize the carboxylic acid (acyl chloride would be best, but there are alternatives) before reacting with the amine. The acid/base reaction between the amine and carboxylic acid is so favorable that direct amide formation is difficult.

@jerepierre to do this i would react it with socl2 right?

@goods That's the most common way.


text:
Bandgap of a semiconductor for photocatalytic water splitting
I'm reading a review article on photocatalytic water splitting (using semiconductors) and I came across the following:
"To achieve photocatalytic water splitting using a single photocatalyst, the bandgap of the semiconductor must straddle the reduction and oxidation potentials of water, which are +0 and +1.23 V vs the normal hydrogen electrode (NHE) at pH =0".
I have a few questions about this statement:

What exactly does it mean when you say that the bandgap has to "straddle" the reduction and oxidation potentials? Does it mean that you need an electron/hole with energy at least equal to 1.23 eV for oxidation/reduction to take place. So, the bandgap has to be greater than 1.23 eV in order to ensure that. Is my understanding accurate?
When using a single photocatalyst, I assume the oxygen and hydrogen evolution take place on the same material. If you had separate oxygen evolution and hydrogen evolution catalysts, what would the condition on the bandgaps of these semiconductors be?

Thanks! Sorry if these questions are too basic, I have little training in chemistry.

The original purported advantage of semiconductor|liquid junctions proposed by those in the field of photoelectrochemistry arose from the hypothesis that charge carriers would have a shorter distance to travel than in solid-state photovoltaic systems, thereby allowing lower quality materials to be used in an integrated solar energy system and still achieve high conversion efficiencies. In theory, one would also be able to avoid forming metallurgical junctions, say those p-n junctions made by using different dopants and creating a well-defined interface to separate charge carriers. Instead of forming these solid-state structures, you can define the electric field by selecting a suitable redox couple in solution. The specific energetics of the semiconductor and redox couple will then define the maximum power obtainable from the given system, so this is why the valence band must be below the energy of the OER and the conduction band above the energy level corresponding to the HER. By natural induction, this means that the bandgap must meet or exceed 1.23 eV, but this is only a thermodynamic constraint.
In reality your system’s bandgap must actually exceed 2.0 eV, this is due to the kinetic constraints based in part on the OER (and to a lesser extent the HER ) and losses encountered within the light absorber. Of course, for a single light absorber there is a tradeoff between the percentage of utilized solar flux and the bandgap.
This explains why many efforts are focused on using dual light absorbers, with one dedicated to the anode and another for the cathode. In this tandem structure they can efficiently utilize more of the solar spectrum. If you were to stack these light absorbers and use a 1.2 eV and 1.8 eV materials (with each one carrying out one of the two half-reactions required to split water), you could in theory get twice the efficiency compared to a single absorber with a bandgap of 2.2 eV.
Now are these semiconductor|liquid junctions going to beat a dedicated photovoltaic solar panel/electrolyzer any time soon? Maybe, but probably not. Is water splitting really worth doing? Maybe, but probably not. Is it a beautiful idea? Definitely.


text:
Organisation in Periodic table
Why do we find big spaces in the periods of the Periodic Table between $\ce{H}$ and $\ce{He}$, $\ce{Be}$ and $\ce{B}$, and $\ce{Mg}$ and $\ce{Al}$?
What is the logic of such organization of the periods?

I believe that you’re meaning groups rather than periods - unless my chemistry is rusty, the groups are the vertical arrangements of elements whereas the periods are horizontal. But I could be mistaken.

related: http://chemistry.stackexchange.com/questions/901/what-are-alternates-to-the-periodic-table-of-the-elements

I don't think that this is really a duplicate of that previous question - at least people react nervously over there that mine is not an answer. Anyway, it is an answer to this question, see: http://chemistry.stackexchange.com/questions/901/what-are-the-alternatives-to-the-periodic-table-of-the-elements/28775#28775

From a historical perspective, this was done to account for the commonalities between how the various elements behaved. In other words, the Alkali metals all exhibited a strong reaction with $H_2O$, the noble gasses all exhibited inertness in reactions, and so on.
This is explained in modern day science via the quantum description of electronic structure which indicates that elements within the same group tend to have the same number of valence electrons in the valence shell. Simply put, the reason why there are such gaps between $H$ and $He$; $Be$ and $B$, and otherwise is because there are simply different quantum arrangements of the electrons which are readily accessible to the element. That is, the ground state of $He$ has both accessible $l=0$ states occupied by electrons, which constitutes the valence shell and so is in an energetically favorable state to remain inert - becoming of the noble gasses. Hence, it’s moved far to the right from $H$ to the 8th group; consistent with the historical structure of the table.
That’s a crude description, but about what’s happening here. I agree however that the chemists may feel more at home discussing the periodic table. I’ll stick to the chart of nuclides.
Also, the Wikipedia article is fairly informative on this topic:
http://en.wikipedia.org/wiki/Periodic_table


text:
Why do we use ethanol in SN2 reaction?
Solomons and Fryhle has an example in which they have showed a very simple substitution reaction. The question was to synthesize methyl iodide. What they have done is taken methyl chloride and produced methyl iodide in the presence of $\ce{NaI}$ and ethanol through an $\mathrm{S_N2}$ mechanism.
Why did they use ethanol? I have read that $\mathrm{S_N2}$ reactions are favored under polar aprotic solvents and clearly ethanol is not that. Also, since it's a methyl and all substitution reactions on methyl go through $\mathrm{S_N2}$; so how would it affect my reaction if I would't use ethanol?

Usually, nucleophilic substitution reactions of anionic nucleophiles occur more rapidly in polar aprotic solvents. In such solvents (e.g. DMSO) the typical relative reactivity order is
$$ \ce{SCN-} < \ce{I-} < \ce{Br-} < \ce{Cl-} < \ce{F-}$$
In protic hydrogen-bonding solvents (e.g. ethanol), anions are solvated. Therefore, the reactivity of nucleophiles is decreased. Since hard nucleophiles are more strongly solvated than soft nucleophiles, the relative reactivity of soft anions is increased in such solvents. The typical relative reactivity order is
$$ \ce{F-} < \ce{Cl-} < \ce{Br-} < \ce{I-} < \ce{SCN-}$$
In the given example, $\ce{-Cl}$ shall be replaced by $\ce{-I}$; and in alcohols, the nucleophilicity of $\ce{I-}$ is higher than the nucleophilicity of $\ce{Cl-}$.

When compared to some polar, aprotic solvents, such as acetonitrile or dimethyl formamide, ethanol has a couple of advantages:

it is rather cheap
it is less toxic
the solubility of sodium iodide is rather high

So you mean to say that it is suited only for this reaction of many other substitution reactions also?

Please see my edit.

Where did you find the statement: "all substitution reactions on methyl go through $S_N1$"? It is obviously wrong. Methyl definitely favor $S_N2$.
Here is the answer. The conversion from $\ce{MeCl}$ to $\ce{MeI}$ go through $S_N2$ mechanism. The best solvent for $\ce{-Cl}$ to $\ce{-I}$ conversion is usually acetone. It is cheap, polar aprotic, good solubility for $\ce{NaI}$ and poor solubility for $\ce{NaCl}$. I think the reason to use ethanol is just because this reaction is too easy to be picky on solvent. It just works in ethanol.

Also what do you mean by "too easy to be picky on solvent"? And what if I don't use ethanol or any other solvent(I mean just NaI and and MeCl). I am sorry if a reaction without that a solvent sounds weird -I have next to zero experience in a lab, so I don't know which one is present in a solid or liquid form!

Neat (no solvent) condition is not possible because $MeCl$ is gas at room temperature.


text:
Mixing solvents, or how does paint thinner work?
I would like to understand how a mixture of solvents works together.
For example, there are two solvents combined, and each one alone can dissolve some material, but one acts faster.
How do they interact when used in combination on the same material?

An example would be paint thinner, for example a mixture of toluene and xylene.
One reason to use multiple solvents in a paint thinner is to make it dissolve more kinds of paints etc. (In German, this is called "universal thinner".)
But does the combination of multiple solvents somehow make it work better dissolving a single compound or mixture?
I had dissolved some polycaprolactone in that universal thinner above, and it seemed to dissolve quicker than in either toluene or xylene alone.
I do not know exactly what is in the thinner, so it does not mean much – but it made me wonder about this!
So, in which ways interact multiple solvents in the context of the example, and are there other mechanisms in general, not covered by the example?

Addendum:
For discussing a simple example, let's just assume the thinner contains only the two solvents above.
But for context, I looked up the Sicherheitsdatenblatt (MSDS) for my example thinner, which shows some more solvents on the list of ingredients:
Compound                                Mass %
----------------------------------------------
Toluol                                  15- 45
Xylol                                   10- 40
Ethanol                                  2- 15
Aceton                                   5- 20
Ethylacetat                              5- 20
n- Butylacetat                          10- 20
aromat. hydrocarbon mixture (C8-C10)     5- 15
aliph. hydrocarbons (C7-C9)              2- 10

Hansen Solubility Parameters (HSP) - is the short answer.
Your observation has very much to do with the energy density of the solvents - with the idea that lesser the difference in the energy densities better is the ability of solvent to dissolve the solute. I am very poor at explaining things and not an expert in the matter at hand (please excuse me for that), so I would instead explain using an example.
In HSP system (Proposed by Dr. Charles Hansen) there are mainly three parameters in consideration, namely: dispersion δd, polarity δp and hydrogen bonding δh. These values exist for solvents as well as solutes. Now we take into consideration two solvents and one polymer as under. The link is here https://www.hansen-solubility.com/HSP-science/solvent-blends.php.
Name                Type        δd      δp      δh
Toluene (T)         Solvent     18      1.4     2
Acetone (A)         Solvent     15.5    10.4    7
Polycaprolactone    Polymer     17.7    5       8.4
T + A (50:50)       Solvent Mix 16.75   5.9     4.5

The energy difference is calculated by the formula:
(Ra)² = 4×(δd2-δd1)² + (δp2-δp1)² + (δh2-δh1)²
Why Ra, I will explain in some while.

For Polycaprolactone in Toluene the value is:
√(4×(18−17.7)²+(1.4−5)²+(2−8.4)²) = 7.367
For Polycaprolactone in Acetone the value is:
√(4×(15.5−17.7)²+(10.4−5)²+(7−8.4)²) = 7.105
However when you mix the above two solvents in 50:50 ratio the value is:
√(4×(16.75−17.7)²+(5.9−5)²+(4.5−8.4)²) = 4.430

So individually solvents can be poor solvents for a solute but their mixture can be a better one.
Now about Ra. The d, p & h values can be considered as three coordinates and through three distinct points one unique sphere exists such that the points lie on the surface. The center of this sphere can be considered as the mean of the three values and Ra is the distance from this centre. The lesser the distance the better is the ability of solvent to dissolve the polymer in question (like dissolves like).
The value of Ra beyond which dissolution is not possible is denoted by R0 (unique to every polymer and calculate by trial and error - reading more on the topic will give you complete detail), so Ra should always be < R0.
Hope I have brought some clarity on the topic.

I had dissolved some polycaprolactone in that universal thinner above, and it seemed to dissolve quicker than in either toluene or xylene alone.

Some of these universal thinners do contain a lot more than just toluene and the xylenes.
If you want to find out about the composition of these mixtures, a material safety data sheet (MSDS, in German: Sicherheitsdatenblatt), provided by the manufacturer gives some information:
As an example, take this MSDS by a swiss manufacturer, which lists the following ingredients:
\begin{array}{lr}
\textrm{compound} & \textrm{amount [%]}\\
\hline
\textrm{toluene} & 25-50\\
\textrm{xylene} & 10-25\\
\textrm{acetone} & 10-25\\
\textrm{propan-2-ol} & 2.5-10\\
\textrm{4-methyl-pentan-2-one} & 2,5-10\\
\textrm{2-methoxy-1-methylethylacetate} & 2,5-10\\
\end{array}
Particularly the latter two might be much better solvents for the polycaprolactone than toluene or xylene.

Indeed, my example thinner also contains more solvents. I'll add details (But regarding the example, I think it's good enough to just pretend there are only two solvents.)

The answer does explain why the thinner worked quicker - but the question is more about all these solvents interact, like when ethylacetate dissolves the compound quickly, and toluene does so slowly, but there is very little ethylacetate, and a lot toluene. (I'm not asking about the actual interactions with these listed solvents - just simplified examples.)


text:
Clemmensen reduction of 1,3 diketones
As part of a synthesis I was considering the following reaction:

How selective is the Clemmensen reduction? I know that this reduction favours ketones adjacent to benzene rings but will it also reduce the other ketone, and/or affect any other parts of the molecule. Are there any other methods that would achieve this transformation more effectively.

Use a weak base/TMSCl to form a 6-cyclic (O-silyl)ketal with the outer carbonyl then use Wolff-Kishner ? Probably not good yield.

A Clemmensen reduction of a 1,3-diketone probably isn't the best idea. The reduction is likely to proceed via a ketyl radical.

In the case of a 1,3-diketone, formation of a dihydroxy cyclopropane, which undergoes opening/rearrangement cannot be excluded.
This reaction has for instance been observed for dimedone, where 2,4,4-trimethyl cyclopentanone was formed.
EDIT 1
Which other viable procedures exist?
If you insist on using the given starting material, the following might (!) work:

regioselective (asymmetric) catalytic hydrogenation of the acetophenone-type carbonyl using a 1,2-diamine-substituted Noyori catalyst
Barton-McCombie reaction of the resulting alcohol

If you are rather interested in the product, you might want to rethink the strategy:
The intended product is an arylbutanone, related to the raspberry ketone. A possible retrosynthesis would be:

If you get your hands on the necessary benzaldehyde, an aldol reaction with acetone, followed by catalytic hydrogenation might give the right product.
EDIT 2
5-tert.-butyl-2-hydroxybenzaldehyde is commerically available, but not really cheap.
However, 4-tert.-butylphenol is cheap as dirt. In case you need significant quantities of your final product, you might want to consider the synthesis of the benzaldehyde. The substitution pattern cries for a Reimer-Tiemann reaction.

Are there any suitable alternative methods for this transformation then?

@bon I've added a possible strategy to my answer.


text:
Interpreting a Phase Diagram?

The correct answer (that is false) is A) however can someone please explain why the other options would be true and why it is in fact possible to compress a solid to produce a liquid?

Do you mean by reference to the provided diagram (from which all the true answers are clearly true)?  As for how you compress a solid to produce a liquid, one clear answer would be that the liquid is denser than the solid forms, and thus is preferred under high enough pressure.  An example is lithium.

Let us go through this step-by-step:

A: A is undeniably false because if we pick a point on the solid side of the graph and add pressure to it, the point eventually
  becomes the liquid.
B: B is true because the triple point temperature is where they all intersect, and the temperature there is less than the critical point
  temperature.
C: You have to understand that C is a can question; it is true because you can form a gas or a liquid.
D: As you can see, where the substance changes phase is below 300 K, so we can assume that this one is also true.
E: The liquid will boil when the  vapor pressure equals the pressure. The line from the triple point to the critical point is
  sloping up. So this is also true.

That leaves us with A as the only false answer.

You might want to review some concepts on phase changes and pressures.
However just to add as a side note, this question kind of seems like a homework question, you might want to add more detail and your process, so that people can take the time to answer you back. Showing some effort, gives a cause for others to care.
Thanks.
Sources: Jon Custer helped find out E.

For E remember that the liquid will boil when it's vapor pressure equals the applied pressure. The line from the triple point to the critical point is sloping up.

Yup, thank you. I will add that into my answer.


text:
What is the product of alpha decay of curium-226?

What product is formed when curium-226 undergoes alpha decay?

Curium has atomic number 96. Since alpha decay decreases the number of protons and neutrons by 2, the product formed has atomic number 94 and atomic mass 222, thus the product is plutonium-222. However this was not an answer choice. The only answer choices were isotopes of radon, radium, and thorium. Either I don't know how to do a very simple question, or there are alternative representations of radioactive isotopes that I don't know about.
I would like to determine, if the question itself is wrong or if I am making a mistake.

curium-226? http://en.wikipedia.org/wiki/Isotopes_of_curium

Are you sure that the question wasn't about $\ce{^{226}Ra}$, which would decay to $\ce{^{222}Rn}$?

curium-226 isn't known and even if it was it would undergo rather beta - decay

The question is flawed (possibly a misprint), your reasoning is sound (given the information provided to you).
Curium-226 does not exist, according to the Web Elements page on the isotopes of curium, the 'smallest' isotope is curium-240, which undergoes $\alpha$ decay to become plutonium-236.
$$\ce{^{240}_{96}Cm -> ^{236}_{94}Pu + ^{4}_{2}\alpha}$$
The most likely candidate of what the question should have been asking about is radium-226 decaying to radon-222,
$$\ce{^{226}_{88}Ra -> ^{222}_{86}Rn + ^{4}_{2}\alpha}$$
which is part of the uranium-series decay chain:

This is the most likely scenario, as the smallest isotope of thorium is thorium-227 (WebElements) and for radium, it is radium-223 (WebElements), both have too high a mass for anything with an atomic mass of 226 to directly decay into by alpha decay (as per the question).


text:
How to visualize or think about spin waves (magnons)?
According to Wikipedia: "A magnon is a quasiparticle, a collective excitation of the electrons' spin structure in a crystal lattice."
I have little pictures in my mind for other quasiparticles. For example, I can picture a phonon as (quantized) vibrations in a crystal lattice, or I can picture excitons as an electron jumping between band gaps (or HOMO-LUMO, etc.), or plasmons as quantized oscillations/sloshing of an electron gas.
I have no mental image for what a spin wave looks like. Can someone help me understand this? What would a collective spin excitation look like? Do the spins just precess faster or something of the sort? Also, how would we measure a spin excitation or spin wave?

In short this is the way I imagine them, as slight directional changes in spins propagating through the system.
The Heisenberg Hamiltonian for the exchange energy associated with magnetic coupling is a pairwise sum over an exchange integral $J_{ij}$ for two sites $i$ and $j$, with spin moments $\hat S_i$ and $\hat S_j$.
\begin{equation}
H_{ex}=-\sum _{i<j}J_{ij}\hat S_i\cdot \hat S_j
\end{equation}
Imagine a periodic array of lattice points each with a spin of two possible orientations $\pm\frac 12$, for a ferromagnetic ground state $(J_{ij}>0)$ with all the spins aligned, denoted as;
\begin{equation} |\uparrow \uparrow \uparrow \dots \uparrow\uparrow\rangle
\end{equation}
A low energy excitation would intuitively be a single flip of a spin somewhere in the lattice e.g.,
\begin{equation}
|\uparrow \downarrow\uparrow\dots\uparrow\uparrow\rangle
\end{equation}
However, this is not an eigenstate of the spin Hamiltonian. Instead low lying excitations have a complex wave character that is akin to a fractional displacement of a spin with respect to its neighbour. Such infinitesimal variation I personally imagine as a single reversed spin distributed over all the spins in the lattice.
Therefore you could view a spin wave as a superposition where all the spins contain a bit of this single inverted spin in the first excited state. It is easy to see that this is the case since we are using ladder operators to change our spins from up to down. Acting as a superposition over the lattice (spin wave) reduces the disturbance between any given adjacent pair, whereas the naive expected eigenstate would include a really big difference in exchange energies at a localised region.
As you may be thinking, how can I have an infinitesimally displaced spin when there are only two spin orientations? Well that is a bit of a misnomer as this oversimplified model has a discrete symmetry of 2. In reality there is a smooth symmetry associated with these spins so that the spin wave is actually a more realistic picture than the 2-fold orientation (Ising model).
Hope that helps :)


text:
Do liquids exist that emit light when vibrational energy is added to the system?
I am looking for a liquid that illuminates when vibrational energy is added to the system.
It does not have to be strictly chemical, it might as well be something else: maybe a crystal in powder form mixed with some kind of oil, or even algae in a thickish liquid, for example.
Does such a liquid exist?

If you want a pure liquid to emit light, you'll probably have to vibrate it at very high frequencies with ultrasound, creating sonoluminescence, which AFAIK is not entirely understood yet but does not seem to be a chemical phenomenon. Perhaps a more exotic possibility would be a piezophototronic solid.

@NicolauSakerNeto interesting. At what frequencies are we talking here?

Well by definition ultrasound means 20 kHz and above. It seems that the exact value used can fluctuate between 20 kHz and 60 kHz (simply because lower frequency ultrasound is easier to produce, I assume), varying with parameters such as the nature of the liquid under sonication and the resonant frequency of the container. I don't really know much about sonoluminescence, but there's a lot you can find on it, if you're curious. The light is incredibly weak and brief, though, so don't expect much.

@NicolauSakerNeto would a more viscous fluid give off more light? I'm thinking that the resistance of the molecules would increase the sonoluminescence effect.

Probably less, actually; the mechanical energy in the system would more quickly be lost in the form of heat while trying to get molecules to scrape past each other.

@NicolauSakerNeto yeah that seems more likely. So a less viscous fluid would be better I guess. But as I mentioned in the question, is there any chemical that illuminates when energy is applied? either by pressure or vibration or anything?

I can't specifically think of one, but I imagine some types of unstable molecules might decompose with the emission of light. Maybe someone knows an example.

Dioxetanes emit light when energy (heat) is applied.  I also think that @NicolauSakerNeto 's comment is right on target.  Sonication (a way of supplying energy) can cause cavitation (bubble formation and collapse) in many liquids.  Upon bubble collapse, light can be released.

You could look up triboluminescence, this would fit your description.

Since the process does not have to be strictly chemical and algae is mentioned as a possibility, the answer is yes, a liquid that emits light when shaken does exist and is a well known natural phenomenon - bioluminescent water.
Many ocean dinoflagellates produce light when disturbed, producing spectacular displays in waters worldwide.  The gif below from Timothy R Fallon via Wikipedia shows surfing on a bioluminescent wave near San Diego.

Many other videos of bioluminescence in agitated ocean water are available, showing the phenomenon in locations as diverse as the Maldives, Puerto Rico, and Wales.
The dinoflagellates produce light in response to mechanical stress as a defense mechanism against predators.
Liquid-filled orbs that glow when shaken or just the bioluminescent liquid itself are commercially available.

Very nice answer after all this time! It's a beautiful phenomenon, both visually and scientifically. Thank you for this contribution.

Find compounds that emit light under photolysis.
As that may involve radical production pathways, one can also try to apply sonolysis to the same compounds as:
H2O (with dissolved O2, N2,..) + Sonolysis -> *H + *OH
Reference source: 'Free radical formation from sonolysis of water in the presence of different gases'), to quote:
"In this case, O2 reacts with •H to form •OH, indicating that the presence of dissolved O2 in water sonolysis gives additional amount of •OH. "
Also:
H2O2 + Sonolysis -> *OH + *OH
Apparently, with applications, see, 'Bactericidal effect of hydroxyl radicals generated by the sonolysis and photolysis of hydrogen peroxide for endodontic applications'.
where it is likely that one may be able to replicate to varying degrees the action of light treatment.
Here is a source example citing the action of reactive oxygen species (ROS), which can be created by the UV photolysis of, for example, H2O2 (or sonolysis per reference above):
"In biological systems much attention is focused on the two free radicals, superoxide anion and hydroxyl radical....The generation processes of ROS can be monitored using luminescence. Because of the very weak native luminescence of ROS, both luminol and lucigenin have been used in the past to give measurable signals. However, the luminescence of these two substances is very small compared to that of Pholasin®, which allows precise analysis on very small samples."
So, perhaps try performing a sonolysis of, say, luminol or lucigenin, in H2O2 might be interesting.


text:
Understanding the hybridization of cyclohexane?
I'm having quite a fair bit of trouble comprehending the hybridization of cyclohexane. I understand that this is a ringed structure with single C-C bonds, therefore all the carbons should have sp3 hybridization right?
Apparently however, this is incorrect, but I am not sure why. I'm reading about how it not planar like benzene is, and adopts a chair/boat conformation which I am having trouble visualizing as well.
If someone could provide some insight on this topic, that would be great!

Well, it is in chair conformation because it's sp3 hybridised.

so all the carbon atoms in cyclohexane are sp3 hybridized?

Yeah, why the doubt?

"doubt"-story of my life...my whole existence is one big uncertainty of myself- but unfortunately there's no stackexchange for my existential crisis, but thank you!

Get a ball/polygon and stick molecule building kit. Build molecules, twist and flip them. Get Geoff Hutchinson's Avogadro and have fun with that!

I'm having quite a fair bit of trouble comprehending the hybridization
  of cyclohexane.

Atoms can be hybridized. Not entire molecules. Just atoms.

I understand that this is a ringed structure with single C-C bonds,
  therefore all the carbons should have sp3 hybridization right?

Yes, all the carbons in cyclohexane can be said to be sp3 hybridized.

I'm reading about how it not planar like benzene is, and adopts a chair/boat conformation which I am having trouble visualizing as well.

In benzene all the carbons are said to be sp2 hybridized because benzene is flat. Carbons which are sp2 are in the same plane with their ligands. Sp3 carbons are not flat with respect to all their ligands.

why are the sp3 carbons not planar (not flat)?

You seem to be adept at identifying sp3 carbons. So I won't go over that.
Within the framework of VSEPR theory -
1) Atoms are made of charged particles, namely, protons and electrons.
2) Protons are at the core (nucleus); electrons are distributed around the protons.
3) Electrons repel each other (like charges repel).
So if we have 4 atoms arranged around a central carbon atom the lowest energy conformation or the configuration which will keep each of the 4 atoms the furthest apart is tetrahedral. The angle between the four ligands in a perfect tetrahedron is about 109.5 degrees. If we constrain ourselves to 2 dimensions and force all the atoms to exist in the same plane then the best angle of separation we can achieve is 90 degrees (imagine carbon to be at the center of a square with each corner of the square representing an atom). Clearly, 109 degrees > 90 degrees so the tetrahedral configuration is favored.

sorry what do you mean by "ligands"?

Whatever that is attached to carbon (or any atom)

are the carbons in benzene sp2 due to the alternating double bonds? Do the single bonds not account for anything?

why are the sp3 carbons not planar (not flat)?

Look at resonance theory because your mental image of benzene is not quite right

oh, so the carbon carbon bonds are all double bonds?

@Nerdingout - No, that's not right either. The Wikipedia article on benzene is useful though.


text:
Why does the black powder that comes off aluminum pans behave the way it does?
When I wet an aluminum pot or cookie sheet and then dry it with a towel, it creates dark gray marks on the towel. This happens more if the aluminum is a little rough. It seems that there's a fine black powder coming off the aluminum. I would guess that's aluminum oxide, but I was under the impression that aluminum oxide forms a hard coating that protects it from corrosion. So is that stuff really aluminum oxide? Why does it come off that way? Does it mean the aluminum isn't protected from corrosion?

If you try rub it with a lemon slice instead will it still leave a dark stain on your rag?  If not then you may have unexpectedly alkaline water and the stain may be an aluminium salt.  I would decline to eat such deposit in large doses or for extended periods as it may lead to some ailments (some research points to it quite strongly).  Baking cookies would not concern me but boiling soup would.


text:
Raoult's law: is the molar fraction the molar fraction in the entire system?
I have two physical chemistry textbooks, one by Atkins and one by Levine, which seem to differ in their notation used when describing Raoult's law.
Atkins talks about the molar fraction term in Raoult's law as if it were of the entire system.
Levine says that Raoult's law has the form $P_i = x^l_i P_i^\circ$, which seems to imply that $x^l_i$ is the molar fraction in the liquid.
So what is it really? And why would Levine use that notation?
Levine also uses this notation for Henry's law.
Thanks.

this is the definition as I was taught: $\frac{p_i}{p_{0i}}=x_i=\frac{n_i}{n}$

Can you let me know the exact things quoted in either of the books? It should be the fraction in liquid that matters.

@SatwikPasani I think I've solved this now. As you say, its the fraction in the liquid. My main source of confusion I think came from the wording in most problems. Problems said "A mixture with $X_a$ and $X_b$... find the pressure in the vapor due to $a$", and that confused me because it doesn't specifically say that is the liquid phase they're talking about. Thanks for your help.


text:
What is the net ionic equation of sodium hydroxide when it dissolves in water?
What is the net ionic equation of sodium hydroxide when it dissolves in water?
For the net ionic equation I got
$$\ce{NaOH(s) + H2O(l) -> NaH+(aq) + OH- (aq) + H2O(l)}$$
but it was wrong.
Then I tried:
$$\ce{NaOH(aq) -> Na+(aq) + OH- (aq)}$$
and it was wrong as well.
Can someone please explain?

Your second answer is close but think about the state of sodium hydroxide before dissolving in water.

Got it! NaOH is a solid!

When you write down the net ionic equation, you have to consider that the educt and the products are in different phases, a solid and a liquid phase. Undissolved $\ce{NaOH}$ is a solid, denoted by the index $\ce{(s)}$, while the dissolved $\ce{Na+}$ and $\ce{OH-}$ ions are solvated in the aqueous phase (index $\ce{(aq)}$ in the equation):
$$\ce{NaOH(s) ~->~ Na+(aq) + OH^{-}(aq)}$$

You don't need the H ion with the Na ion. You were correct the second time. and if there is two of the same product or reactant of on either side of the equation it is eliminated because it is just a spectator.

"NaOH (aq)" is not correct. See Jannis Andreska's answer. Identical products and reactants does not make them a spectator either, acid catalysis is a thing even in secondary school chemistry.


text:
Are ethyl iodophenylundecylate and Ethyl-10-(Iodophenylundecanoate) the same compound?
I have no understanding of chemistry, so these to chemical formulas mean nothing to me. What I am trying to ascertain is are these compounds the same? If not, what differences are there especially if injected into the CSF [which they were]?

Known as Ethyl Iodophenylundecylate:
CH3
I   (CH2)8 COOC2 H5
CH

Known as Ethyl-10-(Iodophenylundecanoate)
CH3
I -CH2 (CH2)6 CH2 COOC2 H5
CH

Hi thanks for your input Beginner and Martin, what I am trying to understand is why would this formula be changed 30 years later why would the developers not just leave it as is.

The formula is still the same. Could you provide your source of information, maybe we can be of better help then. For example, add a picture or a citation.

The "structures" you are trying to post and their respective names do not match, as I cannot spot a phenyl group in there. (Also it is hard to tell from this diagram.) But based on what you gave us, I believe these are the same compounds, where ethyl 10-(4-iodophenyl)undecanoate is the correct IUPAC name and Ethyl iodophenylundecylate is a shortened synonym. I also found trivial names for this compound, for example Pantopaque and Iofendylate.
You can access this molecules record on ChemSpider and you will find a couple more names attached to it. From the same page is the following structure taken:

(source: chemspider.com)

The answer, in principal, is not. But, based on your info, the correct structure is not so easy to determine and the reason is, as Martin states, there is no info related the positions of the functional groups (This is to say the Iodine and the phenyl). However and taken literally the structure name you shared, the structures are:
The structure of the named compound Ethyl 1-( [# ?]-Iodophenyl)-undecylenate is:

The structure of the named compound Ethyl 10-Iodo-phenyl-undecanoate is:

But most probably is Ethyl 10-( [# ?]-Iodophenyl)-undecanoate (Similar as Martin presented):

Related to the second part of the question about their activity into CSF is hard to say based on the incomplete info to determine their respective influence. May be is more easy to look at any vademecum for a more generally medicine.


text:
How much ice will reduce the chicken carcass temperature from 32 to 8 degree C
I am interested to learn through heat claculations the quantity of ice that would achieve a chicken carcass core temperature of 8 degree C.
My chilling tank has a volume of 15,000 liters of flowing water where chicken pass through at a rate of 3,000 carcass per hour having an average carcass weight of 1.4kg and deep muscle temperature of 32 degree C.The flowing water drains at a rate of 5 litres per second.
Supplementary ice added to the chilled water at 250kg per hour while 8 degree C chilled water is fed into the tank at 4800 liters per hour.
How much more ice is required per hour to reduce the core temperature of chicken to 8 degree C? Appreciate a detailed calculation.
I thank you for your time and efforts.


text:
Why do exothermic dissolution reactions occur?
I've always been taught that an exothermic dissolution means that the forming of solute-solvent bonds releases more energy than the energy consumed to break solute-solute bonds and solvent-solvent bonds. However, if the attraction that occurs between solute and solvent particles are mainly based off of intermolecular forces, why would the energy released when it is formed manage to exceed the breaking of the comparably stronger intramolecular bonds between solute-solute particles?

Possible duplicate of Thermodynamics adding salt to water changes the temperature

Obviously the intramolecular bonds are not stronger than the intermolecular bonds if the majority of the solute dissociates when placed in water. There are some salts ($\ce{AgCl}$, $\ce{BaSO4}$, $\ce{ZnS}$) that have extremely low $\ce{K_{sp}}$ values, meaning the majority of the salt placed in the solution does not dissociate, and are therefore considered insoluble salts. In these salts, the intramolecular bonds are stronger than the intermolecular bonds formed with the water, which explains why only extremely small amounts of these salts dissociate in water.
There are, on the other hand, salts that do dissociate in water. Dissolution requires energy, and this energy is supplied by the kinetic motion of the water. This energy is used in breaking apart lattice structure of the salts, thereby lowering the temperature of the water.
The difference between salts with a negative enthalpy of solution and a positive enthalpy of solution, as you stated, is in formation of bonds between the water and the ions of the salt. The formation of a hydrogen bond is an extremely exothermic process, and accounts for the extremely negative enthalpies of solution of $\ce{HF}$, $\ce{NaOH}$, and $\ce{CsOH}$. Weaker intermolecular bonds also form in the dissolution of these salts and many others as a result of the interaction of the water with the electron clouds of the ions. Interactions between the water and the ions always occur, though to a greater extent in some salts than others.


text:
How much sulphuric acid can be obtained from 5 kg of sulfur

How much sulfuric acid $(\ce{H2SO4})$ can be obtained from 5 kg of sulfur $\ce{S8}$?
answer: 15.312 kg

To  make 1 molecules of $\ce{H2SO4}$ we need 1 atoms of S.
Now coming to $\ce{S8}$. This gives us 8 atoms of S. so from this we can obtain 8 $\ce{H2SO4}$ molecules if we take 1 atom of $\ce{S8}$ and use it to make $\ce{H2SO4}$ molecules. So,

1 molecule of $\ce{S_8}$ $\ce{->}$ 8 molecules of $\ce{H2SO4}$
0.019 moles of $\ce{S_8}$ $\ce{->}$ $8 \times 0.019=0.15$ moles of $\ce{H2SO4}$

Formula mass of $\ce{H2SO4} = 1(2)+32+16(4) = 98$
we know the formula
for $$\text{Number of moles} = \frac{\text{Mass}}{\text{Formula Mass}}$$
So from this $\text{Mass}_\ce{H2SO4} = \text{Moles} \times \text{Formula Mass} = 0.15 \times 98=14.7$
Now you see that my answer doesn't match to the answer. What did I do wrong?

I'm also getting 14.7 kg$. Maybe the answer given is wrong

Do you Think my method of doing this question is 100% correct?

Yes it is. The stoichiometry and application of unitary method all look good to me.

then our answer is correct.

Too much approximation, both in your answer and the answer provided. $8\cdot0.019$ is not equal to $0.15$… If you pile up approximation, your result will start to diverge. In their answer $15.312$, they also used approximation with $M(\ce{O})=16$ instead of $15.999$ for instance, and $M(\ce{S})=32$ instead of $32.06$. The best answer given are indeed the one from ringo and Klaus.

Your result isn't wrong it is just inaccurate.
If you add an additional digit behind your values we get:
0.0195 instead of 0.019 and 0.156 instead of 0.15 and finally 15.288 instead of 14.7

$\ce{S6}$, $\ce{S8}$, $\ce{S12}$ – does it make a difference or is it just a trick to make the question more complicated than it is?
What you know for sure is:

$M(\ce{S}) = 32.065\ \mathrm{g}\cdot \mathrm{mol}^{-1}$
$M(\ce{H2SO4}) = 98.079\ \mathrm{g}\cdot \mathrm{mol}^{-1}$

$\frac{5000\,\mathrm{g}}{32.065\,\mathrm{g}\cdot \mathrm{mol}^{-1}} = 155.93\,\mathrm{mol}$
$155.93\,\mathrm{mol} \cdot 98.097\,\mathrm{g}\cdot \mathrm{mol}^{-1} = 15.29\,\mathrm{kg}$, which is close enough to the answer.

I prefer simple dimensional analysis as a method of arriving at the answer. I'm sure it will simplify many of your future calculations.
${5 \hspace{.5mm} kg \hspace{1mm} S_8}·\frac{1000 \hspace{.5mm} g}{1 \hspace{.5mm} kg}·\frac{1\hspace{.5mm}mol\hspace{1mm}S_8}{256.528\hspace{.5mm}g\hspace{1mm}S_8}·\frac{8\hspace{.5mm}mol\hspace{1mm}H_2SO_4}{1\hspace{.5mm}mol\hspace{1mm}S_8}·\frac{98.07948\hspace{.5mm}g\hspace{1mm}H_2SO_4}{1\hspace{.5mm}mol\hspace{1mm}H_2SO_4}·\frac{1\hspace{.5mm}kg}{1000\hspace{.5mm}g}=15.2934\hspace{.5mm}kg\hspace{1mm}H_2SO_4$
The discrepancy in the answer is likely as a result of me using very precise values for the molar mass of the atoms used in the calculation, but clearly the answer is correct.
If you are not familiar with dimensional analysis, the goal is to cancel the units via dividing with fractions. You will notice that all the units in the above equation cancel to leave us with our desired answer, with the units $kg\hspace{1mm}H_2SO_4$.

Expressions for units shall contain nothing else than unit symbols and mathematical symbols. Write ‘$\mathrm{kg}$’, not ‘$\mathrm{kg}\ \ce{H2SO4}$’.

@Loong yes, that is an SI guideline. And those guidelines were written with physicists and engineers in mind, not chemists, who often have to deal with several quantities of different substances, all with the same unit. Ask yourself: Would deleting the H2SO4 at the end make this answer more clear or less clear? Would you advocate deleting all the species from the working, too, which would lead to it being indecipherable?


text:
Finding concentrations in a voltaic cell

A voltaic cell with $\ce{Ni/Ni^2+}$ and $\ce{Co/Co^2+}$ half cells has the initial concentrations of $\ce{Ni^2+}$ $\pu{0.80 M}$ and $\ce{Co^2+}$ $\pu{0.20 M}$.
  (a) Find initial cell potential;
  (b) Find concentration of $\ce{Ni^2+}$ when $E_\mathrm{cell}$ reaches $\pu{0.03 V}$

In part (a), I correctly found the initial $E_\mathrm{cell}$ to be $\pu{0.05 V}$.
In part (b), the book doesn't have any other examples that are explained showing how to get one of the concentrations, it's supposed to be $\ce{0.50 M}$, but I'm not sure how they get to that. From my tables, I found $E$ half cell for $\ce{Ni}$ to be $\pu{-0.25 V}$ and $E$ half cell for $\ce{Co}$ to be $\pu{-0.28 V}$, and the overall then to be $\pu{0.03 V}$.
Tried plugging in everything I have into the Nernst equation, but that still leaves me coming up short.

Are we to assume that the temperature is $25^oC$?

Idea: the Initial cell potential is 0.05V, the potential drops to 0.03V. Since the overal reaction is equimolar and half cells potential similar,  Ni ions concentration drops by 0.03/0.05

I think you were on the right track!
You were right that you need the Nernst equation, which must be how you determined the initial $E_\text{cell}$.
$$E_\text{cell}=E_\text{cell}^\circ-\frac{RT}{nF}\ln Q$$
You correctly identified that your $\ce{Co|Co^2+||Ni|Ni^2+}$ cell has the cell standard potential
$$E_\text{cell}=E_\text{cathode}^\circ-E_\text{anode}^\circ=-0.25\ \mathrm V--0.28\ \mathrm V=0.03\ \mathrm V$$
So the question is essentially asking what concentrations are needed for your cell to have the standard potential. Again looking at the Nernst equation, you can solve for the reaction quotient $Q$ when $E_\text{cell}=E_\text{cell}^\circ$
$$E_\text{cell}=E_\text{cell}^\circ-\frac{RT}{nF}\ln Q$$
$$E_\text{cell}-E_\text{cell}^\circ=0=\frac{RT}{nF}\ln Q$$
$$\ln Q=0$$
$$Q=1$$
At this point, you should be able to solve for the concentration of $\ce{Ni^2+}$ with simple algebra.


text:
Addition of HBr to Vinyl bromide
The addition of $\ce{HBr}$ to $\ce{CH2=CHBr}$ produces $\ce{CH3-CHBr2}$. Why is this so?
According to me, halogens have a stronger $-I$ effect than a $+M$ effect (for example a halogen substituent on benzene, deactivates the ring), and hence the $\ce{CH3-CHBr+}$ carbocation should be more unstable as compared to $\ce{CH2+-CH2Br}$. Where am I going wrong?
Edit : I have seen this question but it doesn't answer my query.

What are the reaction conditions ? Addition of HBr proceeds via a radical mechanism if intitiated by photolysis.

@J.LS I'm not sure of the conditions, but I'm talking about the ones that will cause the reaction to proceed through a carbocation mechanism.

How does the linked question not answer your's? What are you still confused about?

@bon As I have written in the question, my understanding is that halogens destabilise a carbocation due a stronger $-I$ effect. However this does not reflect in the product formed on addition of $\ce{HBr}$ to vinyl bromide. The link does not talk about why halogens are stabilising the carbocation instead of destabilising it.

@BinaryGeek Ok sorry I misread your question slightly.

The reaction of $\ce{HBr}$ with olefins is a relatively fast reaction, in other words, the activation energy required is small.  Further, formation of the carbocation intermediate is the rate determining (slow) step in the overall process.  Considering both of these facts and applying the Hammond postulate we can say that the transition state leading from the reactants to the carbocation intermediate will be an early transition state, it will look more like the reactants than the carbocation.
This suggests that resonance and inductive effects need to be examined with regard to the reactants, rather than the transition state, in order to determine how the molecule will react with an electrophile (the proton).

As the above diagram illustrates, both resonance and inductive effects (in an early transition state that resembles the reactant) suggest that the beta carbon will be more nucleophilic than the alpha carbon.  Therefore addition of the proton (electrophile) will occur at the beta carbon.  This will be followed by rapid addition of $\ce{Br^{-}}$ at the carbon already bearing a bromine atom (the alpha carbon) to yield a geminal dibromide.

According to me, halogens have a stronger −I effect than a +M effect (for example a halogen substituent on benzene, deactivates the ring

The -I effect of a bromo substituent does indeed reduce the rate in a further electrophilic aromatic substitution.
However, the +M effect of the same substituent is ortho/para-directing, as to be seen in the bromination of bromobenzene.
This effect, namely the stabilisation of a positive charge through interaction with the lone pair of the halogen ($\ce{H3C-CH+-Br <-> H3C-CH=Br+}$) controls the addition of $\ce{HBr}$ to vinyl bromide.

I'm still not able to understand. I follow that the $+M$ effect will stabilise the positive charge. However halogens exhibit a $-I$ effect which is stronger than the $+M$ effect. So why does the $+M$ effect control the addition of $\ce{HBr}$ to vinyl bromide when the $-I$ effect is stronger?

An additional point that I wanted to state from this source.
The same source has an image which shows the reaction mechanism:

Note the 2nd reaction in the above image.
I quote some relevant text:

Note that the carbocation in this case bears a chloride ion. Since carbocations are electron poor, and chlorine is quite an electronegative element, it’s interesting to point out that the electron releasing ability of the alkyl group [and the ability of chlorine to donate a lone pair to the carbocation] “win out” here over the electron-withdrawing character of chloride ion.

This is the genaralized case while in your case the R=H and it no longers qualifies as an alkyl group. But it is important to note that in the actual pathway, 3 H contribute in the hyperconjugation while in the other pathway , only 2. This effect, combined with the +M of the Cl overpower the -I effect.


text:
Which plastic type is used inside dishwashers?
We have a drinking bottle made partly from polypropylene that wraps when washed in a dishwasher. We would like to change the plastic type so that it can withstand in dishwasher.
I know for a fact that my dishwasher at home has a lining inside made from polypropylene. After googling this  I have found a few sources that suggest that dishwashers would use a filler material in the PP called "calcium carbonate".
I would like to hear from someone who actually with confidence can tell us:
What is the usual plastic type used inside dishwashers? How is this plastic different from "normal" PP? Will we be able to use the same mold for this kind of plastic as "normal" PP?

One possible reason for the different behaviour of the drinking bottle, compared to the lining of the dishwasher may be indeed the different composition of the polypropylene deployed.  The part of poly indicates that many hundreds and thousands smaller propylene molecules are brought together to constitute the polypropylene.  Yet there are different ways this may happen, as this picture (source) depicts:

Along the polymer chain, there a methyl groups (marked as $\ce{CH3}$) occupying more volume than hydrogen atoms (marked by $\ce{H}$). It is imaginable that all of these methyl groups are attached on one side of the chain only (the upper picture), or consistently "once on one, the next time on the opposite side" (the lower picture).  Not shown here, there may be a randomized distribution, too, where you can't predict on which side the next methyl group will be attached (not shown here) as an intermediate situation.
Such alterations in the molecular structure may alter how individual polymer molecules interact with each other.  It influences if the polymer becomes more crystalline, or more glass-like; and by the same token, it influences the melting point of the material.  Indeed, the reference states melting points of 130 degree Celsius (266 Fahrenheit), as well as 171 degree Celsius (340 Fahrenheit).
Now think about the production of the bottle: it is done by (blow) molding.  A manufacture aims to produce a high number of bottles per unit of time, and preferentially at a low cost to spend, for example to heat the polymer that it becomes flexible enough to form it.  So it is better to select a polymer with a lower melting point, and by admixing / blending with other polymers to polypropylene, the melting point probably may be lowered further.
On contrast, a manufacturer of a linings of dishwashers equally takes into account the processes while the dishwasher is used: chemicals (detergent), hot water during the washing, and heating to eventually dry the tableware.  So the selection prefers a polymer that tends to be more crystalline and hence exhibits a higher melting point.
You may encounter situations showing you that this comes with an disadvantage, too:  The more crystalline, high melting polypropylene liner often is more brisk than the more glassy, low melting polypropylene of your bottle.


text:
Transistors: How does the gate work?
Two questions:
First, how does the gate provide a positive charge by running a current through it? I really just don't see how this positive charge could be achieved, unless you had some sort of battery in the gate itself and attached a wire to donate some of the electrons flowing through to the cathode of some other battery. Is it possible to get electrons out of the gate by some other means?

perhaps this should be asked on Physics.SE

Is there a significant $I_{GD}$ in FETs? In contrast to BJTs, the former aren't current-, but voltage-controlled. A conducting channel between source and drain is typically induced by applying a (positive) voltage between gate and source.

@KlausWarzecha Applying a positive voltage on the gate means that electrons are moving into the gate, right? But wouldn't that make the gate more negative?

Why wouldn't the electrons move to the positive voltage source? Instead of Physics SE I might suggest Electrical Engineering.  But, you seem to have a number of concepts mixed up with each other, so a little more thought should be placed before re-asking the question.

The gate of a FET works much like the grid of a radio tube (valve). The gate is insulated from source and drain; the only current flow (if the insulator were perfect) is to charge the intrinsic gate capacitance. Even a MOSFET has some leakage, but it's pico- or femtoamperes. See this discussion on capacitance and leakage.

It is really not charge injection, but charge separation. When a potential is applied to the gate with respect to the SD, the charges in the gate dielectric separate. This separation creates a layer of positive or negative charges just next to the SD channel.
In ideal cases there is no current through the gate.


text:
Is it possible to dope Ca ions in an NaCl crystal to produce a vacancy defect?
I was just reading about defects and doping in solids and read about the defect in  ionic crystal ($\ce{NaCl}$). There were examples of $\ce{SrCl2}$. Could this also occur with $\ce{CaCl2}$?

Absolutely. If you analyze even "pure" NaCl you'll find all sorts of trace amounts of other cations and anions. The defect is of course caused by the +2 charge on a calcium ion, whereas the lattice wants a +1 charge. Not sure about the relative size of the ions which can cause disruptions in the lattice too.

A peek at the NaCl - CaCl2 phase diagram (say at FactSage) will show you that NaCl can hold quite a reasonable fraction of CaCl2 in the NaCl crystal. In contrast, CaCl2 has very limited to no solubility of NaCl.


text:
Why are multiple oxidation states useful for a catalyst?
Transition elements are good catalysts because they have multiple oxidation states?Why multiple oxidation states matter?

Think about what a catalyst does, and how it might do it, then edit your question to add your thoughts - it's easier to explain something when we have a better idea of what you already know.

The answer has a lot to do with "what is a catalyst in the first place?"
A catalyst is a compound that can affect the rate of a chemical reaction
by providing an alternative and lower energy profile or pathway. That is,
it is a substance that makes it so that A -> B at a lower energy cost. But
it only has an effect on the rate of the reaction. That is, it only changes
the cost of the activation energy. It is not related to the thermodynamics
of the process and hence, the final product distribution. That is, if the
above process gives 90% B via a non-catalyzed pathway in 4 days, then a
catalyst will help you get 90% B in say 4 hours.
Okay, so a catalyst affects the transition state and activation path. How
does it do this? Typically, by complexing one of the reagents. Complexation
by transition metals affords access to a wide variety of oxidation states
for the metal. This has the property of providing electrons or withdrawing
electrons from the transition state of the reaction. That is, if the
transition state is electron rich, then the transition metal might hold
some of that electron density and those prevent too much from building up
on the reagent. This would then facilitate the reaction. Or the transition
metal might undergo formal oxidation/reduction to achieve electron transfer
to a substrate, thereby allowing a reaction to occur. This is "complexation
and electron storage" taken to the extreme but is a common mechanism in
organometallic chemistry. Indeed, a variety of catalytic pathways rely on
a two electron transfer between the metal and the substrate (e.g.
hydroformylation). It is the ability of the transition metal to be in a
variety of oxidation states, to undergo facile transitions between these
oxidation states, to coordinate to a substrate, and to be a good
source/sink for electrons that makes transition metals such good catalysts.
Of course, this is for transition metal complexes. Most industrial used
catalysts are the transition metal in a bed, as a metal or bound structure.
The above considerations are important but also the physical properties of
absorption/adsorption and the electron band structure of the material.


text:
Why do we insert a numeric multiplier after group positions when naming an isomer?
Question: Why do we insert a numeric multiplier after group positions when naming an isomer?
Take 2,3 dimethylbutane for example. Why do we add the numerical multiplier di when we've already explicitly mentioned two locations? Why don't we just use 2,3 methylbutane? This just doesn't feel intuitive.

take 2,3-dimethylbutane

In the case of different substituents on different or the same atom, we use a combination of the position and the name of the substituent, such as in

2-chloro-3-methoxy...
3-cyano-4-bromo..., etc.

Following this rule, 1,2-dimethoxyethane would be 1-methoxy-2-methoxyethane. Using the multiplier di, together with the positions, just shortens and simplifies the name and still is unambiguous.
The multiplier di explicitly denotes that there are two identical substituents at the given positions and not just one that bridges two centres.

An example for such a situation would be 1,4-methanonaphthalene (1). Note that even in this context, di can be used, as in 1,4:5,8-dimethanonaphthalene (2).

Why not just 1,2-methoxyethane?

@VatsalManot Because that does not explicitly state that there are two substituents and not one, that is bridging two centres, such as in 1,4-methanonaphthalene.

Take 2,3 dimethylbutane for example. Why does one add the numerical multiplier di when one has already explicitly mentioned two locations? Why doesn't one just use 2,3 methylbutane?

Actually, it is the other way round. Multiplicative prefixes (di, tri, tetra, etc.) are always used in names to denote multiplicity of identical features in structures, whereas locants (1, 2, 3, etc.) may be omitted under certain conditions when there is no ambiguity, e.g.
‘trichloromethane’ (instead of ‘1,1,1-trichloromethane’) or
‘heptafluorobutanoic acid’ (instead of ‘2,2,3,3,4,4,4-heptafluorobutanoic acid’).

Why did you omit 2,2,3,3,4,4,4 ? I thought only 1 can be omitted.

@A---B because it is not ambiguous. Only one molecule with the name heptafluorobutanoic acid exists. In other words, those numbers are implicitly understood.