text:
Is this carbocation rearrangement possible?
Is this rearrangement possible in carbocations?

https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange    see that link once.

Not a homework question. Was learning about rearrangement , just thought if it is possible for something like this to happen

@SR810 its fine whether it is or isn't a literal homework question, but you should include you thoughts on why you think it should/shouldn't occur and what research you have attempted to support that conclusion.

Oh! I was learning about rearrangements and saw 5 membered to 6 membered ring expansions. So I thought if 4 membered to 6 membered was possible.

Possible duplicate of Carbocation Rearrangement - Till which neighbouring carbon is it possible?

I'd say it's almost surely impossible.
First of all, a carbocation on primary carbon is extremely unstable, so the question becomes how would you generate that. It is possible under some very extreme conditions, but then those conditions and the surroundings of the newly formed carbocation would be more controlling the reaction than the carbocation on its own. This newly generated carbocation would probably instantly react with solvent or with the leaving group thus reversing to the compound it was formed from.
Secondly, if we assume that the primary carbocation formed and this is some idealized situation where there are no other molecules around then the more probable scenario is that you will observe hydrogen transfer from the neighbouring secondary carbon, which would then undergo another hydrogen transfer, eventually forming a relatively stable tertiary carbocation.

What I was thinking was that formation of a 6 membered ring will reduce ring strain to a great extent hence making the carbocation more stable. So maybe in a reaction where the 4membered ring carbocation cation is formed as an intermediate  the product is formed as a result of the ring expansion into a 6 membered ring. Like it happens for 5 membered ring Carbocations.

When it occur in five-membered rings the carbocation the original carbocation is just next to where expansion take place - you can imagine it as if the carbon forming bond slid along the bond of carbocation and tertiary carbon; or imagine it with a transition state where all three carbons are connected by partial bonds. When it comes to rearrangements remember that the bonds don't actually exist like sticks holding atoms - a line is just a metaphor for two atoms being held close by the cloud of two electrons. But when there are more atoms around (and charges), this cloud can deform.

Anyway, if that is for your own knowledge then I would recommend you to forget about carbocation rearrangements and move on. There are beloved by chemistry professors because they can trip students, but they have almost none relevance in actual chemistry. All those primary carbocations from exams and tests exist almost only on paper. Only carbocations to actually care about are the stabilized ones - like tertiary, allylic, benzylic.


text:
Do plasma globes generate ozone?
My kids and I were playing with a small plasma globe like this:

After touching it we all noted a faint smell like that created by other high voltage static discharges, which I always understood to be the smell of ozone.
So I guess this is actually two questions:

Is the smell associated with static electric discharge in fact ozone?
Do plasma globes like this create ozone?

(Presumably if the answer is "yes" then it would not be a good idea to leave one of these operating for an extended period in a closed space like a child's bedroom!)

I think the answer to this one is already summarized in Wikipedia's article on ionisers. Whether the ozone and the products of oxidation are dangerous, is still controversial; but I think there is just not enough ozone produced by the properly working plasma globe to be a real problem.

@andselisk – Yes, given the controversy, we don't need to get into my parenthetical health comment.

Small amounts of nitrogen oxides may also be generated, and they also have distinctive odors. However, ESD created by pulling fabric from a clothes dryer or peeling adhesive tape also makes a tiny bit of these reactive chemicals, and I don't know of any ill effects.

Ozone has a pretty distinct smell (although some people say it smells like chlorine gas, which obviously couldn't be produced by the plasma globe) and humans have a good memory for smell. So if you've smelled ozone before and think that's what your globe smells like, then you've got a pretty good positive test for ozone production. The threshold for ozone to be detected by smell is about 100 ppb, which is about the same as the threshold for ozone to cause damage to the respiratory system, so I wouldn't dismiss safety issues out of hand (although this isn't the place to discuss such things).

Plasma globes do indeed produce small quantities of ozone when they are touched.
This is because the high-voltage within the globe is at very high frequency and this can drive high kinetic-energy electrons through the glass to produce ozone generating micro-discharges at your fingertips.
Nox generation is highly unlikely at the low energies used.

Any reliable references for that?

source: Stack Exchange

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

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

sort: votes


text:
Does silver oxidise in a pH 1 solution?

Consider submerging a plate of pure silver in a solution with $\mathrm{pH} = 1$. Does the redox reaction with hydrogen happen spontaneously?
The composition of the solution is not given.

My attempt:

I considered the standard potential for $\mathrm{pH} = 0$. Since the silver must be oxidised, the reaction happens at the anode.

$$
\begin{align}
\ce{Ag+ + e- &-> Ag} &\qquad E^\circ &= \pu{0.80 V} \\
\ce{2 H+ + e- &-> H2} &\qquad E^\circ &= \pu{0.00 V}
\end{align}
$$

At this point, it is obvious that the reaction cannot be spontaneous since the potential for this cell is less than zero. In fact:

$$\Delta E = E_\mathrm{red} - E_\mathrm{ox} = \pu{-0.80 V}$$

From Nernst equation, the potential for the hydrogen at $\mathrm{pH}=1$ must be negative. Thus the reaction is all the more non spontaneous ($E = \pu{-0.059 V}$).

However, this seems to be wrong. I must be applying the $\Delta E$ wrong, because the solution for this problem states clearly that it is given by:
$$\Delta E = E_\mathrm{red} - E_\mathrm{ox} = \pu{-0.059 V} - (\pu{-0.80 V}) = \pu{0.741 V}$$
This is driving me crazy. Where does the minus sign come from?

I would love it if the downvoters explained. I think the question is clear, and my attempt to solve the problem is given.

The question looks fine to me, I edited it a little. Sometimes one gets downvoted for no good reason. Feel free to check out this page, this page and this one on how to format your future posts better with MathJax and Markdown.

@andselisk Thanks, however I can't see the formulas after the edit.

Let me guess, you are using a mobile app? If so, refer to the web version of stack exchange (with a web browser) since the mobile app is severely handicapped and doesn't render many things properly.

@andselisk yes, I am using the mobile app. That's too bad, I'll check it out from PC browser. Thanks again.

No prob, I only use mobile app to get notifications. The Stack Overflow app is great, but they don't really care much about other sites on SX network since there is not much traffic in comparison.

The answer given in the solution was wrong. It has been corrected, and it eventually proved my point.

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


text:
Why are HCl fumes white?
I just came to know concentrated HCl forms fumes because it dissolves in water vapors present in air. But why are these fumes white, HCl solution is transparent.

Water is transparent too; guess why fog is white?

something related to refraction maybe.

Tiny particles scatter light.

All right so i get it now. Thanks! Water droplets scatter all wavelengths equally so they appear white. Am I right?

Also, any ammonia in the air (common in labs) reacts with HCl to make a powdery precipitate of NH4Cl.

HCl is a colorless gas; it forms tiny droplets with moisture in air , as noted -fog. I have read that it( HCl )  is used  for "skywriting".

Same reason as https://chemistry.stackexchange.com/questions/26749/why-is-snow-white?rq=1 and for all white things.

I used to work in a chemical factory in the engineering dept. where we had a diazo (blueprint machine) which developed the prints using ammonia gas. On the days when the HCl truck delivered we had a dense fog in there.

Given that the fumes are scattering droplets, I would say that polar HCl acts as centre of nucleation for water vapour.
This is about the same way by which nitric acid containing small amount of nitrogen dioxide fumes white (in this case first the red-brownish oxide reacts with moisture to form the colorless acid, the latter in turn works as nucleus for water condensation and acidic droplets formation, ie white fume).
A more detailed or perhaps better answer should be under paywall:
Chang, O.-K. Why does concentrated HCl fume? J. Chem. Educ. 1985, 62 (5), 395. DOI: 10.1021/ed062p395.


text:
Correct way to draw a carbene
What is the correct way to draw a carbene? On most/all material I have seen, both singlet and triplet carbenes are depicted as so:

With 2 "dots" representing the non-bonding electrons on either a singlet or a triplet carbene.
However in a class today my tutor was adamant that a singlet carbene should be drawn with a positive and a negative sign on the carbon. His argument as such was that "you have an empty and a full orbital and so the empty orbital corresponds to the positive sign, the full orbital corresponds to the negative sign".
So my question is - How is a carbene correctly drawn? As above or with a carbon bearing both a positive and a negative sign?

You may wish to see Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) especially GR-5.3 where examples of singlet and triplet carbenes are given. This is probably worth an answer but I'm not free now, sorry.

There are two types of carbenes: singlet and triplet. Triplet ones should be drawn with two clearly separate dots, while singlet ones should be drawn with an electron pair and often with a vacant orbital (unless it participates in a strong conjugated system).

Put simply, there's no precedent for this unusual depiction of a singlet carbene.
Your comment about most carbene depictions, for both singlet and triplet carbenes is true, but a bit lazy. Yes, many chemists, frequently just draw two electrons as a lone pair on the carbon for both. In principle, with a moment's thought, you can often deduce the likely spin state.
As described, by the comment above, the standard for 2D chemical structure depiction is the IUPAC guide Graphical representation standards for chemical structure diagrams

In short, there's a difference in representing a triplet (two unpaired electrons), which should show preferably two separate electron "dots."
For a singlet, the lone pair is indicated, emphasizing the connection between the electron spins.

There is simply no place for an argument. It is about as insane as if someone forgot the the shape of the Cyrillic letter "Ы" and tried to deduce it from the corresponding sound, or from the shapes of other letters. The information is simply not there. The shape of a letter is a matter of convention, not deduction. Ditto for the way of drawing a carbene.
As for the convention, the one accepted by your tutor seems to differ from the one I'm familiar with. This is not much of a problem, as long as you remember when to use which. So when you are in your class, use the plus/minus form; when you are elsewhere, use two dots, and you'll be understood just fine, Google be my witness.

"Ы" is just a Russian way of typing "lol".

In Germany, I would argue that the most common convention is to write a line to symbolise two electrons — which would only be correct for singlet carbenes. (Rather than two dots, an electron pair is drawn as a straight line in all cases.)

BTW, the triplet/singlet question is another thing entirely. When you discuss it, you are typically going to draw phallomorphic appendages pierced by two arrows.

While I agree with your response, I added a slightly more .. er, non-judgemental alternative. :-)


text:
Stereochemistry of eliminations similar to E2' elimination
In the E2 reaction, the proton and the leaving group are anti-periplanar typically, so as to attain best overlap of the C-H bonding orbital and C-LG anti-bonding orbital.
In a similar reaction, with a hydrogen alpha to a C=C and a leaving group alpha the other side, we have a syn-elimination, which can be explained by a tau-bond model. Eg:
Eliminates to give:
Demonstrating that the groups eliminate from the same side in a so called E2' reaction.
My question is what happens to the required stereo-chemistry as a we increased the number of alkenes units in between - does the stereo-chemistry flip-flop, or do we get a fixed stereo-chemistry requirement. Or is the stereochemistry requirment, as I would theorise, relaxed as we increased the number of alkenes?

Unfortunately, allylic strain probably determines the product here. Alignment is less important since it would need to be mediated by a $\pi$ bond that can't overlap perfectly with the breaking $\sigma$ bonds anyway.


text:
Molarity of silver(I) nitrate solution after the reaction with magnesium

$\pu{3 g}$ of $\ce{Mg}$ are placed in $\pu{500 mL}$ of $\pu{0.625 M}$ $\ce{AgNO3}$. When the reaction is complete, what is the molarity of the $\ce{AgNO3}$ solution?

I'm thinking you need to use $M_1V_1=M_2V_2$, but I'm stuck on how to use it.

Try to think of what the definitions of M and V are, what do the subscripts mean? Units are always a good indicator.

Whenever possible, start with writing down equation(s) for the chemical reactions taking place in the system and don't use random formulas. Magnesium as a more active metal is going to reduce silver in the solution, thus lowering its molarity:
$$\ce{\overset{0}{Mg} (s) + 2\overset{+1}{Ag}NO3 (aq) -> \overset{+2}{Mg}(NO3)2 (aq) + \overset{0}{Ag} (s)}$$
Final concentration $c_2(\ce{AgNO3})$ is pretty much defined by the remaining silver(I) nitrate when magnesium is depleted in the reaction:
$$c_2(\ce{AgNO3}) = c_1(\ce{AgNO3}) - \Delta c(\ce{AgNO3})\label{eq:1}\tag{1}$$
$c_1(\ce{AgNO3})$ is the known initial concentration; $V$ is the volume; $\Delta c(\ce{AgNO3})$ is the change in concentration:
$$\Delta c(\ce{AgNO3}) = \frac{\Delta n(\ce{AgNO3})}{V}\label{eq:2}\tag{2}$$
where the unknown amount of silver nitrate $\Delta n(\ce{AgNO3})$ can be found knowing stoichiometry of the reaction (assuming the reaction is complete and is irreversible), mass and molar mass of magnesium $m(\ce{Mg})$ and $M(\ce{Mg})$, respectively:
$$\Delta n(\ce{AgNO3}) = 2n(\ce{Mg}) = \frac{2m(\ce{Mg})}{M(\ce{Mg})}\label{eq:3}\tag{3}$$
At this point we can rewrite \eqref{eq:1} using \eqref{eq:2} and \eqref{eq:3} since all the variables are known:
$$c_2(\ce{AgNO3}) = c_1(\ce{AgNO3}) - \frac{2m(\ce{Mg})}{VM(\ce{Mg})} = \pu{0.625 M} - \frac{2\cdot\pu{3 g}}{\pu{0.500 L}\cdot\pu{24 g mol-1}} = \pu{0.125 M}$$


text:
glycine hydrochloride salt how I can get it
I do not understand what Glycine•HCl salt is (concretely). Is it a mix between glycine and HCl salt (I don't know if HCl salt is existing...) or something else ?
I understand that glycine is an amino acid and in solution glycine HCl is like the following picture

But what happens when the solvent evaporate? What is the meaning of the point • between Glycine and HCl ?

Glycine•HCl salt is an ionic salt, much like NaCl. It is one single compound and not a mix of any sort.

When the solvent evaporates you are left with solid glycine hydrochloride

Thank you for your both answers. I think it's cear now. It was strange to figure HCl like salt... Thx

Just a comment on nomenclature. The nature of glycine hydrochloride would be much clearer if its more systematic name (and according to the real contents of the salt), glycinium chloride, was used.

In general, the dot in the formula indicates some kind of association compound; the prime examples of such dots are crystal hydrates such as $\ce{CuSO4 . 5 H2O}$. In that, five molecules of water are present for each pair of copper(II) and sulphate ions.
In organic chemistry, these dots are usually used in a somewhat misleading way including the somewhat nonintuitive description as a ‘hydrochloride salt’. What this means is that the most basic position of the molecule is protonated and this cation forms a salt with a chloride anion. The effect is equivalent to adding hydrochloric acid to a solution of the molecule and then removing the solvent. In glycine’s case, while the compound is often called glycine hydrochloride salt its systematic name would be glycinium chloride; indicating both the fact that the proton is somewhere on the glycine molecule (where depends on the original solvent) and an additional chloride ion forms a salt with this cation.


text:
Is glycine strong or weak field ligand? If yes, how?
The fact that it contains both oxygen and nitrogen atoms is confusing me since N is generally strong ligand field while O has a weak ligand field

A man can be a strong sword fighter if he uses his right hand, and at the same time a poor fighter with his left hand (or vice versa). Same thing here.

@IvanNeretin Except that we can’t really classify either atom as strong or weak field …


text:
Why does pH change with temperature?
Why does pH change with temperature?  I recently read up on some chemistry notes, and found out that the higher the temperature of distilled water, the lower the pH.  Why?  Does this apply to other fluids too?  Does this also mean dipping a litmus paper into 2 beakers of water of different temperature would yield different results?

The equilibrium  $\ce{H2O <-> H+ + OH-}$ changes with temperature and hence so does the pH. As to 'other fluids', they may not dissociate at all in any meaningful sense, e..g benzene. (The equilibrium constant for any reaction changes with temperature to a greater of lesser extent, look up the Van't Hoff isochore)

Water always contains a certain amount of $\ce{OH-}$ and $\ce{H3O+}$ ions. The pH is the negative decadic logarithm of the $\ce{H3O+}$ concentration (pH = - log(c($\ce{H3O+}$)). As chemical equilibria might change with temperature so does the concentration of $\ce{H3O+}$ and therefore the pH.
Although I never tried, I suppose this could affect measurements using pH paper. On the other hand the pH changes might very well be to small to be detectable using this method.
How this affects other fluids / equilibria strongly depends on how each equilibrium is affected by temperature. At higher temperatures equilibria are shifted in the endothermic direction and at lower temperatures they are shifted in the exothermic direction. You will need to look up the details for each equilibrium I suppose.

What about for other substances?

Hmm... that is very very brief

@XcoderX better?

...the higher the temperature of distilled water, the lower the pH. Why?

$$\ce{2H2O(l)<=> H3O+(aq) + OH-(aq)}$$ ΔH = +ve
The autoionization of water is endothermic therefore the increase in temperature will cause the equilibrium to shift to the endothermic direction, to oppose the increase in temperature.
This increases the amount of H3O+ ions in the system.
since pH=-log [H3O] this decreases the pH.

Does this also mean dipping a litmus paper into 2 beakers of water of different temperature would yield different results?

Yes it does, if you have very good litmus paper and a sharp eye!, if one beaker was 273K (pH 7.47) and another at 373K (pH 6.14) the difference in pH would be 1.33.
For this very reason, a worker would have to take temperature into account when doing pH measurements.

A common oversight is when you take a sample from a process tank and make the pH measurement in the laboratory. At that time, you are probably not measuring pH at the same temperature as the temperature in the process tank, this means that you will not have the correct pH value for the sample. Thus a pH value without a temperature value is meaningless. This can be simply overcome by testing pH on site and at the source of the sample.


text:
Decarboxylation of alpha-amino acids to amines
In this reaction, what are the ideal circumstances?

Which base would be a great choice?
How much time does it take for tryptophan to fully decarboxylate?
What is the ideal temperature for this type of reaction?

Just heating with base is ineffective.
Laval and Golding [1] describe a general procedure for the decarboxylation of α-amino acids using N-bromosuccinimide and Ni(0) in aqueous solution. The yields are pretty good.
Reference

Laval, G.; Golding, B. T. One-Pot Sequence for the Decarboxylation of α-Amino Acids. Synlett 2003, No. 4, 0542–0546. DOI: 10.1055/s-2003-37512.

Thank you, btw heating with base why doesn't effective? What would happen to tryptophan?

I think you would just get a very hot soln of the sodium salt of tryptophan.

I added a complete reference; however, it doesn't look like authors mention Ni(0) at all, instead they used nickel(II) dichloride hexahydrate. IMO it would be beneficial for others if there were a bit more details from the paper itself, including reaction schemes and the reported yields (authors give exact numbers for each reaction, not just "pretty good"). Otherwise it leaves an impression that the paper haven't been read beyond the abstract.

The authors use Nickel(II) Chloride plus sodium borohydride which generates Ni(0) in situ

The reaction of Laval is two steps. NBS effects oxidative decarboxylation and oxidation of the resultant imine to a nitrile. NiCl2/NaBH4  effects reduction of the nitrile to an amine.


text:
Calculation of total pressure given ratio of pressure equilibrium constants

The equilibrium constants $K_{\mathrm{p},1}$ and $K_{\mathrm{p},2}$ for the reactions $\ce{X(g) <=> 2 Y(g)}$ and $\ce{Z(g) <=> P(g) + Q(g)}$, respectively, are in the ratio of $1:9$. If the degree of dissociation of $\ce{X}$ and $\ce{Z}$ are equal then the ratio of total pressure at these equilibria are?

My Approach: Let the degree of dissociation be $\alpha$
So $K_\mathrm{c,1}$ for the first reaction is
$K_\mathrm{c,1} = \frac{(2\alpha)^2}{1+\alpha}$
Similarly, $Kc_2$ for the second reaction is $K_\mathrm{c,2} = \frac{(\alpha)(\alpha)}{1+\alpha}$
I don't know how to proceed from here and relate these to the ratio of $K_{\mathrm{p},1}$ and $K_{\mathrm{p},2}$. I need help here.

What are the states of matter in those reactions?

Nothing is given. I assumed them to be gas.

The equilibrium expressions for these reactions are:
$$ K_{\text{p},1} = \frac{p_{\text{Y}}^2}{p_{\text{X}}}$$
$$ K_{\text{p},2} = \frac{p_{\text{P}}p_{\text{Q}}}{p_{\text{Z}}}$$
Since $p_\text{A} = x_\text{A}p_\text{T}$ for component $\text{A}$, we can rewrite them as:
$$ K_{\text{p},1} = \frac{(x_\text{Y}p_{\text{T},1})^2} {x_\text{X}p_{\text{T},1}} = \frac{x_{\text{Y}}^2p_{\text{T},1}}{x_\text{X}}$$
$$ K_{\text{p},2} = \frac{(x_\text{P}p_{\text{T},2})(x_\text{Q}p_{\text{T},2})}{x_\text{Z}p_{\text{T},2}} = \frac{x_{\text{P}}x_{\text{Q}}p_{\text{T},2}}{x_\text{Z}}$$
Taking the ratios:
$$ \frac{ K_{\text{p},1} }{K_{\text{p},2}}
 = \frac{x_{\text{Y}}^2x_\text{Z}}{x_\text{X}x_\text{P}x_\text{Q}}\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}\label{eqn}\tag{1}$$
We're asked the ratios of the total pressures, and we're given the ratios of the equilibrium constants, so we just need to find the molar fractions ratio in that equation. We know that the degree of dissociation, let's call it $\alpha$, is the same for both reactions.
Assume the first equilibrium starts with $m$ moles of $\text{X}$:
$$\begin{array}{cccc}
  &\ce{X(g) &<=> &2 Y(g)}\\ \hline
\text{I} &m & &0\\
\text{C} &-\alpha m & &+2 \alpha m\\
\text{E} &m - \alpha m & & 2 \alpha m
\end{array}
$$
Then, the molar fractions are:
$$x_\text{X} = \frac{n_\text{X}}{n_\text{T}} = \frac{m - \alpha m}{m - \alpha m + 2 \alpha m} = \frac{1 - \alpha}{1 + \alpha}$$
$$x_\text{Y} = \frac{n_\text{Y}}{n_\text{T}} = \frac{2\alpha m}{m - \alpha m + 2 \alpha m} = \frac{2\alpha}{1 + \alpha}$$
Similarly, for the second equilibrium, assuming we start with $r$ moles of $\text{Z}$:
$$\begin{array}{cccccc}
  &\ce{Z(g) &<=> & P(g) &+ &Q(g)}\\ \hline
\text{I} &r &    &0     &  & 0\\
\text{C} &-\alpha r &    &+\alpha r     &  &+\alpha r\\
\text{E} &r - \alpha r &    &\alpha r     &  & \alpha r
\end{array}
$$
And we calculate the mole fractions of each species:
$$x_\text{Z} = \frac{n_\text{Z}}{n_\text{T}} = \frac{r - \alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{1 - \alpha}{1 + \alpha}$$
$$x_\text{P} = \frac{n_\text{P}}{n_\text{T}} = \frac{\alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{\alpha}{1 + \alpha}$$
$$x_\text{Q} = \frac{n_\text{Q}}{n_\text{T}} = \frac{\alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{\alpha}{1 + \alpha}$$
Plugging the found molar fractions into $\ref{eqn}$, we have:
$$\frac{ K_{\text{p},1} }{K_{\text{p},2}}
 = \frac{\left(\frac{2\alpha}{1 + \alpha}\right)^2 \left(\frac{1 - \alpha}{1 + \alpha}\right)}{\left(\frac{1 - \alpha}{1 + \alpha}\right)\left(\frac{\alpha}{1 + \alpha}\right)\left(\frac{\alpha}{1 + \alpha}\right)}\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$
Which simplifies to:
$$\frac{ K_{\text{p},1} }{K_{\text{p},2}}
 = 4\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$
Since $\displaystyle \frac{ K_{\text{p},1} }{K_{\text{p},2}} = \frac{1}{9}$, we get:
$$\frac{1}{9}
 = 4\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$
And finally:
$$\frac{p_{\text{T},1}}{p_{\text{T},2}} = \frac{1}{36}$$
So the total pressures are on a ratio of $1:36$.


text:
How is the number of successful collisions in a time-step distributed?
In a spatial homogeneous reaction system with two different molecules, how is the number of collisions which will lead to a reaction in a fixed time-step distributed?
My guess: Poisson distribution, as collisions are Bernoulli-experiments, usually have a low probability for success and the number of trials are huge.

It's OK to use a Gaussian approximation so long as the time step is not too small. The Gaussian noise is a small perturbation on the dynamics of the reaction. But it can be important to consider in microtubes or cells of living things.

I want a very detailed description, so I need a discrete distribution. I also do not know how to calculate the variance, if I would use a Gaussian. A Poisson would solve that because I know the mean number of successful collisions. Is it reasonable to use a Poisson?

I think so. But in case the result does not agree with experiment or expectation, you still can allow the variance to be an adjustable parameter. The Poisson distribution gives you a guideline of how big the variance can be. Going beyond the Poisson theory would be difficult.

With "how big it can be" you do not mean an that Poisson variance is an upper limit but an estimation, right?

Right. Let's say the true std should be within an order of magnitude from the Poisson std ($\mathcal{O}(\sqrt{N})$).


text:
Convert the lowest IR wavelength to energy in J, frequency in Hz, and wavenumbers in cm-1
Look up the range of wavelengths in nm in the IR region of the electromagnetic spectrum. Convert the lowest IR wavelength to energy in J, frequency in Hz, and wavenumbers in cm-1.
I think I’m going at this the wrong way but if anyone can explain or help with steps that would be really helpful.
$$
E = \frac{(6.676 \times 10^{34})(\pu{3.0 \times 10^8 m/s})}{(\pu{700 \times 10^{-9} nm})} = \pu{0.28 \times 10^{-19} J}
$$
$$
V = \frac{\pu{0.28 \times 10^{-19} J}}{6.626 \times 10^{-34}}= 4.28 \times 10^{13}
$$

Well, you took $\lambda=700nm$ which is correct.

if anyone can explain or help with step

There isn't really any explanation involved here. The question is very straightforward. Recall these three formulae from your chemistry lectures:
$$\text{Energy}=hv=\frac{hc}{\lambda}$$
and$$\text{wave number}=1/\lambda$$
and $$\text{frequency}=c/\lambda$$
Substitute the necessary values of the constants in them, and now you are good to go!
PS:  $1nm=10^{-9}m\neq10^{-9}nm$! Be careful with how you inter convert the units.
Hope it helps!


text:
Order of acidity of water, an alcohol, an amine, and acetylene

What is the order of acidity of $\ce{H2O}$, $\ce{ROH}$, $\ce{RNH2}$, $\ce{C2H2}$?

I began by forming the conjugate base of each. The conjugate bases are as follows $\ce{-OH}$, $\ce{-RO}$, $\ce{-RNH}$, $\ce{-C2H}$.
Between stability of $\ce{-OH}$ and $\ce{-ROH}$, $\ce{-ROH}$ is more unstable. Also, in $\ce{-RNH}$, the negative charge is on nitrogen atom making it more unstable than the other two groups.
In $\ce{-C2H}$ the pi elctron cloud will face repulsion of with the negative charge on the carbon atom.
So by this logic, decreasing order of acidity should be:
$$\ce{H2O > ROH > RNH2 > C2H2}$$
But the $\mathrm{p}K_\mathrm{a}$ values suggest the very reverse order. What is the fault in my reasoning?

If you just start with being able to rationalize the difference in acidity between ammonia and water, you're more than half way to the answer.

@Zhe Sois my answer correct

Unless I'm mistaken, acetylene should be more acidic than a primary amine.

According to
$$
\begin{array}{lc}
\hline
\text{Compound} & \mathrm{p}K_\mathrm{a} \\
\hline
\text{Water} & 15.7 \\
\text{Methanol} & 16.0 \\
\text{Methylamine} & 36.0 \\
\text{Ethyne} & 25 \\
\hline
\end{array}
$$
the order of acidity is
$$\text{water} > \text{methanol} > \text{ethyne} > \text{methylamine}$$
You logic is absolutely correct, but only there is an exception in case of electronegativities of $\mathrm{sp}$ carbon and $\mathrm{sp^3}$ nirogen.
The $\mathrm{sp}$ carbon is more electronegative than $\mathrm{sp^3}$ nitrogen and hence ethyne is more acidic here. The $\mathrm{p}K_\mathrm{a}$ values also tell you the same story.

pKa values are -logKa.
Ka is the dissociation constant of the acid
Smaller the Ka value, weaker is the acid.
Therefore, pka values will go on decreasing from strong acid to weak acid.

I think everybody knows that

@starunique2016 I wouldn't assume that. But I agree that this may not answer the OP's question. It's not clear to me what they're confused about...

I don't know the pka values of the given compounds, and I'm assuming that the person asking the question has got his Ka and Pka mixed up, which is why I wrote the answer.

Yep but certainly this is not an answer with definite reasoning.

I can't post an answer with definite reasoning, because I don't know the Pka values, and so, I can't examine the trends in them. I think the OP's reasoning behind the acidity order is right, so the only confusion is in the Pka values. I can't comment due to lack of reputation, otherwise I would've done that

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


text:
Galvanic/Voltaic Cell: Does Cell Voltage Change When A Resistor Is Connected?
I tried making a lead acid battery. After charging for about 10 minutes, I connect the voltmeter in parallel to the cell and see that there is around a 1.2V (which is quite lower than what I predicted). However, when connecting to a lightbulb, the voltage seems to be smaller. Is this supposed to happen? Shouldn't the voltage be the same with the cell providing an appropriate current to match the resistance of the bulb?

This is how is supposed to be. For instance refer to https://electronics.stackexchange.com/questions/190857/confusion-about-why-voltage-of-a-cell-decreases-as-its-used-up-and-how-it-relat

@Alchimista So does that mean if I put a lightbulb with a higher resistance, the voltage would be even lower because a higher current will be drawn out and therefore more voltage drop due to the internal resistance of the cell?

No . More resistance means less current is drawn. To the limit of unconnected electrodes when current is zero and V is at maximum (Voc).

"voltage seems to be smaller" => you mean it is measured smaller by the voltmeter, or the bulb's brightness appears to you less?

I mean that the voltage reading on the voltmeter decreases

Yes, this happens because there is an internal resistance within the battery. Look up how batteries are modeled within circuits.


text:
Ice Crystal Matrix measurements of H2O molecules?
I want a mathematically generated 3D STL copy of this image, however it's represent algorithmically.
If I do all the H2O's as individual pieces with female connections prepared for chopped bike spokes, are all the H2O's the same and what angles do they have?
Is there a written array of positions and degrees of rotation of the water molecules of a regular ice matrix?

http://www.crystallography.net/cod/1011023.html

Matrix transformations for space group $P\mathrm{6_3cm}$ (# 185).

What do you need STL for? You won't be able to 3D-print that assembly of water molecules anyway.

I may print 3 atom H2O's molecules individually, including female sockets for bike spoke segments representing hydrogen polar links, then would the H2O's be all the same 3d model ? else i'd print it all as one using small links rather than bike spokes.

I have absolutely no experience with 3D printing, but I guess one can convert CIF file (first link) to XYZ format (e.g. with Mercury), then feed that XYZ to Blender, do fine tuning with 3D modeling there, and finally export to STL. If you really want to do it hard way, then all the necessary matrix operations are described by $P\mathrm{6_3cm}$ group itself so that the structure could be replicated.

I decided to summarize all my comments in a form of answer and add some illustrations. Note that I have no experience in 3D printing, so I mainly focus on crystallographic part.
Since you asked for"a mathematically generated" 3D structure, you ought to know that water crystallizes in $P\mathrm{6_3cm}$ (# 185) space group having a hexagonal symmetry. Detailed information about symmetry generators and matrix transformations can be found online or in International Tables for Crystallography [1, pp. 582-583]:

There is already enough information to create a 3D model. If you want a more convenient way, you can get a CIF file for ice (COD-1011023) (which already embeds all the information above), and load it with Mercury (free, available on Windows, Linux, MacOS). From here on it's just a screencast of what's to click to get the desired molecular pattern like the one on your animated GIF:

Load the structure (drug-n-drop CIF file, or via Ctrl+O). What you see now is an asymmetric unit, in simple terms – a seed which an entire crystal structure can be grown from:

Let's expand the structure beyond those 6 atoms you've stuck with. Go to Calculate > Packing/Slicing.... Tick Pack option, and click + 0.5 boxes next to $a$ and $b$ axis:

Rotate the grown structure approximately like it's shown below:

Note that you have 4 water molecules on each side that you don't need. Those can be deleted by right-clicking on them and selecting Delete this molecule:

At this point you should have exactly the same 3D representation of the crystal structure:

Now you can print a 3D model directly from Mercury (File > Print in 3D...). The one with supporting framework looks like this:

6.1. Alternatively, you can save the structure as XYZ (Ctrl+S, choose Xmol files) and do some post-processing work in Blender if needed.

Reference

International Tables for Crystallography: Space-group symmetry, 1st ed.; Hahn, T., Ed.; Fuess, H., Hahn, T., Wondratschek, H., Müller, U., Shmueli, U., Prince, E., Authier, A., Kopský, V., Litvin, D. B., Rossmann, M. G., et al., Series Eds.; International Union of Crystallography: Chester, England, 2006; Vol. A.

very awesome information, i'll complete that and i'll have send an h20 ice kit in the comments. 3D printers are advancing nicely they are about 150 for smaller ones, it's a fun hobby.

You can produce STL (and VRML) files directly from Mercury - see https://www.ccdc.cam.ac.uk/Community/blog/post-56/
I'm not sure how best you'd go about including female sockets for bike spoke segments, but be aware that for some 3D printing types you're charged for the total volume of material used, so you might be better off trying to create a model with multiple molecules in a smaller area with less empty space.

Thanks, Mercury is very cool. I can't find a good ice file, I found a .cif file with 6 atoms only. How can I find ice/H2O from the cabridge/other databases?


text:
Classical Least Squares in UV-Vis Absorbance Experiment
How do I find the minimum number of standards needed to calibrate a spectrophotometer using CLS? We never talked about CLS in class so I'm not sure of my approach. Would 1 standard solution per analyte be enough? We'll be plotting absorbance vs. concentration then finding the residuals from the best fit line.

That depends on the nature of the analyte and matrix. Ideally (Beer's Law) A should be linearly proportional to concentration but that may not be the case. As an example of this in the water industry barium chloride is often added to samples to determine how much sulfate ion they contain. Spectrophotometers are used to determine how much barium sulfate is in suspension. The relationship between A and sulfate concentration is not linear in this case.
What you should do is make up a series of standards spanning the range of concentrations you are interested in, measure the absorptions and plot A vs. c. Now start fitting curves, starting with linear and progressing to higher order until doing so does not make the residuals look any noisier. That gives you the order of the response curve you need and the number of standards required is one more than the order.
Ideally, the matrix wont interfere and Beer's law applies. In that case when c = 0, A = 0 and you only need one standard.


text:
Sodium Chlorite vs. Chlorine Dioxide
I ordered Sodium Chlorite in the mail and received Chlorine Dioxide. Can they be used interchangeably, or are they different? My background in chemistry is limited so laymen's terms would be appreciated. I appreciate any responses or feedback.

https://en.wikipedia.org/wiki/Chlorine_dioxide I find it hard to believe someone delivered ClO2 It would be very dangerous, especially if you don't know much about it.

If you are 100 % sure it's ClO2 you received, then claim for refund and blacklist this supplier. If they cannot tell a difference between sodium chlorite and chlorine dioxide, they are not worth your time and money.

If you have no background in chemistry, you should not be handling either. https://www.fda.gov/news-events/press-announcements/fda-warns-consumers-about-dangerous-and-potentially-life-threatening-side-effects-miracle-mineral

How is it possible to make such a mistake ? Sodium chlorite is a white solid, looking like salt. Chlorine dioxyde is a reddish gas. A gas !!

Sodium Chlorite is a white powder, which dissolves in water.
Chlorine Dioxide is a gas. This gas also dissolves in water to some degree and gives a yellow color.
Sodium Chlorite is used to make Chlorine Dioxide in the laboratory.
I strongly doubt, that someone has sent you Chlorine Dioxide, as it is to dangerous to store and to send.

Chlorine dioxide also tends to explode.

Warning: Both sodium chlorite and chlorine dioxide are toxic. There are numerous reports of nausea, diarrhea, severe vomiting, and life-threatening low blood pressure caused by dehydration as a result of chlorite poisoning. Sodium chlorite is a strong oxidizing agent and upon slight contamination would lead to explosion.

Chlorine dioxide is the product formed from the action of chlorine on sodium chlorite (laboratory process of making chlorine dioxide). This reaction is relatively safe. The reverse reaction of converting chlorine dioxide to sodium chlorite is dangerous and I highly recommend against it. This reaction is a highly oxidizing one and may ignite or explode if something goes wrong.
Chlorine dioxide can be revert back to sodium chlorite by reacting it with sodium peroxide.
$$\ce{Na2O2 + 2ClO2 → O2 + 2NaClO2}$$
But sodium peroxide can enhance combustion if contact with any combustible substance. Also, it causes redness of skin and pain. Don't try this.
Other preparation include reacting with sodium carbonate or sodium hydroxide but it tends to form a mixture of sodium chlorite and sodium chlorate.
So my opinion would be to return the chlorine dioxide. Don't try to store it or convert it to sodium chlorite as the reactions are dangerous.

simply marketing.
OP didn't receive Chlorine Dioxide gas, likely didn't receive Chlorine Dioxide solution of any concentration, and either received technical grade sodium chlorite flakes 'marketed' as chlorine dioxide which was misleading, or a solution of sodium chlorite labeled ACD/SCD "chlorine dioxide" which while industry practice, confused OP.
"Acidified Chlorine Dioxide" or "Stabilized Chlorine Dioxide" are both industry terms for a Sodium Chlorite solution. See Closys mouthwash ingredients as an example:

Purified water, stabilized chlorine dioxide, trisodium phosphate, etc.

I would assume OP received a solution. If you received the powder though, my answer is the same. The seller is marketing the end product you would use the chlorite salt to produce. Nevertheless you should be able to ask for the MSDS/SDS[^0] and inquire as to the concentration. Sodium Chlorite (NaClO2) is generally sold at 80% technical grade.
some of the newer Closys competitors utilize a 2 bottle system where you mix two solutions, usually one consisting of sodium chlorite and another of phosphoric acid yielding their proprietary concentration of Chlorine Dioxide with leftover TSP[^1] (same as in Closys). The ClO2 is usually in the range of 20-50ppm. There are many such products like this now.
As for topics brought up in other answers/comments. Chlorite is an oxidant, so is Chlorine Dioxide. Chlorine Dioxide is also a "free radical." Other common oxidants include Sunlight, Ozone (O3) Hydrogen peroxide (H202), and Nitric Oxide (NO), the latter also being a free radical.
Chlorine Dioxide is a more selective oxidant than Hydrogen Peroxide which one could argue makes it safer, and at appropriate concentration solutions (0.3%/3000ppm) is nearly as easily stored and used.[^2] (air-sealed container in the fridge)
If the Chlorine Dioxide gas is red you are in danger. Red would be highly concentrated. the color ranges from invisible to yellow, green, brown, red. The appropriate storage concentration of 3000ppm in solution is a neon yellow which could indeed be shipped with proper precautions taken.
all you might want to know about chlorite and chlorine dioxide. Both are used regularly and safely in water treatment, food industry, medicine
I was going to make this a comment, but I don't have the rep for that. So here's an "answer" instead.
[^0]: "(M)SDS: (Material) Safety Data Sheet - chemical information on the product. when you ask for this the provider may also give information on the process used to derive the product and what concentration or grade it is."
[^1]: TSP: Trisodium Phosphate - a rinsing agent which used to be in dishwashing detergent (still is in the commercial stuff) and is commonly sold at hardware stores for painters who use it to wash/rinse walls before painting.
[^2]: Chlorine dioxide and chlorite solutions are used for wound healing (just like H2O2 used to be commonly used) in many countries both in medical and veterenary practice
hm, I see this markdown doesn't do footnotes.

¯_(ツ)_/¯

sunlight is an oxidant? Agreed it is capable of ejecting electrons but has some trouble in receiving them. You just invented photonium, get thee to the patent office!! You neglected a common oxidant, the one the concept is named for, oxygen, O2, also a free diradical.  ClO2 is not particularly selective but more brute force; it can react by both radical and ionic mechanisms and oxidizes almost anything since in the process almost every lower oxidation state of chlorine is available.

thanks for the excellent input @jimchmst. I meant to leave the obvious O2 off the list actually.
By 'selective' I meant to imply ClO2's limited reactivity to organic compounds as compared to chlorine. perhaps I can word that differently somehow.
and feel free to edit the answer as you see fit, my friend. =D

ClO2 is reasonably inert to alkyl groups we found that ppm solutions with sodium lauryl sulfate were reasonably stable, less pure surfactants reacted. It does not seem to be readily activated by room lighting.


text:
How to remove nitrose gasses from ozone gas mixture?
I produce ozone gas with an ozone generator based on corona discharge principle.
These kind of ozone generators have the disadvantage that if $\ce{N2}$ is present, a bit of $\ce{NO}$ and $\ce{NO2}$ will be produced too.
To produce oxygen I use a oxygen concentrator, with an output of about $90\%$ $\ce{O2}$ and about $10\%$ $\ce{N2}$.
For that reason I have a little quantity of $\ce{N2}$ entering the ozone generator, which subsequently lead to unwanted side reactions like $$\ce{N2 -> NO -> NO2}.$$
The $\ce{NO2}$, even in that small quantity as it is, does influence the following processes in an negative way.
The easiest way to circumnavigate this issue would be to take pure oxygen – however for cost reasons I cant use LOX, so I have to stay with the concentrator.
I am searching for an reagent or an catalyst, which is not effected by the ozone but does absorb or disintegrate the $\ce{NO}$, and especially the $\ce{NO2}$.
To stripe $\ce{NO2}$ from a normal gas is not so difficult, but the ozone does make trouble.
I thought about dry $\ce{NaOH}$, soda lime, or dry $\ce{KOH}$, but I have the feeling that these chemicals will react with $\ce{O3}$.
Any suggestions?

I have a feeling they won't.

Well it is difficult to find information about it. I have found that KOH will react with O3 to KO3 at room temperature. Don't know about NaOH - I guess I will have to try it.

@Andreas why don't you make the experiment by yourself to find how much of them react or not and then show us what you found? :-)

that is exactly my plan for next weekend ;-)

There is no need to use caustic since $\ce{NO2}$ reacts with water itself. As per [1] it is even better absorbed in water than in $\ce{NaOH}$.
Ozone is soluble in water depending on the concentration of $\ce{O3}$ in the gas phase. It is much more soluble than oxygen so you will loose some, but it will have much lower concentration of $\ce{NO2}$. Of course, the washed ozone will be moist. If you need it dry, you can use a column of molecular sieves or some other desiccant.
[1] Absorption of Nitrogen Dioxide by Aqueous Solutions,
F. S. Chambers Jr., T. K. Sherwood,
Ind. Eng. Chem., 1937, 29 (12), pp 1415–1422


text:
Potassium citrate in water
I take potassium as a supplement in the form of potassium citrate. I dissolve a small amount in tap water and I drink it. Usually, I need to stir a lot, and sometimes I warm the water a little to make it easier to dissolve the potassium.
Recently, I bought some sparkling water (San Pellegrino, but I don't think that makes a difference) and when I put the potassium in it, it reacted making it bubble a little and it dissolved immediately.
Why is that? Did the potassium react with the CO2 or something else in the sparkling water?

It was influenced (at least to some extent) by physical effect: powders  make bubbles as nucleating agents and bubbling mixes the powder. BTW you use pure potassium citrate? I don't think it's such a good idea.

@Mithoron Why not? I use a pinch (like a pinch of salt) before going to bed.

Bubbles need nucleation points to form.
The growth of bubbles is limited by the surface tension of the liquid. The surface tension acts stronger on smaller bubbles and less strong on bigger bubbles.
Adding a solid salt to sparkling water means adding a lot of new nucleation points, which subsequently will lead to bubbling.
The small bubbles will combine to bigger bubbles, these bigger bubbles are not so much effected by surface tension and so even more gas is produced.
In other words, adding the salt will trigger a kind of avalanche effect, which weakens either if the dissolved CO2 gets towards used up or the salt gets dissolved completely.
It is not an reaction of dissolved potassium ions with the CO2, nor is it a reaction of dissolved citrate ions with the CO2.
What happens here is a physical reaction and not an chemical reaction.

so any salt would do? or even sugar?

Any solid with high surface area would do this


text:
How isotopes atomic mass was defined?
So we say that 1 Amu (u) = 1/12 of the mass of a Carbon 12 atom.
Im not sure about this, but I believe that first we measured  the mass of 1 carbon 12 atom (This includes the mass defects that occur during the formation of nucleus) and then defined that 1 atomic mass unit has a value of 1/12 of the carbon 12 atom mass... Therefore one Carbon 12 atom has exactly 12 atomic mass units.
Until here everything is fine.
Now my question is... To define all the other elements isotopes atomic mass did we first measured each isotope atom mass and then said " If 1/12 of the Carbon 12 atom is equal to X grams and this X grams are equal to 1 atomic unit the then my isotope atom which has Y grams must have Y atomic mass units " did we work it out this way or did we use calculation to define other isotopes mass based on carbon 12 atomic mass ?
Basically what I want to know is, do we need to have the value of an atoms mass in grams or kilograms before we know how many Amu´s it has, or can we calculate it ?

Yes we do need to measure each isotope of each element. I guess you've heard of mass spectrometry.

One other odd point. We know the mass of the individual isotopes much more accurately than we know the average atomic mass of the element. The problem is that the isotopic ratios have variation.

@IvanNeretin I wont say that I completely understand it but yes I´ve heard about it.  Thank you for making it clear ;)

Not entirely sure what you're getting at here. As far as I can tell the process of "calculate isotope mass from ratio with C12 amu" and "calculate isotope mass by first converting to grams" are mathematically equivalent. What do you expect to differ between the two?

Historically, the relative masses of the isotopes were measured relative to O=16.
There is a good historical explanation in A New Mass-Spectrograph and the Whole Number Rule (1927).
At this time (1927), though it was understood that many elements had more than one isotope, only one isotope of oxygen was known.


text:
Why the pH of xanthan gum in aqueous solution is 7 (neutral) when it has glucuronic and pyruvic acids in its chemical structure?
Since xanthan gum has both glucuronic and pyruvic acids in its chemical structure, it should release $\ce{H+}$ ions upon solvation in an aqueous solution and make it acidic. However, the meausured $\ce{pH}$ is $7$.

Pyruvic and glucuronic acids are both pretty strong acids (pK's of respectively 2.5 and 3.28) but that does not mean that the polymer of xanthan gum will be. Those protons are still there but they may be more tightly bound in the polymer than in the free acids. To find out try titrating a solution with strong base. Eventually you should start prying some of those protons off and you should see an inflection point. OTOH as there are so many carboxylic protons in the polymer the inflection points would probably smear and not be visible but you should see  some buffering above some pH if my hypothesis is correct.
I should make clear that this is purely speculation on my part but it sounds like an interesting experiment.

If this is the case (protons are tightly bound in the polymer), does it mean that xanthan gum is not negatively charged (since it does not release protons to become negatively charged)?

As another comment here points out the answer is no - at least not at netural or acidic pH. Raise the pH and the protons come off  leaving a net negative charge

Xanthan gum disperses difficultly in water; the dispersion process can be speeded up by raising the pH to about 10-11. This implies that acidic groups on the polymer are not much ionized, but addition of alkali puts a lot of negative charge on the polymer, helping it to unfold.
Measuring pH of gums, goos, slimes and muds is difficult, especially when the concentration of protons or hydroxide is low. The more xanthan gum you put into water, the goopier it gets; 1% is quite goopy. A measurement of 7 could be +/- 0.25 at least.


text:
Electrophilic addition to alkenes
Consider the addition of HBr (in absence of any radical initiators or environment that promotes radical mechanistic pathway), to 3-bromocyclopentene.

The first step is the attack of $\ce{H^+}$
On the exposed π electron cloud. But my query is whether it forms a classical carbocation like this one to oppose the negative inductive effects of the halogen atom:

Or does it proceed with the formation of bromonium ion like this one to partially complete its octet using lone pairs on Br.

These two would lead to different products but I'm interested in the major one.
The first one would lead to 1,3-dibromocyclohexane, while the second one would lead to 1,2-dibromocyclohexane.

Yeah, there should be bromonium here, but it's not what is called non-classical carbocation.

But sir I have been told that the inductive effect is so strong that it lead the the 1st pathway. But I feel it should be the one with bromonium ion. As in anchimeric effects during substitution reactions. Is there a strong justification for that?

Since it is electrophilic addition, it can go via 2 ways. Either by forming classical carbocation or cyclic carbocation. Since the electrophile will be H+ it will go with classical carbocation. Therefore product will be 1,3 dibromo cyclopentane. Also taken in account the mesomeric effect of bromine atom

Why would it go with the classical carbocation pathway? You have not given the explanation for that, please elaborate a bit.

And it's not a cyclic carbocation btw. It's bromine that's carrying the positive charge (if you are referring to this).


text:
Separation of solvent from residue of material.
I have a residue of 2kl in which 1.8kl is solvent (EDC) and rest material is residue. First i was thinking to go with normal distillation to recover our solvent. But i heared somewhere that there is some kind of RESIN by which if it got passes solvent will be pass only and residue will collect on another side of that resin. if there is any concept like this please help me.
or there is any other to do?

Just to make it crystal clear: are there 2 kiloliters of residue to process?

Dichlorethane has a boiling point of about 84°C - If I would have to separate EDC from a product / residue, I would use a rotovap with water bath configuration. This distillation is a very gentle way to separate solvents from product / residue.
Set the temperature of the water bath slightly higher than the boiling point of the EDC.
If you want to get fancy to make your solvent super pure again after the rotovap distillation, you could perform a fractional distillation on it - however in most cases I think this might not be necessary.

sir. if there is any way other than distillation. like any RESIN by which i can pass it through and it got separate.

That will depend on the natural of the residue. Is it acidic? Polar? Non-polar?

Distillation is the way to go. Especially if you have to do it on industrial scale (if we have understood well, you have 2000 L to process).
Normally it yields recovered solvent with low level of impurities. The operation is easy to perform and there are many installations available to do it.
Resins are not an option unless the only impurities in the solvent are acidic or basic, and preferably strong ones. Then you can use  ion exchange resins.
Otherwise the situation is just on the reverse side of what you want to do.
Since your impurities are soluble in an apolar solvent, they are apolar too.
There are resins for adsorbing these kinds of compounds, but then they must be in a polar solvent. Usually water is the option because even in alcohols the impurities do not get adsorbed. At least not completely.
If your impurities are not very apolar, they might be adsorbed on silica, but it is not likely that all of them will do so and probably none will do it completely.
Further, there is the economic aspect. Resins and silica are much more expensive than energy, so distillation is always preferred unless unfeasible. Moreover, for using adsorbents, the most efficient is to have them in a column with specific dimensions and a well designed bottom plate. Since the adsorption is an equilibrium, working in a column reduces the residual impurities much more. Unfortunately, these kind of installation are much less common than distillation units.
If you do not have such an installation yourself, it is quite easy to externalize the operation to some contract manufacturing unit.


text:
Changing Temperature of a Galvanic/Voltaic Cell: Should I wait for the electrodes to also change in temperature?
I'm conducting an experiment where I change the temperature of the sulfuric acid in a lead acid storage cell. My one question in controlling the variables is that should I let the electrodes also change in temperature, or should I keep them at room temperature and measure the cell potential? Does it even matter at all?
What I'm guessing is that a change in temperature affects the resistance of a metal, which would then affect the ability of the electrodes to transfer the electrons involved; therefore, the electrodes should be kept at room temperature.
If I should do one or the other, could someone explain to me why?

Andreas, thank you so much! I'm a student experimenting on how temperature will affect electrical power generated by a lemon battery. I have one question though. Do the ions of the electrolyte move more freely with the increase in temperature because kinetic energy increases alongside the temperature? Thank you so much...

In general the higher the temperature of the electrolyte, the more power you can draw from the cell.
The reason for this is, because the ions of the electrolyte can move more freely when the temperature is higher and chemical reactions do in general happen quicker with increased temperature.
On the other hand side the hotter a metal gets, the bigger its resistance will become. This is because electrons can’t move as freely when the temperature is increased.
However this effect usually is of importance when temperatures are quite high, like in an light bulb.
If you just heat the electrodes for 30-40°C, I doubt you would be able to even measure the difference in resistance (even though it does exist).
So to sum it up, the output power of a cell in relation to the temperature will mainly be altered by the temperature of the electrolyte.


text:
What is the correct IUPAC name of this compund?

I thought C=C is the "functional group" here and Br is a substituent so the numbering should start from the top left C, making the name 4-Bromo-3-methylbut-2-ene, but my book says 1-Bromo-2-methylbut-2-ene. Which is correct?

Overall smaller locators. Not sure as I studied nomenclature long time ago

According to P-14.4 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-14.4 Numbering
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
[...]
(e) saturation/unsaturation: [...]
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;

In this compound, unsaturation, in the form of the double bond, has the priority for low locants. Since both numbering schemes (shown below) assign the locant '2' to the double bond, criterion (e) is not sufficient for the purposes of deciding between the two options.

However, criterion (f) does allow for a choice to be made. The detachable prefixes in this name are 'bromo' and 'methyl'. Since the locant set (1,2) is lower than the locant set (3,4) (note that both sets are ordered in increasing numerical order as stipulated), the preferred IUPAC name (PIN) will use the locant set (1,2).
Note, however, that your double bond also contains configurational information. According to the Cahn–Ingold–Prelog rules (see P-92 of the Blue Book), the double bond in your molecule as drawn is (E)-configured. Stereodescriptors such as 'E' are cited together with the locant describing the position of the stereogenic unit (see P-91.3).
Hence, the PIN of the molecule as depicted in your question is (2E)-1-bromo-2-methylbut-2-ene.

Note that contrary to existing answers, there is no inherent preference between "bromo" and "methyl" for low locants, unless the application of criterion (f) does not lead to a difference. In that case, criterion (g) must be invoked. Since 'bromo' is earlier than 'methyl' in alphanumerical order, it will be the prefix cited first in the name, and will therefore have priority for a lower locant. For example:

You are correct that double bond gets preference over halogen. But after that, we seek to give small locant to bromo and methyl. Since in second name, both substituents get smaller locants, it is the correct name.
That is correct name is 1-Bromo-2-methylbut-2-ene
Had the question been this instead:

Then the name would have been 1-Bromo-3-methylbut-2-ene because in this case as orthocresol comments:

The name 1-bromo-3-methylbut-2-ene is preferred over 4-bromo-2-methylbut-2-ene because the locant set (1,3) is lower than the locant set (2,4). Alphabetical order only comes into play when the locant sets are the same.

The first part is correct now, but your second part is not quite right: the name 1-bromo-3-methylbut-2-ene is preferred over 4-bromo-2-methylbut-2-ene because the locant set (1,3) is lower than the locant set (2,4). Alphabetical order only comes into play when the locant sets are the same; I have an example in my answer.

I better stay away from chemistry.

Well, the IUPAC nomenclature system is complicated (the 2013 Blue Book has >1,500 pages) and is often taught based on outdated recommendations, or simply taught wrongly. I wouldn't fault you.

ortho - Ok so as I understand, this concept of  locant set means all numbers in set $(1,3)$ are lesser than corresponding set $(2,4)$ so there is no confustion. Am i right there

The numbers must be compared one by one from the first one onwards. Since 1 is smaller than 2, the locant set (1,3) is lower than (2,4). If one were to compare (1,3,5,7) with (1,3,5,8), then the first three numbers would be ties; however, since 7 is smaller than 8, (1,3,5,7) is lower than (1,3,5,8). Similarly (1,5) is lower than (2,3) because 1 is smaller than 2. You can see the link on your other deleted answer for some real life examples (I made up all the above).


text:
Alkane free radical halogenation

I do not understand how we obtain the product in option one.
According to Free radical stability order tertiary carbon radical is more stable than secondary carbon radical which is more stable than primary carbon radical. So hydrogen selectivity should also follow the same order. Then why is it that we get option(1) instead of option (2).
I’m only in high school so please try to keep it simple.

This question is based on relative rates of reaction of different $\ce{H}$ with halogens. In the free radical mechanism of chlorination of alkane, relative rate of reaction of $\ce{3^\circ : 2^\circ : 1^\circ}$ $\ce{H}$ is in order $5:3.8: 1$.
Using this, we get an idea about product amount. To get a rough estimate of relative amounts of different products, we multiply number of identical hydrogen and its rate of reaction with chlorine.
So there is only $1$ $3^\circ$ $\ce{H}$ and its rel. rate is $5$. So relative amount is $5$. For the $2^\circ$ $\ce{H}$, there are $2$ of them and their relative rate is $3.8$. So its relative amount is $2 \cdot 3.8 = 7.6$. Similar calculation for rest of the hydrogens will show you why given answer is correct.
Refer: Wikipedia

Them number of primary hydrogens is 9. So the relative amount should be 9*1=9. Then percentage of primary chlorine should be (9/21.5)*100= 41%. Percentage of secondary chlorine should be (7.5/21.6)*100= 35%. Then shouldn’t primary monosubstitued chlorine be the major compound?

Raheeel No here s a catch. There are two types of one degree  hydrogen present.

So I shouldn’t consider primary hydrogen on substituent carbon ie side chain. Am I right?

I am saying products 3 and 4 are not identical as hydrogen's replaced by chlorine in them are not identical

You take into account 3 and 4 as different product. Then find relative amount

Ok. So that means treat it as a completely different product. Which means product 4 will have a percentage of (3/21.6)*100= 13.8%. Am I right?

Rachel yes treat as completely different product. Btw product 3 will have that percent! Product 4 will have 6/21.6 of tot. fraction

Oh yes. Product 3. My bad. Thank you very much for clearing that up.


text:
Conductivity variation on Caustic (NaOH) Dilution System
I am reading the drawing for an automatic caustic dilution system where 50% NaOH is diluted down to 20%.
I was expecting to see a density sensor on the outlet to provide feedback to the dilution pumps for ratio-ing. i.e. To keep the density at 20% and prevent it from moving to say 19% or 21% by adjusting the amount of water.
However, the vendor seems to have used a conductity sensor on the outlet instead. I am skeptical whether the conductivity of NaOH will vary a lot between 19% to 21%. Isn't it already fully ionized?
Thoughts? Will this work?

Yes 19-21% solutions are fully ionized, but the relative amount of water is different so the conductivity changes.

Thanks. Any idea about what would be the percent variation in conductivity between say a 19% vs 20% solution?

Relax as for conductivity measurements allow for quantitative analysis of electrolytes. It will work

Relax as for conductivity measurements allow for quantitative analysis of electrolytes. It will work

As conductivity and concentration are linearly related over small regions I'd guess that in going from 19 to 20% the conductivity would increase by a factor of 20/19.


text:
What is the mass of (NH4)2CO3 need to add to 250 g water in order to lower the freezing point approximately 5.5 °C?
What is the mass of $\ce{(NH4)2CO3}$ need to add to $250\ \mathrm g$ water in order to lower the freezing point approximately $5.5\ \mathrm{^\circ C}$?
$K_\mathrm f=1.86\ \mathrm{^\circ C\ kg/mol}$
What would be the boiling point?
$K_\mathrm b=0.51\ \mathrm{^\circ C\ kg/mol}$
What is the osmotic pressure (the solution in the first question) in $14.5\ \mathrm{^\circ C}$?

The van 't Hoff factor  $i$ of $\ce{(NH4)2CO3}$ is $3$ in water. So $\Delta T_\mathrm f = ik_\mathrm fm$ can be used.
$$\Delta T_f = 3 \cdot 1.86 \cdot \frac{w}{0.25 \cdot M_{\ce{(NH4)2CO3}}} = \frac{3\cdot 1.86 \cdot w}{0.25\cdot 96} = 5.5$$
Solving gives $w = 23.66\ \mathrm g$.
Similarly you can use $\Delta T_\mathrm b = ik_\mathrm bm$ to get:
$$\Delta T_b = 3 \cdot 0.51 \cdot \frac{w}{0.25 \cdot M_{\ce{(NH4)2CO3}}} = \frac{\Delta T_f \cdot 0.51}{1.86} = 1.51$$
For osmotic pressure use $\Pi = C R T$ where $C$ is molarity and is equal to approximately $$C = \frac{w}{M_{\ce{(NH4)2CO3}}}\cdot \frac{1}{V}$$

Please do not omit the units – not only when answering homework questions; it is generally not permissible to omit the units.


text:
Which orbital lobes are occupied?
In the p-orbitals, there are 3 dumbbell-shaped lobes, each of which can contain 2 electrons. In the diagrams of the d- and f-orbitals, there are two or more lobes (or orbital shapes) or each value of m.  In all of these cases, which lobes/shapes do the electrons occupy for any given element?

The question implies a misunderstanding about orbitals and their representation as dumbbell-shaped lobes. These plots show the relative orientation of orbitals to each other by showing regions that represent some threshold probability (say, 90%) of finding an electron contained therein. The underlying probability distribution has the same symmetry as the orbital itself because it defines the orbital. Any (spatial) orbital can hold up to two electrons$^1$, regardless of its shape. It thus makes little sense to ask in which lobe (or, further, which half of the dumbbell, since there is a nodal plane in between the halves) an electron is.
Moreover, any single atom is spherically symmetric in the absence of external fields, meaning that the shapes of the orbitals are not actually important to understand electron distribution in a free atom. The shapes are useful tools to understand certain bonding concepts such as crystal/ligand field theory nevertheless.
$^1$ a spin orbital holds one.

doesn't answer the question, and kindly avoid insulting comments.

to clarify:  every question posed on every stackexchange site, and every lecture hall, implies some level of misunderstanding by the questioner.  Otherwise no need to ask a question.  Perhaps the answer you're looking for, albeit not necessarily correct, would be: "We cannot be certain in which of several possible orbitals any given electron resides, because of the XYZ Principle.


text:
Coordination isomerism variation
The school book and Wikipedia state that coordination isomers are:

In a coordination isomer the total ratio of ligand to metal remains the same, but the ligands attached to a specific metal ion change.
For example, a solution containing $\ce{[Co(NH3)6]^3+}$ and $\ce{[Cr(CN)6]^3−}$ is a coordination isomer with a solution containing $\ce{[Cr(NH3)6]^3+}$ and $\ce{[Co(CN)6]^3−}$.

So in this definition, consider $\ce{[Co(NH3)5(CN)][Cr(CN)5(NH3)]}$. Is it coordination isomer of $\ce{[Co(NH3)6][Cr(CN)6]}$ or not?
My teacher says yes, but I'm not quite sure according to Wikipedia.

Yes, it is.
In both the cases, the number of ligands remains same, which is according to the definition in Wikipedia.
For any type of structural isomerism, the premise is that the primary valence and secondary valence (as defined by Werner) of central metal/ ion remains same for all the isomers.
In this case, the primary valence is 3 and secondary valence is 6, which remains same for all isomers.
Other possible coordination isomers are:
$\ce{[Co(NH3)4(CN)2]+[Cr(NH3)2(CN)4]-}$
$\ce{[Cr(NH3)4(CN)2]+[Co(NH3)2(CN)4]-}$
$\ce{[Cr(NH3)5(CN)]^2+[Co(NH3)(CN)5]^2-}$
(cation is always written first followed by anion)
I don't think $\ce{[Co(NH3)3(CN)3][Cr(NH3)3(CN)3]}$ can exist because, now both parts become neutral.


text:
Why In This Reaction Acetic Acid is strong acid and NH3 is strong base ?please explain in details and thanks for answer

the acetic acid and amonia reaction why the reaction is towards completion and why nh3 and ch3cooh are strong

"Weak" and "strong" are relative terms.  In the example shown in the question, ammonium ion is stronger than the acid on the other side of the equation, which is just water.  So it looks like a strong acid.  But if you rendered the following reaction instead things appear different:
$\ce{H_3O^+}+\ce{NH_3}\leftrightharpoons\ce{H_2O}+\ce{NH_4^+}$
Now the ammonium ion is compared with a hydronium ion, which is a much stronger acid than either plain water or ammonium ion.  Suddenly ammonium ion looks like only a weak acid.
We really should be using comparative terms.  In the reaction between ammonia and plain water ammonium ion is a stronger acid than water.  But in the reaction between ammonia and a mineral acid, ammonium ion is a weaker acid than the hydronium ion.


text:
How to balance equation of alcohol oxidation with potassium permanganate?
I need to balance the equation for the oxidation of primary alcohol. The media is neutral to start out with, but the product will be basic. I assigned oxidation numbers and added stoichiometric coefficients accordingly:
$$\ce{3 R-\overset{-1}{C}H2OH(aq) + 4 \overset{+7}{Mn}O4^-(aq) -> 3 R-\overset{+3}{C}OO^-(aq) + 4 \overset{+4}{Mn}O2(s)}$$
The next step is balancing the charge(?): currently, net charge on the left is −4, and −3 on the right. From what I know, a reaction like the above will be balanced in a neutral media by adding $\ce{H+}$ or $\ce{OH-}$ on the right side of the equation. To balance that out, you will instead add $\ce{H2O}$ on the left side.
We, therefore, need to add $\ce{OH-}$ on the left side of the equation:
$$\ce{3 R-\overset{-1}{C}H2OH(aq) + 4 \overset{+7}{Mn}O4^-(aq) -> 3 R-\overset{+3}{C}OO^-(aq) + 4 \overset{+4}{Mn}O2(s) + OH^-(aq)}$$
Now that it is balanced, we add $\ce{H2O}$ to equal… oh wait, that's not possible. We are missing $\ce{H}$ on the right side, and we can't add more $\ce{OH-}$ since that would make the charges(?) not be balanced.
I need an advice on this with an explanation.

It might be worth invoking acid on one side and water on the other

Not sure what you mean here. The equation should be solvable as above. Anyhow, I figured it out on my own. I didn't know you could balance with OH- and H2O on the same side; which means in the above example the answar would be: 3R-CHOH-R + 2MnO4- → 3R-CO-R + 2MnO2 + 2OH- + 2H2O

3CH3CH2OH + 2MnO4- ---> 3CH3COOH + 2MnO2 + 4H+ + H2O

@TheSmartestNoo please consult [meta] for some information on how to apply chemistry markup with MathJax. We are using the mhchem extension here.

The equation proposed by deLange is interesting, but it is not an answer, because it requires $\ce{H+}$ ions to proceed, and Oliver, the question's author, states that the reaction should not occur in an acidic solution.
So the $\ce{H+}$ ions have to disappear from the deLange's equation, which I repeat here.
$$\ce{4H+ + 4 MnO4^- + 3 RCH2OH -> 4 MnO2 + 5 H2O + 3 RCOOH}$$ The only way of getting rid of these $\ce{H+}$ ions is to add $\ce{4 OH-}$ ions on both sides of deLange's final equation. This will transform the $\ce{4 H+}$ into $\ce{4 H2O}$ on the left-hand-side. On the right-hand side, this will transform $\ce{3 RCH2COOH}$ into $\ce{3 RCH2COO- + 3 H2O}$ and the fourth $\ce{OH-}$ ion will remain in the solution, which becomes basic. The obtained equation will be$$\ce{4 H2O + 4 MnO4^-  + 3 RCH2OH -> 4 MnO2  + 5 H2O +  3 RCOO- + 3 H2O + OH-}$$which can be shortened according to  $$\ce{4 MnO4^-  + 3 RCH2OH -> 4 MnO2  +  3 RCOO- + 4 H2O + OH-}$$If the initial solution was neutral, it becomes basic at the end.

The half reaction for the reduction of permanganate is
$\ce{3e- + 4H+ + MnO4^- -> MnO2 + 2H2O}$
The half reaction for the oxidation of an alcohol to an acid is
$\ce{RCH2OH + H2O -> RCOOH + 4e- + 4H+}$
Add $4$ times the reduction half reaction to $3$ times the oxidation
$\ce{16H+ + 4MnO4- + 3RCH2OH + 3H2O -> 4MnO2 +  8H2O + 3RCOOH + 12H+}$
Cancelling things that appear on both sides
$\ce{4H+ + 4MnO4- + 3RCH2OH  -> 4MnO2 + 5H2O + 3RCOOH}$
and Bob's your uncle. Writing it this way also shows that adding acid pushes it to the right, IOW this reaction should be carried out in acid solution.


text:
What type of polyester is most commonly used for clothing?
Many polymers have the name of the monomer in their names, e.g. polyethylene, polyvinyl chloride, polypropylene, polystyrene etc. But the ester in polyester refers to a group of chemicals (having an hydroxyl group replaced by an alkoxy group1) rather than a single chemical compound.
This article from Wikipedia states that polyethylene terephthalate (PET or PETE) is the most common type of polyester. PET is the plastic from which almost all plastic soda and beverage bottles are made. This leads me to wonder if the polyester from which clothes are commonly made are PET, and if not, what types of other polyesters (e.g. polyglycolide, polycaprolactone, polyethylene adipate, polybutylene succinate, etc.) clothes may be made from.I'm aware that most plastics can be stretched into long thin filaments/strands and wound into threads in a process similar to how natural-fiber threads are made. My question is specifically whether or not PET is commonly used as the polyester in clothes, and if not, what is? I've done some Google searches but can't find any references; maybe I just haven't gotten the search string right. Thanks.

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

I'm not sure putting it on hold as too broad is fair. There is a clear and unambiguous answer with good explanations as to why fabrics don't have the same characteristics as beverage bottles.

Based on several sources (shown below), it appears that polyethylene terephthalate (PET) is in fact used in the production of fabrics, including textiles. PET is marketed under names such as Dacron, Terelene, or Lavsan by various companies depending on your location. As to if it is the only textile polyester, I was unable to be completely certain. However, it appears to be a major textile polyester at the very least, as 39 million megagrams were produced in 2008 for fabrics alone.
Though it was not part of your question, I thought that I should include the following, as I found it interesting:
As it turns out, not only is PET used as a fabric, but it also makes a very nice fabric that is highly resistant to wrinkling. This is due to the presence of an aromatic ring within the monomers, which makes it quite strong/stiff; especially in this fibre form. These wrinkle-resistant fabrics are appropriately known as "permanent press fabrics."
To find out more information, I have included some sources below:

https://en.wikipedia.org/wiki/Polyester
https://en.wikipedia.org/wiki/Polyethylene_terephthalate
https://www.britannica.com/science/polyethylene-terephthalate

Why I didn't think of Googling "polyethylene terephthalate" is beyond me. (Too obvious?) I tried, "polyester used in clothing" and "...apparel" but didn't find anything useful. I did find the first article you reference but don't recall seeing that paragraph about PET being used in clothing. As far as Wikipedia being frowned upon, I think it depends on what's being cited. Pseudoscience claptrap often finds its way into medical articles, but articles concerning the physical sciences tend to be fairly accurate.

Polybutylene terephalate (PBT; for example Unifi Profiber(TM)) and polytrimethylene terephthalate (PTT; for example Dupont Sorona(TM)) are also used in some apparel, but in much smaller volumes that PET.


text:
Dropping Borosilicate glass mug
I recently got a glass tea mug. It's has double wall construction with a vacuum insulation (advertised as such). I noticed on the label, it specifically mentioned it's made of borosilicate glass, and to "discard if dropped".
I understand the purpose of making it out of borosilicate glass (thermal expansion when adding hot tea). But what would be the reason for the directions to link the material to needing to discard the mug if dropped (and not cracked or broken)?
Does the fact that it's made of borosilicate mean it's structural or thermal properties might be compromised if it's dropped?

Chances are the glass is perfectly fine. The inner space in the double wall is evacuated, so it's going to implode if it breaks. The evacuated volume is actually very small, so this thing is actually not going to send glass shards all over your room. But, if you have it if front of your eyes when it breaks, you might get injured. So, to be sure you don't sue them, they put this sentence in the label text.

@Karl That would make sense, but the way they worded the warnings, it made it seem the two were linked (the material and the drop warning)

I absolutely agree with @Karl; also, it might be helpful if there were a photo of the mug of a link to manufacturer. Borosilicate glass is a sturdy material.

I think not, you just cannot make dewar mugs out of anything than borosilicate (except quartz, if you have the money). That sentence is from a lawyer, not a chemist or physicist. ;-)

@andselisk I can't upload any photos right now, and even if I could I don't have the warning label anymore. The mug looks like any other double walled thermos, just made of glass.

OK, there is sort of a danger to borosilicate glass. Or to you, rather. Normal kitchen glassware is made to be under a lot of inside tension, by cooling it a bit too fast in production (roughly speaking). If it breaks, it shatters into thousands of harmless small pieces, like car windshields (which are made the same way). Borosilicate cannot have a lot of inner tension, because the thermal expansion (i.e. shrinkage in the end of production) is so low. So it breaks into a smaller number of shards, which are rather more dangerous.

Pyrex glass labware can be dropped, scratched, even used if cracked (I've started filling a beaker, then noticing that there was a crack - but it still seemed OK).
Borosilicate glass for commercial (non-lab) use is often tempered, so that a scratch is a potential hazard - the item might fail unexpectedly. The issue is not so much the borosilicate but the tempering.


text:
Mixing Supplements - Will the Distributions Remain?
I am combining various vitamins and minerals into a food supplement for my dog and I currently measure each powder individually for each batch of food each week.  The supplements are in powder form but have varying grain sizes.  Examples include copper, zinc, dicalcium phosphate, kelp powder, table salt, Vitamin E, Vitamin B-12.
If I were to make a giant batch of the supplements above using the proper proportions can I safely measure out of that large batch the appropriate volume each week or will the various consistencies / densities of the powders unevenly distribute making the weekly scoops proportionally unbalanced?
Is there some piece of equipment that could be used to mix the powders guaranteeing even distribution for batch usage?
Thank you in advance for any advice.

It's hard to maintain homogeneity for the mix of powders. The bigger the batch, the less homogeneous it's going to be. Ideally you want to disperse solid ingredients; there are several techniques to do that. Two of the most popular ones are:

dissolving all the components in a non-reactive solvent and they dry out or precipitate the powder;
using a ball mill.

With supplements, however, it's somewhat trickier, as they may react with each other over time, with the solvent, or be contaminated in grinder. Also they are often granulated for a good reason: this reduces the contact with atmosphere and prolongs shelf life.
If you decide to avoid the dispersion part, then due to Brazil nut effect the largest powder particles will remain on top of the batch, the smallest – on the bottom. This is especially noticeable when the batch has been shaken, e.g. after being transported in a car trunk, or just being kicked a few times lying on the kitchen's floor.
All in all, I wouldn't recommend doing this. The probability of suffering from an imbalanced diet is too high. There are some multivitamin and mineral complexes designed for humans, maybe look for something similar for your pet.

@airhuff Thanks for the edit, this was indeed a typo.

No problem, and +1 for the answer. Also, I agree with the bottom line of "All in all, I wouldn't recommend doing this." However, I'm not clear what advantage your pre-grinding step of dissolving (or attempting to dissolve) all the components would bring to the table. Is the idea to further reduce the size of the granules in the initial powders? My first thought is that that would be a messy, time consuming step that wouldn't buy you much. It of course would also require a drying step before grinding. Just my 2 cents.

@airhuff Oh, I didn't mean to make the grinding and ball-milling look like those are sequential steps. Those are two unrelated techniques, I'm going to edit the answer to make it clearer (if I fail, feel free to do it for me:) )

Ah, I get it. It's definitely clear in your edit. :)

Thank you for your answer!  I suspected this was the case but did not want to assume given my lack of chemistry knowledge!  THANK YOU SO MUCH!

@BenE. You are very welcomed, and thank you for an interesting question!


text:
Which is discharged?
Why Na+ ion is discharged at cathode rather than H+ ion when Mercury is used as cathode in electrolysis of aqueous NaCl solution ?

Because there is much more Na+ than H+.

@IvanNeretin That is not correct. Replace cathode with platinum and there will still be much more sodium ions than hydrogen ions but sodium will not form.

Yes, that too. I didn't intend to write a complete answer.

Mercury has a high overvoltage for hydrogen evolution; that means a barrier exists so that molecular hydrogen is not formed easily. Evolving hydrogen requires two protons to be neutralized, get together to form a molecule and clump up with a few more hydrogen molecules to form a gas bubble. If a sodium cation is neutralized right next to mercury liquid, it can quickly amalgamate/dissolve into the mercury, where its activity is much reduced. Sodium amalgam reacts with water, but much slower than metallic sodium reacts with water.
Similarly, ammonia amalgam can be made by electrolysis, and the mercury will foam up - it's not very stable.


text:
Bromonium ion or Mesomeric effect ( intermediate during addition to alkenes)
It's mentioned in Peter Sykes , A Guidebook To Mechanism In Organic Chemistry , 6th edition, 13th impression by Pearson, Pg. No. 185 , that on addition of $\ce{HBr}$ to bromoethene the only product obtained is 1,1-dibromoethane. It's due to the positve mesomeric effect of bromine to stabilize the carbocation in this case, however the rate is remarkably slower as compared to addition to ethene.
1
Another reaction corresponding to addition of $\ce{Br2}$ to ethene proceeds with the formation of a bromonium ion intermediate.
2
My question is, why can't the first reaction just proceed with a bromonium ion intermediate, like (28) above can form a bromonium ion and give 1,2-dibromoethane as a product (there may be a possibility that the carbocation may be formed away from bromine to counteract the strong negative inductive effect and in that process it may interact with the lone pairs of bromine to form bromonium ion)? And if bromonium ion is unstable (that is if it can't be formed in 1st reaction) why does it form at all in reaction 2? Why can't the Bromonium ion in the 2nd reaction just undergo a 1,2- hydride shift and convert into the "more stable" ion like 29b. It should also lead to formation  of geminal dibromide .It's a circular argument.
It's clearly mentioned that bromonium ion and carbocation are canonical forms of each other , which means you expect them to convert into bromonium ion whenever there's a positively charged carbon near a bromine atom.
3
Here's a similar example of phenonium ion.
[4]: https://i.stack.imgur.com/0j5Ww.jpg .
The carbocation doesn't stay the way it is in structure 4. It rather forms 3. Notice the bond between phenyl ring and the carbon atom on the right. It's not a completely solid bond.

Surely, you can also formulate bromonium formation for the first reaction. But clearly, the carbocation in alpha position to the brom atom will be better stabilized due to the mesomeric stabilization.

If you then consider that formation of the carbocation is likely the rate determining step, it makes sense to assume that essentially only the pathway via the alpha-carbocation is populated.

If bromonium ion can't be formed in first reaction why does the second reaction form it? It should also favour carbocation on the alpha position.

But still it's encouraging to hear that Bromonium ion can be formulated formulated for the first reaction.

Anything can be formulated only question if it is reasonable. In the second reaction though no alpha cation can be formed.

Can you please elaborate why can't bromonium ion break and undergo 1,2 hydride shift to form alpha carbocation?

Now we are getting somewhere. So you propose a 1,2 hydride shift after the bromonium ion has been formed. Thats a good question. I dont know for sure why this doesnt happen. Maybe the bromide is just the better nucleophile.

In other words the hydride shift is likely slower than the bromide attack.

That may be the possibility. But I'm glad you understood what I was trying to say. Thank you.

After some consideration I would say that the bromonium ion is likely even more stable than the alpha-bromonium ion you proposed after a possible hydride shift, which would explain all observations.

You are not going to see the formation of a primary carbocation—the activation energy is too high. Even though it is true the following are resonance structures of one another:

the geometry in which the primary carbocation is formed does not allow for the simultaneous formation of the bromonium ion:

The bromonium would only be able to form after a rotation of the $\ce{C-C}$ bond, so we need to consider the primary carbocation sans-bromonium as the real product of the protonation of the alkene. This is a very high energy product, so it will be slow to form at low temperatures and is out-competed by the other reaction. Since positive charge on carbon-1 can immediately be delocalized to the bromine, this product of protonation is more stable (meaning it forms faster).

First of all, structure (9)-bromonium ion intermediate and structure (10)-the open intermediate aren't canonical forms of each other simply because "In most resonance, $\sigma$ bonds are not involved(except in hyper-conjugation but what we have here is not hyper-conjugation), and only the $\pi$ or unshared electrons are put in different ways and canonical forms don't really exist(its our imagination) but reaction intermediates like bromonium ion and structure (10)-type intermediates have been isolated under special conditions."
Had structure(9) and (10) been canonical forms of the same molecule, we would've never been able to isolate structure(9) and (10). The positive charge in structure(10) might be stabilized by an attraction for Br but that does not involve a full bond.(see the image below)

The possibility that "the carbocation may be formed away from bromine to counteract the strong negative inductive effect and in that process it may interact with the lone pairs of bromine to form bromonium ion" sounds unreasonable to me because the carbocation will always form on the bromine side as the +M effect of bromine that stabilizes (29a) & (29b) is way more than its destabilizing -I effect.
In no way is bromonium ion intermediate less stable than structure (10).
Ab initio molecular orbital studies show that bromonium ion intermediate is more stable than the intermediate in reaction 1(structure-(10)).
In the addition of HBr to bromoethene, it follows the convention: "the electrophile attacks the substrate before the nucleophile does". Electrophile need not actually be a positive ion but can be the positive end of a dipole or an induced dipole, with the negative part breaking off either during the ﬁrst step or shortly after. So, the formation of the carbocation after the attack of the proton is inevitable.
But due to the fact that in reaction 1, the electrophile is hydrogen ion, the reaction takes place through the mechanism involving structure (10) and not the one in reaction 2 involving the bromonium ion intermediate(9).
I hope you realize that (2) never forms (4). Your statement saying that (4) forms (3) interconvert among themselves is not correct rather (3) and (4) are separate species formed through different set of mechanisms. None of the structure is a canonical form of the other.

The above reaction clearly shows that (12) and (9a) are not really canonical forms of each other. I really don't get the sense in which that statement on pg. 180 has been written but it definitely doesn't specifically mean that (9) and (10) are canonical forms of each other.
Don't get confused by the reaction that I mentioned above, Peter Sykes says This neighbouring group participation by bromine does not ofcourse prove that addition to alkenes proceeds via cyclic bromonium ions, in the past,the reaction was just used to show that such species(cyclic bromonium ions) are not just mere assumptions. That's all!

Structure 10 or 28 and bromonium ion intermediate are not canonical forms of each other...didn't i say that earlier??...because two canonical structures of the same compound have the same number of sigma bonds...that's a basic thing dude...

Can you please tell me the problem that you are still facing

@gowreeshmago I've edited the answer, hope its understandable now.

Its mentioned in Peter sykes that they are canonical forms. It's not me saying that.

Can you tell the page number.

Pg 180 its written bromonium ion is an alternative canonical form of carbocation 10

give me a min..

edited the answer again.

You yourself showed bromonium ion can be formed if there's a carbocation at beta position.

Look at the comments of logical×2 .

didn't you notice the reaction conditions, its a special reaction that has occurred under specific conditions so that the desired cyclic bromonium ion can be showed that it is actually formed in some reactions(I'll specify about it in my answer), but the point is in reaction 1 of your question the cyclic bromonium ion won't be formed because the $\beta$ carbocation is formed in a very less concentration, as compared to (29a). There are many proposed ways for a reaction to occur but most of them form the minor products (like the cyclic bromonium ion way does for reaction 1)

And you can rely on my statements, most of them have been taken from Jerry March which is kind of the most trustworthy book for organic chemistry. Believe me

I saw you asking "If bromonium ion can't be formed in first reaction why does the second reaction form it? It should also favour carbocation on the alpha position." I guess you didn't read my answer properly, I've mentioned why that is not feasible(or better if I say "won't count for the major product").

And the bromonium ion won't really undergo a 1,2-hydride shift because bromonium is itself more stable than the species it will form after the shift.


text:
Is it possible to have anions of metals?
I've been wondering if it is possible to have negatively charged metal ions. Most ions tend to have electron configuratons of noble gases. So then is it possible to obtain metal ions from metal atoms by changing their electron configuration to that of the next noble gas in the periodic table? (e.g. something like Be(6-))

https://chemistry.stackexchange.com/questions/27407/can-metals-have-a-net-negative-charge

Yes its common in coordination compound

Yes.  And, ironically, the most clear-cut examples involve alkali metals.  In these compounds you have the cation, often an alkali metal itself, stabilized in a complex so that electrons can be transferred to the recipient alkali metal (Na or heavier, based on today's known compounds), and the latter is a full-fledged alkali metal anion.