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What is the Purpose of an Electrolyte in a Galvanic Cell?
I understand how a galvanic cell works and the purpose of erverything such as the ssalt bridge and the electrodes, however I don't understand why an electrolyte is required. For example, consider this scenario.
I have a copper electrode in one beaker, connected by a wire with a voltmeter to a graphite electrode in another beaker. The copper electrode is immersed in water while the graphite electrode is immersed in lead ions. There is also a salt bridge connecting the two beakers. Now my question is, will there be a reading in the voltmeter?
I believe that there should be one since copper is more reactive than lead so copper ions will form while solid lead will form on the graphite electrode. However my school teacher said it won't work as there is no electrolyte in the first beaker. Why does it make a difference if there is an electrolyte or not?

Pure water without any electrolytes doesn't conduct electricity (or only does so extremely poorly), so how can you expect a current to flow?

@orthocresol why does the current need to flow through the water. Forgive me for sounding stupid, but doesn't the current only need to flow from the copper electrode to the lead solution?

You can't have a current going from A to B that doesn't go back to A. This is simple physics - you need a closed circuit for any electricity to flow

Just because you have a potential difference (i.e., a voltage reading) does not mean you'll get much current. Without supporting electrolyte, the water will be too resistive and (as @orthocresol said) very little will happen.

@orthocresol it is my understanding that the copper electrode continuously supplies electrons to the graphite electrode via the wire. Hence why is there a need for a close circuit for electrons to continue circle around if there is already a supply of electrons?

LOL this just hits my funny bone. " What is the Purpose of an Electrolyte in a Galvanic Cell? I understand how a galvanic cell works..." If you understand how a galvanic cell works, why are you asking about the purpose of the electrolyte?

Because if your electrons only move from point A to point B, there will very soon be a buildup of excess electrons and negative charge at B. Electrons don't want to move to a place where there's too much negative charge, so you need some way for negative charge (in a galvanic cell - this role is performed by anions in solution) to go back from point B to point A. If there are no charge carriers, then that doesn't happen.

I understand what you mean, but isn't that why there is a salt bridge? My teacher told our class that the purpose of the salt bridge is to prevent build up of excess charge in both beakers. This is achieved by dipping the salt bridge in $\ce{KNO3}$, thus the K ions go to the cathode while the NO3 ions go to the anode. If that this true, I don't see why an electrolyte is required.

Since copper has greater potential, won't it deposit on the copper electrode whereas lead would dissolve?

Two issues are at play here.
1) You will measure a potential without any supporting electrolyte. As soon as you have everything connected, a potential will develop. Because a self-respecting voltmeter doesn't require much current (ideally zero), the potential will be measured.
2) As soon as you try to pass any current, as comments have mentioned, that current has to be carried from the electrode through the solutions and the salt bridge. Without any electrolyte in the solution, the solution resistance will be huge(as Geoff mentioned) and the amount of current passing will be cell voltage/cell resistance. This will essentially limit the current passing through the solution to very low values.

Thanks for your explanation, however I still don't get why the electrons need to come back to the anode to form a complete circuit if the anode electrode is  continuously supplying electrons. Also as the reaction proceeds wouldn't the solution in beaker A become an electrolyte anyway as Cu ions are being formed?

Electrons don't need to come back to the electrode, but overall electroneutrality has to be sustained. Otherwise, charge will buildup, and the current will not flow.

@Ulgut But as I mentioned before, isn't that why we have the salt bridge. It is dipped in KNO3, and the K ions go to the cathode solution while the nitrate ions go the anode solution  to prevent charge build up (this is what my teacher told me).

But as told before, that would take a very long time, hence the very high resistance.


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Color imparted to flame and color of coordination compounds
When we heat a ionic substance/compound, the ionic bond gets weakened and the metal is ionized. Is the color imparted to the flame due to absorption of light? i.e. Do we see the remaining wavelengths? Or is it due to the falling back of electrons to their ground state?
Shouldn't we be seeing two colours one after other? Since the electrons take up energy of particular wavelengths (so we see the remaining colours) and then the wavelength which got used up earlier returning back when electrons return back to the ground state?
The coordination compounds should also show same dual-color property. Since when electrons absorb energy they go from t2g to eg and when they return, they should give back the wavelength.
This could also be said for halogens.
Why do we see only one colour? Which colour is it in which case? The absorption one or the emission one?

You seem to be mixing up a term. Let me get that out of the way first.
Ionisation is the name for a process creating an ion from a neutral species. However, if you heat an ionic substance and break the ionic bond, you are not ionising but deionising (although that term is usually used for a different process). The reaction equation for sodium chloride would be the following:
$$\ce{NaCl ->[\Delta] Na + Cl}$$

Now to the actual question. The perception of colour is based on one of two principles (as you noted):

out of white light, certain wavelengths are absorbed leaving behind a remaining spectrum and causing the perceived colour to be complementary to that of the absorbed light
light of a certain wavelength is emitted, directly creating the impression of said colour.

As you also noted, this could mean that we see up to two colours in the same electronic system: One due to the absorption and the second, complementary one during emission. So what is it that we see?
For coloured compounds at room temperature — that includes, but is not limited to transition metal salts and coordination complexes — we typically see absorption colour. Take for example the hexachloridoferrate(III) anion $\ce{[FeCl6]^3-}$: It has an intense yellow colour which can be explained by the absorption of very short visible wavelengths. The shortest visible wavelength is a blue one, so the perceived colour is yellow. The colouring is due to ligand-to-metal charge transfer transitions. In the corresponding fluoridocomplex $\ce{[FeF6]^3-}$, the orbital energies of the fluoride ions is predicted to be lower, giving a higher transition energy and shifting that transition just into UV wavelengths — the hexafluorideferrate(III) complex is colourless.
Also in organic chemistry, there is a number of yellow-ish compounds which occur because they just absorb at high enough wavelengths to shift out of the UV range. (UV absorption is very common for organic compounds containing a phenyl ring.)
We only see one colour, however. This is because the absorbed energy — which immediately excites an electron into a higher orbital — is liberated into the surroundings as vibrations and ultimately heat. The electron does not fall back into the ground state in a single transition but via a number of intermediates loosing energy all the way; hence we do not see a corresponding emission colour.

In flame colouring, one typically takes white (or colourless) compounds and heats them in a flame (the flame ideally being colourless, too). Only in the heat of the flame do they show their colour — and in a typical ion-lottery (qualitative inorganic analysis) lab, this is done in a dark corner of the lab, because there is actual emitted light. What is happening here?
Through the heat of the flame we are transferring energy into the compound, first causing the bonds to break (liberating atoms) and then also causing these atoms to vapourise so that we end up with gaseous metal atoms and gaseous non-metal atoms. Further heat is introduced which — if it is enough — will finally cause the transition of an electron into a higher orbital. This is, of course, unstable, but here the relaxation is immediately back down into the ground state for a significant percentage of these atoms; each of them releasing a photon with exactly the wavelength of the corresponding transition. And what we see are those photons, corresponding exactly to the wavelength of the energy difference between two orbitals.

And finally, just to confuse you, I am going to quote an experiment I saw in my school’s biology course. We had isolated chlorophyll from leaves of a plant and had it in a solution in an Erlenmeyer flask. It appeared green. Our teacher held it into the strong light beam of an overhead projector and it appeared as red. The experiment was explained as follows:

Red light is absorbed by chlorophyll, causing a green impression under normal conditions.
in the strong light beam, so much red light was absorbed and so many chlorophyll molecules were excited that the energy had ‘nowhere to go’; thus, the same red light was emitted to all directions and we were able to see that red colour.

So to sum it up:

in coloured salts at room temperature, we see the complementary colour of an absorbed photon. We do not see the emission (and thus the actual colour) because it is too unfrequent; most energy absorbed is released as vibrations and heat.
in flame colouring, we have highly excited atoms and we do not actually see the absorption since the energy necessary is transferred by heat. However, we do see emitted photons of exactly the energy difference wavelength from a significant number of electrons falling back into a lower energy state. We see the actual colour but not the complementary one.

you should add 'black body radiation color' and color temperature part to this post, I believe.

@permeakra Unfortunately, I don’t know enough about it =C

Just to check if I got it right - when we provide too much energy, we see emission color but when the compound is at normal conditions, we see absorption color? Right?

@Quark Simply put: Yes.

@Jan Sorry to bother  But in many coordination compounds, colour of complex is due to d-d transition (or so I was taught). Since d-d transitions can only be $$t_2g\rightarrow eg\text{ and } eg\rightarrow t_2g$$ the $e^-$ should only absorb and emit a single frequency of light. And in $t_2g\rightarrow eg\rightarrow t_2g \rightarrow.........$ and so on, the same frequency photon should be absorbed, emitted, reabsorbed, reemitted and the net effect should be zero( or white colour) right? What am I missing ?

@Aurelius this relates to the biology experiment I noted near the bottom of the answer (chlorophyll). You will only ever see one: either the colour of absorption under conditions that allow for absorption (this is the standard case); or the colour of emission. Under absorption conditions, the energy is typically not lost in form of a single photon but as a combination of vibrations, rotations and others. So there are no single emitted photons that could cancel out the absorption colour.

"Under absorption conditions,.......single photon........combination of vibrations, rotations and others..."- But for d-d transition there are no intermediate states between $t_2g$ and $eg$, right? So if electron doesn't emit the same photon it absorbed,  what will happen to it as it loses its energy? Which state (/energy state) will it go to?

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How to differentiate Zn 2+ and Al 3+ on the result of blue lake test
While performing test for basic radical in analyzing a salt qualitatively. I found out that the salt gave the blue lake test and since we were taught that only Al 3+ gives blue lake test, I thought it was Aluminium ion and wrote it the same. Later, my teacher told me I was wrong and it was Zinc ion. But then how did it give the blue lake test. How do I tell if the ion is Zinc or Aluminium?
If I could check the ion by any other reagent?

What’s the Blue Lake Test?

@Jan You add H2S in presence of acid to the salt solution. Then if the ppt it white ( which shows it maybe Aluminium )  you add dil HCl to the ppt and then add blue litmus solution (which is turned red ) and then NH4OH is added dropwise. So that precipitate of Al(OH)3 is formed (which floats) and the blue litmus formed from red ( formed by excess OH-  ions maybe) appears like it is floating on the Al(OH)3.

I thought aluminium does not precipitate under $\ce{H2S}$ conditions; it belonged to the $\ce{NH4S}$ group in my qualitative inorganic analysis course (the ion lottery), meaning that it precipitated as $\ce{Al(OH)3}$ under basic conditions …

Chemists usually take a course "Qualitative inorganic analysis" which gives a decision tree of tests to decide which cations and anions are present. There are few "specific" tests that will react with only one cation (Ni and DMG is the only one that comes to mind).

@MaxW Both Zn 2+ and Al 3+ give the blue lake test

@Jan Sorry I mean NH4OH and NH4Cl. Very sorry. I'm just trying to remember all the tests so...

Quark - I don't know what chemistry course you're taking, but I'd guess that an inorganic qual scheme that uses the blue lake test would have you do you'd do 3-4 other tests in tree and then use blue lake test to "confirm" Al. The previous 3-4 other tests would have eliminated a slew of cations. So I'd wonder if only Zn and Al would give a positive blue lake test.

@MaxW  Yes, there are 3-4 other tests which we perform before coming to test for Al or Zn ( Zn is much later than Al)  and Zn does give Al's test.  If it did. I'd think it is Al and stop there. Why would I want to continue if I think i've done the confirmatory for Al.

@Jan  I'd really be grateful if you help me with this one  too cause I have my practicals tomorrow.

You can use Thénard’s Blue (or Rinman’s Green in the zinc case; they are called Cobalt blue/green in English, apparantly). Also, aluminium displays characteristic fluorescence with morin (make sure there are no sodium ions present) and I used a different test which I forgot.

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Effects on Stretching Frequencies in Organometallic Complexes
If you added a negative charge to $\ce{Mo(CO)_6}$ how would this effect the $\ce{CO}$ stretching frequency in an IR spectrum?
My thoughts are that it would lower the $\ce{CO}$ stretching frequency because the metal would be able to donate electron density to the $\pi^*$ orbitals of $\ce{CO}$. Does this reasoning sound plausible?

That is a reasonable answer. You may want to explicitly mention why back-bonding into $\pi^\ast$ would lead to a lower IR stretching frequency.


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Find the concentration of the HCl solution when mixed with 1.265g of ACES-K+

$\pu{1.265 g}$ of N-(2-acetamido)-2-aminoethanesulfonic acid potassium salt ($\ce{ACES^-K+}$, $M=\pu{220.29 g/mol}$) is dissolved in $\pu{88.42 mL}$ of water. $\pu{27.59 mL}$ of $\ce{HCl}$ is added to the solution, resulting in $\mathrm{pH}$ of $6.54$. Calculate the concentration of the $\ce{HCl}$ solution. The $\mathrm pK_\mathrm a$ of $\ce{ACES}$ is $6.85$.

I tried setting up the equation as follows:
$$\ce{ACES + H3O+ <=> HACES +H2O}$$
The moles of ACES initial is $\pu{0.0057 mol}$ $(1.265/22.290)$ and we are trying to find $x$ the amount of moles of $\ce{HCl}$ $\ce{(H3O+)}$ we have. Therefore, using an ice table $\ce{ACES}=0.00574-x~\pu{mol}$ and $\ce{HCl}= x~\pu{mol}$.
When you plus into the equation
$$\mathrm{pH}=\mathrm pK_\mathrm a+\log(\text{base}/\text{acid})$$ I did $6.54=6.85+\log(0.00574-x/x)$ and got $\pu{0.0038 mol}$ of acid divided by total volume = $0.03321\ \pu M$. Is that correct?

Let us call the potassium salt "A-" and its corresponding acid for "HA".
The total volume of the final solution is 88.42 + 27.59 = 116.01 ml.
The final concentration of [HA] + [A-] = 1.265/(220.29 *0.11601) = 0.049499 (M).
[K+] = [HA] + [A-] = 0.049499 (M).
The equilibrium equation gives:
6.54 = 6.85 + log ([A-]/[HA]) => [A-]/[HA] = 0.48977 => [A-] = 0.48977 * [HA].
=> [HA] + 0.48977 * [HA] = 0.049499 => [HA] = 0.03322 (M) and [A-] = 0.016279 (M).
The charge balance is:
[H3O+] + [K+] = [OH-] + [Cl-] + [A-]
At pH = 6,54 [H3O+] = 10^-6.54 and [OH-] can be neglected in comparison with [A-].
=> [Cl-] = [H3O+] + [K+] - [A-] = 0.03322 (M).
The amount of chloride ions in the final solution will be 0.03322 * 0.11601 = 0.003853 mole. Thus, the concentration of the hydrochloric acid used will be:
(0.003853 * 1000) /27.59 = 0.1396 or about 0.140 M.

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Why do we scratch the walls of test tube in test for magnesium or calcium basic radicals?
In test for calcium ion, we add ammonium carbonate in presence of ammonium hydroxide and then white precipitate (of $\ce{CaCO3}$) is formed which is dissolved in acetic acid and then reacted with ammonium oxalate to form $\ce{CaC2O4}$. But it says you have to scratch the walls of the test tube to get the precipitate. Why is it so?
Similarly, for magnesium ion we add ammonium hydrogen phosphate to get white precipitate. For this reaction too, the walls of the test tube are to be scratched to get the precipitate.
I read somewhere that scratching creates a surface for recrystallization of the precipitate. I just don't really understand what does it mean. Can someone explain it?

Scratching helps the pppt nucleate.

@MaxW Would you mind elaborating it as an answer?

Scratching causes microscopic shards of glass to chip off the scratched surface which can break the surface tension of the liquid, allowing the solute to precipitate/crystallize from of the supersaturated solution.


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Theory of volume contraction in solution
When a solid is dissolved into a liquid, or two solids are amalgamated, their mutual volume will contract. For example, a crystal of sugar having a volume of one cubic inch dissolved into 10 cubic inches of water will have less than 11 cubic inches of volume.
Is there any theory that explains exactly the degree of contraction, and can correctly predict the contraction for any given solute and solvent?

Note that this is, of course, not always true, particularly for alloys where the new phase may have a different crystal structure.


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Stability of MnO4- in basic vs acidic conditions
Why is $\ce{MnO4-}$ more stable in basic than acidic conditions? I know that electrode potential values suggest this, but what is the underlying chemical explanation for the phenomenon?

$\ce{{MnO_4}^{-}}$ reduction starts as an electron transfer, which happens easier if the ion is protonated first..

The obviously feasible and spontaneous disproportionation reaction can be explained by considering the standard electrode potentials (standard reduction potential) involved (quoted as half–cell reductions, as is the convention).

$\displaystyle \ce{MnO4- (aq) + e- -> MnO4^2- (aq)} \quad\quad (E^0 = +0.56~\mathrm{V})$
$\displaystyle \ce{MnO4^2- (aq) + 4H+(aq) + 2e-  -> MnO2(s) + 2H2O(l)} \quad\quad  (E^0 = +1.70~\mathrm{V}, \text{in acid solution})$

2. has the more positive potential, so this will be the reduction half–cell reaction.
1. has the less positive potential, so this will be (reversed) the oxidation half–cell reaction.
$$E^0_\text{reaction} = E^0_\text{reduction} – E^0_\text{oxidation} = (+1.70~\mathrm{V}) - (+0.56~\mathrm{V}) = +1.14~\mathrm{V} \gg 0~\mathrm{V}$$
Therefore, the reaction is very feasible.
Incidentally: Given the two half–cell reactions, you get the complete balanced equation by adding 2. plus 2 x 1. reversed.
The greater stability of the manganate(VI) ion in alkali can also be explained by considering the electrode potential for 2. in an alkaline media.

$\displaystyle \ce{MnO4^2- (aq) + 2H2O(l) + 2e-  -> MnO2(s) + 4OH- (aq)}\quad \quad  (E^0 = +0.59~\mathrm{V}, \text{in alkaline solution})$

so, re–calculating gives
$$E^0_\text{reaction} = E^0_\text{reduction} - E^0_\text{oxidation} = (+0.59~\mathrm{V}) - (+0.56~\mathrm{V}) = +0.03~\mathrm{V} > 0~\mathrm{V}$$
Therefore just feasible! but on the basis of an equilibrium argument, here, the far lower $E^0_\text{reaction}$, suggests the $\ce{MnO4^2-}$ ion is far more likely to exist, i.e. more stable, in a very high pH solution and in practice it is stable for a few hours in alkali.

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That's an empirical argument based on comparing electrode potential values. I would like to know, in terms of coordination complex formation, why is the MnO4- complex less stable in acidic than in alkaline conditions?

@kane9530 It would be bad to think of $\ce{MnO4-}$’s stability in terms of complex stability. Ligand exchange is not an option, the oxygens are bound much too tightly to the manganese.


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Cis/trans interconversion of crotyl-boron compounds
Why is the following interconversion slowest for crotyl boronates and fastest for crotyl boranes while crotyl borinates are somewhere in-between?

The lack of an octet and the empty p-orbital on boron make trialkyl boron compounds quite reactive.  If we replace one of the alkyl ligands with an alkoxy group (borinate), we can now draw a resonance structure where oxygen shares one of its lone pairs with boron.  In such a resonance structure the boron atom has achieved an octet and the empty p-orbital on boron has been filled.  Consequently borinates are less reactive than the corresponding boranes.  Boronates, with two alkoxy substituents, have an even higher contribution from the resonance structure with the $\ce{B^{-}=O^+}$ double bond and are even less reactive.


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Which is the specific heat of human body?
I'm studying about heat transfer and I wanted to know if there exists a specific heat of the human body or how one studies the human body's heat transfer.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. You cannot really measure a constant specific heat of the human body since it is a mixture of so many compounds — but considering we are water to a large percentage, a body’s specific heat is rather close to water’s ($4.2~\mathrm{\frac{J}{g\ ^\circ C}}$).

Thank you :), but one question, what is the g and the c?

grams and Celsius; one of the usual ways of describing heat capacity.

Does exists any bibliography that I can read for this?

@Jan Incorrect. The body is only 70% water and the other 30% could change the specific heat of the body drastically if its own specific heat is very great or less.

Also, I don't think we can define a specific heat for the body since it would burn up if heated too much, and then it's not exactly the human body anymore.

@Lolsillymortals. Well, I said ‘a large percentage’ and everything above $50~%$ is usually considered large. My guess was a little away from the actual heat but it’s not absolutely terrible. Re your second comment: of course we can, because heating it by a fraction of a degree is sufficient to determine the specific heat (albeit with a large error)

A specific heat for humans typically quoted in the literature is 3.5 kJ/(kg K) (for example, pg 16 of Herman's "Physics of the human body", 2nd ed).   This is below the value of water (4.2 kJ/(kg K)) primarily due to body fat, which has a lower specific heat than water.
Despite its title, the article:
Faber and Garby, "Fat content affects heat capacity: a study in mice", Acta. Physiol. Scand., 1995, 153, 185-187, https://doi.org/10.1111/j.1748-1716.1995.tb09850.x
contains a brief summary of the literature concerning the average specific heat of humans.  The authors find that the specific heats of lean and fat mice agree reasonably well with a relationship originally proposed for humans:
$$ c_P = (3.72 - 1.84 \: m_f ) \text{ kJ/(kg K)}$$
where $m_f$ is the fraction of body fat by weight.  Wikipedia  ( https://en.wikipedia.org/wiki/Body_fat_percentage ) indicates that body fat percentage (by mass) is 17-21% and 25-30% for "average" men and women respectively (both neither "fit" or "overweight").  Substituting $m_f = 0.22 $ into the formula above gives 3.3 kJ/(kg K), in rough agreement with typical value of 3.5 kJ/(kg K) quoted in the literature (the literature value corresponding to $m_f = 0.1$, a value typical for a male athlete).
There is some uncertainty in my mind about the accuracy of the formula above, as it is based on a heat capacity of fat significantly smaller than recently reported (see database reference below).  However, I have trouble imagining a situation where the average specific heat capacity of a human would need to be known to great accuracy (it is needed for the artificial, but instructive Problem 1.54 in Schroeder's Introduction to Thermal Physics).  On the other hand, the heat capacity of various components of the human body could be of significant utility (the influence of localized heating etc...).  These specific heats (blood, bone, etc...) are tabulated (with citations) in the "Tissue Properties Database"  http://dx.doi.org/10.13099/VIP21000-03-0 (V3.0).

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Halides of group 15 elements
Why are the pentahalides of group 15 elements more covalent than the trihalides?

This is a part of a much more general trend, neither confined to halides nor to group 15: the higher you go in oxidation state, the less ionic compounds you tend to get.

Yeah but there must be a reason.

Fajans' Rule is  used to predict whether a chemical bond will be covalent or ionic.
Polarization will be increased by:
1) High charge and small size of the cation
2) Ionic potential
3) High charge and large size of the anion
4) The polarizability of an anion is related to the deformability of its                                            electron cloud (i.e. its "softness")
5) An incomplete valence shell electron configuration
In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides.


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How to calculate enthalpy change of neutralisation - what mass?
How to calculate enthalpy change of neutralisation - what mass?
I made an experiment where I reacted 100ml HCL 2M with 100ml NaOH 2M in 1300ml Water, and got a temperature change dT.
My question is do i use 1500g in the formula, Q=mcdT ?
c (specific heat capacity) will be calculated from another experiment where I measure the work required to heat the same solution to ~dT. In wheere I also need this mass
Do I include the small amount of HCl and NaOH?
Many thanks!

You should use the same total mass as in the neutralization experiment.  And, in your work experiment, the total mass of solution and the concentration of salt should be the same as for the final state of your neutralization.


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How to determine the empirical formula given mass percentages of the elements and the molecular weight?

A compound has $67.3 \% \rm~C,4.62 \%~N,6.93 \%~ H$ and O. Determine the molecular formula of the compound knowing its molar mass is $283\rm~\frac{g}{mol}$.

My solution : The percentage for the oxygen is $$100-( 67.3+4.62+6.93)=21.15$$ percent oxygen. Lets determine the number of moles for every substance.
For the carbon we have $$n_C = \frac{67.3 \rm~g}{12\rm~\frac{g}{mol}} =5.6~\rm mol$$
We do the same for N, H and O. We have 0.33 atoms of N,6.93 atoms of H and 1.32 atoms of O.
So the ratio is $5.6 : 0.33 : 6.93 : 1.32$.
We divide by $0.33$ and we have $17 : 1 : 21 : 4$. The empirical formula is $\ce{C17H21NO4}$.
Its molar mass is 303 g/mol. Now we find the quotient:
$$K= \frac{\text{Molar mass of the real formula}}{\text{Molar mass of the empirical formula}} =\frac{283}{303}\approx 1$$
So the molecular formula is $\ce{C17H21NO4}$. Is this correct?

Rather than starting out with 100 g of unknown, I would start out with 1 mol of unknown, i.e. 283 g. That way, you can directly calculate how many moles of each element are in one mole of compound:
$$ n = 67.3 \%  \cdot \frac{\pu{283 g}}{\pu{12 g mol-1}} = \pu{15.86 mol}$$
Or to get the stoichiometric coefficient $\nu_C$ directly, take the percentage and multiply it by the ratio of compound molar mass to atomic mass:
$$ \nu_C =  67.3\% \cdot \frac{\pu{283 g mol-1}}{\pu{12 g mol-1}} = \pu{15.86}$$
If you do that for C, N, O and H, you get the formula:
$$\ce{C_{15.86}H_{19.4}N_{0.93}O_{3.80}}$$
You can try rounding this to the nearest integers:
$$\ce{C16H19NO4}$$
This compound has a molar mass of 293 g/mol. Playing around with the formula, you can find coefficients that match the molar mass exactly, e.g. $\ce{C16H13NO4}$ or $\ce{C17H17NO3}$. In any case, neither one of these nor the formula suggested by the OP matches both the molar mass and the mass percentages.

Is this correct?

$\ce{C17H21NO4}$ is a good answer, but the data are not consistent, so it is not the complete answer. I would say depending on whether the errors on the molar mass determination or the elemental composition are larger, you would have a different top candidate, and should report the numerically exact formula with non-integer coefficients. That way, the reader can critically evaluate the meaning of the data themselves.

I suspect there is something fishy about the molar mass: if it were 303 g/mol, not 283 g/mol, then $\ce{C17H21NO4}$ would be a perfect fit. If you google the percentage, you find that originally it was a part of bigger problem where a student was supposed to determine molar mass on their own, so probably OP or a person who adapted this problem made a mistake in determining $M_\mathrm{r}$.

@andselisk Yeah, or they were on crack after doing the experimental analysis...


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What would Kp be if there is no gas reactant
What would $K_\text{p}$ be if there is no gas reactant but the reaction produces gas. I found something confusing: $$K_\text{p}=\frac{P_\text{gas product}}{P_\text{gas reactant}}$$ If there is no gas in the reactant side, then the partial pressure of the reactant gas is 0 so the $K_\text{p}$ would be infinity.

Kp by definition is the product of the partial pressures of the gaseous products divided by the gaseous reactants raised to the power of their respective stoichiometric coefficient in the balanced chemical equation.
To answer your question, this definition by itself isn't enough. You need to go a little deeper. Since your reactants are not gaseous they got to be either liquid or solids. Compared to the gaseous products, the active mass of the liquid or solid does not change much. We assume the concentration of such substances to be constant value hence we shift them to the Kp to get a new Kp (actually we consider this Kp to be the correct constant). Or in other words, we can safely substitute 1 in place of all non-gaseous reactants while writing the Kp formula.
Therefore, Kp for your question is
= (P.Pressure of gaseous productA)$^a$ (P.Pressure of gaseous productB)$^b...$


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What is the oxidation state of carbon atoms in Ethanoic acid?
The oxidation states of $sp^2$ and $sp^3$ carbon in ethanoic acid  are $-3$ if we do not consider the electronegativity difference of these two carbons due to their different  hybridisation  but this is not the case with the thiosulphuric acid ($\ce{H2S2O3}$); here the oxidation states of sulphur here are +6 and -2. So why are we considering the electronegativity difference of two $sp^3$ and $sp^2$ sulphurs here?

Welcome to Chemistry! This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Typically when figuring oxidation state of an atom, electronegativity of different atoms of same element is considered equal. This means that in thiosulphate oxidation state of central sulfur would be somewhere around 4.
However!
Let us condider a very tricky case: ozone. Technically it consists of atoms of same element and the oxidation state of oxygen atoms should be zero. But ouch, auch, the central atom bears full positive chanrge in all feasible structures.... So, should ozone, disregarding electron shifts in covalent bonds, have oxidation state of atoms +1 (for central atom) and -1/2  (for two other atoms)? But this doesn't make much sense. This is a place where oxidation state abstraction breaks.
So, there are situations when oxidation state abstraction doesn't make much sence.
Thiosulphate is somewhate similar. The central sulfur behaves very muc like sulfur in sulfate, so by analogy it should be in +6 oxidation state. However, formally it is not the case. Here a clear conflict between formal abstraction and real life occurs, so people cope with it variously. One possibility is to arbitrarily say that the central sulfur of thiosulfate is in +6 oxidation state (and it makes sence). Another is to go formally (and it also makes sense). I saw both options in the wild.
TL;DR : formally the central sulfur in thiosulphate is +4; but people sometemes disregard it as the ion is very similar to sulphate with central sulfphur atom being +6.

@KumarGaurav It's not so much widespread, but it happens. For example, PH3, AsH3 and SbH3  are typically considered as 'reduced' -3 compounds, despite that hydrogen have electronegativity slightly above that of the elements in question.

@KumarGaurav ethanoic acid  is not the case, though. It is typically considered to have -3 and +3 carbons.

An sp2 carbon is more electronegative than an sp3 carbon but we generally don't consider the electronegative difference of the carbons while calculating the oxidation state.
Carbon is more electronegative than hydrogen hence the sp3 carbon will take the electrons from the hydrogen.
Oxidation state of sp3 carbon = $3(-1)= -3$
The sp2 carbon has a double bonded oxygen which is more electronegative than the carbon hence the carbon atom will lose 2 electrons to the oxygen. The oxygen of the OH group will also steal an electron from carbon hence giving a total of +3 oxidation state to the carbon.
Oxidation state of sp2 carbon =  $2(+1) + 1(+1) = 3$

exactly , just opposite in the of H2S2O3 (thiosulphuric acid) ,we consider the electronegativity difference of these two sp3 and sp2 sulphur ; why this kind of irregularities is practised  in carbon?


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How to measure corrosion rate of an iron nail in salt water?
I want to measure rate of corrosion of an iron nail. One way to calculate the corrosion rate is to use the weight loss of the iron nail. However rust does not come off easily. Do have any other ways I could measure the rate of corrosion? Can I measure the weight lost somehow?

If you assume that all the rust sticks to the nail, maybe you could measure the weight gain of all those extra oxygen atoms.

There is an ASTM standard(G01) that gives detailed descriptions of cleaning procedure for corrosion measurements.
For iron and iron oxide, the recommended way is to leave the sample in a cleaning solution(e.g a dilute acid solution), followed by a light brushing repeatedly until the weight stabilizes.
You could also try sonication in a solvent to see if that helps.
In any case, to make sure you are not losing significant amounts of the base metal, you should repeat the cleaning procedure on some uncorroded control sample. This will help you dial in the proper procedure for your samples.

Thank you for your answer. I have access to a high school chemistry lab. What would you suggest as a cleaning solution? Could vinegar or baking soda be used? Any other simple solutions?

Baking soda could be a good start. As I mentioned, this will be a trial and error to dial in.


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Is it true that human cells grow in levorotation?
I remember reading this in a popular science magazine, some 20 years ago: that in most of the living organisms, the proteins (in fact, the amino acid molecules that make the proteins) are levorotatory and very few such molecules are dextrorotatory.
Is that true, at least for humans?

Homochirality has been asked about before here and isn't off topic imo. It may well be a duplicate though.

I remember reading this in a popular science magazine, some 20 years ago: that in most of the living organisms, cells grow in levorotation and very few grow in dextrorotation.

There is a kernel of truth in what you wrote, but either the science fiction magazine or your memory has distorted things a fair bit.  Cells do not grow "in levorotation" or "in dextrorotation".  Chirality is not a property of cells, it is a property of the individual molecules that comprise the cell.  Many different types of molecules are essential for life and are found in all living cells.
Two examples of these ubiquitous, essential molecules are carbohydrates and amino acids.  In all organisms on Earth, most carbohydrates are found in what is known as the D configuration.  The "D" stands for "dextrorotatory" but the nomenclature is confusing.  For carbohydrates, the configuration is defined as relative to glyceraldehyde, a simple sugar.  Thus D-fructose or D-glucose have the same configuration as D-glyceraldehyde, but only D-glyceraldehyde is guaranteed to be dextrorotatory.
Amino acids in life on Earth are predominatly "L", meaning they share an absolute configuration with L-glyceraldehyde.

Is that true, at least for humans?

So humans, like most life on Earth, is made of a mixture of levorotatory and dextrorotatory molecules.  Some types of the molecules in humans are predominantly dextrorotatory and some are levorotatory.

Awesome answer, thanks! I am sorry to answer so late, I asked the same question on biology SE, as advised, and I wasn't looking for answers here until now. Yes, that's what I wanted to ask, if the molecules inside living cells are levo-rotating, sorry for not being clear enough. Thanks for explaining about carbohydrates and amino acids, how about fats then? Do they expose levo-rotation and dextro-rotation too? Are a group of them preponderant in the living cells?

Amino acids in life on Earth are predominatly "L" - in which way? Like in general, the living cells have some 90%  L amino acids and the rest of 10% of amino acids are D, OR, for the majority of organisms, they contain 100% L, and just some minority of specific plants or animals have 100% D amino acids? Where can I learn more about this? I find the topic extremely interesting.


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How do I control the negative temperature in an experiment?
If one were to adjust the positive temperature in an experiment, they would use a heating plate, or a stovetop.
How would one adjust negative temperatures in an experiment? (Not keeping the temperature constant, but decreasing and increasing the temperature at my will in specific amounts).
What I tried so far: "Spraying" liquid nitrogen at my work surface until it cools to the desired temperature.  It is really inaccurate though.

There are various ways of cooling. What do you mean by negative? Lower than 0 C?

@Mithoron Yes, temperatures below 273 K.

Salt+ice bath, for example.

@IvanNeretin I considered an ice bath, but it didn't seem possible to adjust the temperature accurately.

Making fine adjustments to the temperature "at will" is not a simple task without an apparatus such as a low-temperature bath recirculator (i.e. very expensive equipment).

Depends on what you want to do, cooling bath, thermostats or chriostates  are used.

Peltier element
Here's another idea: it's quite practical to pump heat just by passing electrical current through a device, which is usually just a plate with two wires:

I just love these things.
What's especially cool about them is that by simply reversing the current you reverse the heat flow! This just begs for a trivial control loop via thermocouple feedback.
A quick google query reveals commercial offers of heat baths built on this promising whopping 0.01 °C precision of regulation!
The elements are easily available in electronics shops, so with a good power supply it wouldn't be too hard to build your own low-cost precision heat bath.
Hope that helps!

Nice answer.  I haven't seen Peltier modules get down to very low temperatures, especially at sub-zero. They are used a lot with glycol as the cooling medium for home brewing, but it would be nice to experiment with other liquids, and perhaps multistaging to see how far down they could go. Do you know of any low temperature examples?

Well, FWIW, the one SparkFun sells specifies [-50, +180] Celcius range. Not awfully wide, but still decent I think. Certainly more useful multistaged (stacked). Here they modestly boast of  reaching 145 K.

Interesting... only $12 too.  Too bad they're out of stock right now :(

@scriptHero I never feel awkward when showing people how to search on the internet ;) Here look http://www.ebay.com/bhp/peltier-module

@ulidtko No, it's just that I prefer buying materials from reliable stores like SparkFun.  Searching on ebay isn't my type of bargain hunting ;)  Also because I'm Canadian, and everything is in the US.

I can't agree with "it wouldn't be too hard to build your own low-cost precision heat bath" - turning a bare Peltier element into a practical laboratory chiller/heater would not be trivial either from an electrical or mechanical perspective (see e.g. http://electronics.stackexchange.com/questions/28634/how-to-drive-a-peltier-element). You can have low cost, or precision, or not too hard, but not two or more of these three...

Peltiers of ~40-75W (power consumption) are commonly used for portable refrigeration, so are widely available online.  They're not incredibly efficient but they are effective.  I've also worked with a CCD cooled by a 4-stage Peltier stack, with vacuum insulation, down to ~200K.

For temperatures below room temperature cooling baths are used.
These baths consist of a solvent + something that has a low temperature already (ice, dry ice, or liquid nitrogen). A comprehensive list is given below:

Although all of these temperatures are technically achievable, in practice we rarely prepare most of them due to either cost (acetonitrile is quite expensive to be using for a cooling bath), toxicity (a benzene/CO2 bath is not ideal) or difficulty (a methanol/LN2 bath forms a very thick slurry which is difficult to work with).
Most reactions below room temperature are therefore conducted at 0 (ice/water) or at -78 (dry-ice/acetone), with in-between temperatures achieved by adding dry ice slowly to acetone until the desired temperature is reached.
In addition to using this kind of cooling baths, there are machines that will allow you to dial up a given temperature and they will cool the solvent to within a few degrees. These are incredibly useful but their size and cost restricts them from everyday use (most labs don't have one, and those that do won't have enough for everyone to use one everytime they need cryogenic conditions)

How can I make a DIY cooling bath?  Do I just place my experiment on top of the cooling solution?

Inside it, not just on top.

You place your reaction flask inside the cooling bath (i.e. so just the top of the flask is poking out). Ideally the thing you make the cooling bath in is insulated (a dewar) to prevent it warming up too fast. Also be aware that at low temperatures moisture condenses so if your reaction is in any way sensitive it needs to be tightly sealed

I disagree with Acetonitrile being too expensive. Any lab that does HPLC will have enough acetonitrile standing around that half a litre can be stolen quietly without anyone noticing ;) Whoops, did I just give myself away …?

Well, most labs have acetone that isn't >99% stuff for doing their washing up etc. The MeCN we use for HPLC is ultra high grade and quite pricey (I'm sure you can get cheap stuff, we just don't, and there is something a bit strange about the pricing/availability/market for MeCN) admittedly its fine for a reaction in a 50mL flask, but repeatedly making up large baths gets expensive quickly (MeCN/CO2 baths also seem a bit crap anyway).

Temperature control for the home setup can be very difficult to control precisely. For high temperatures, heating your flask held within an oil bath is acceptable - never heat directly on a hot plate. One can achieve very accurate temperature regulation this way by adjusting the heater power.
For low temperatures, a cold bath may suffice, and NotNicolaou has provided a very nice answer in this regard, along with a very comprehensive table of components for reaching designated temperatures. However, the crux of your questions asks for a method that allows you to decrease and increase the reaction vessel temperature at will. Not easy. A cold bath is not likely to allow this, and so you need to engineer a solution based on how the 'real' labs do this.
A lot will depend on the scale of your reaction - micro vessels can be easily regulated with cooled gas flows. A regulated flow of evaporated liquid nitrogen is perfect for this, but not easy to implement. This is how most spectrometer temperature apparatus work, and is great for small vessels. If you are using a bigger reaction scale (anything in a round bottom flask for instance), as is most likely, then a liquid cooled bath is preferable.
To do this, you may need to consider having your cooling source external to your reaction bath, and applying a thermal regulation stage prior to passing it your reaction bath. Below is a rough schematic for this. Your cooling fluid is maintained at its lowest temperature (keep adding dry ice!). It is pumped through a heat exchange vector (such as a condenser) that has a high temperature medium (room temperature water, heated gas etc), and this heats your cooling medium to your desired temperature which is then passed into the reaction bath. This is ideally recycled back the source bath. The pump is typically a peristaltic pump design to prevent corrosion of the pump.

The critical element of the design is the heat exchanger, and some thought needs to be given to this. The simplest heat exchanger that the chemist uses is the condenser, although you might have to trial suitable diameters to manage heat exchange most effectively.
You can now regulate your reaction bath temperature up and down. You have two points from which to control the temperature of the final reaction bath: flow of cooling medium through the heat exchanger and the temperature and flow of the heat exchange medium. The slower the flow of your cooling medium through the heat exchanger, the greater the heat transfer. You can get very fine control of temperature using a setup of this nature, and it is essentially the same design of many temperature controllers.

Nice idea, very tempted to give this a go to see how it compares to commercial solutions.

The easiest - though not necessarily the cheapest - way to achieve a stable temperature below ambient is to use a refrigerated bath, together with a suitable bath fluid. Ethylene glycol - water mixtures can be used for moderately low temperatures; specialist silicone oils are better but more expensive.
If you can't conveniently immerse your experiment in an open-topped refrigerated bath then you need a circulating bath, which can pump the bath liquid around an external circuit. Cheaper ones have a single pressure pump so the external circuit needs to be closed, e.g. a jacketed vessel; more expensive ones can have balanced output and return pumps so they can circulate to an open container. With an open system you need to be aware of the possibility of siphoning when the circulator is switched off. A circulator should have the facility to connect an external temperature sensor, which could even be inside your reaction vessel, and the instrument will try and maintain the set temperature as measured by the external sensor. You may need to be aware of the possibility of condensation or icing, both on the chilled surfaces and in the bath fluid itself which may accumulate water over time.
Refrigerated baths aren't cheap, but they are fairly common pieces of equipment so if you are on a budget you could try looking for one at used equipment suppliers.
If you already have a temperature-controlled heated water bath or circulator then it may be more cost-effective to use it in combination with an immersion or flow-through chiller, which removes heat at an approximately constant rate allowing you to set a temperature below ambient on the heated bath. For a one-off or relatively short experiment you might be able to achieve a similar effect by running the circulating fluid through a loop immersed in a low-temperature mixture as described by NotNicolau - you'd just have to monitor the temperature of this mixture and top it up with the cold component as necessary.


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Why does shaking a match put the fire out?
Move a match slowly and nothing happens but if you shake it violently the fire will extinguish. Oxygen makes fire grow so why does waving a flame through the oxygen rich air put the fire out? Does this primarily have to do with a decrease in temperature to the burning materials or is it something else?
Also what about forest fires, do high winds spread or kill the fire?

Crossposted to http://physics.stackexchange.com/q/232989/2451

Combustion of small materials, such as a match or birthday candle, actually involve the release of volatile vapours, which themselves burn. It is not the solid material that burns. There needs to be a minimum amount of volatile material present in this combustion zone (just above the burning match) for the ignition to occur. As the combustion process continues, heat is given off, and more volatile materials are released, which in turn continues the combustion cycle. Now, if you shake a match or blow on a candle, you rapidly disperse these volatile fuels from the combustion zone, and there is no longer sufficient fuel to ignite. It is effectively removing the fuel component of the fire triangle, for just a brief moment.
Large fires can reignite because there is sufficient heat left in the fuel to further release volatile fuels, which can either self-ignite or ignite through the presence of embers.
Blowing gently on a small wood fire increase the oxygen content into the combustion zone, without dispersing the fuel. Similarly, if you experiment with a small enough wood fire, you will see that blowing on different parts of the fire will have a different outcome: blowing on the base of the fuels will increase oxygen content, and not affect volatile fuels. Blowing on the top of the fire (where the flame starts to burn) will probably put the fire out.
Will this put out a small paper fire? That will depend on the heat retained by the burning fuel. A single piece of A4 paper if shaken hard enough will extinguish. A ream of A4 paper that has burned half way down the page will be put out this way, but could easily reignite if the paper pages are pulled apart to allow oxygen into the released volatile fuels.
Generally, forest fires are accelerated by strong winds. Winds affect forest fires in a number of ways, and increase the rate of spread significantly. This is a topic for a whole other question.

Interesting. Does this also apply to paper fires? If I have no other options should I try to shake a paper fire out?

You might want to consider that shaking a burning paper will actually distribute burning embers around the room, which is not desirable. You are much better to try other methods of extinguishing, such as smothering.

My high school "experiments" with paper and magnifying glasses and sunlight seem to support the idea of a ream of paper being harder to put out. When I folded the paper before lighting it, it was significantly harder to put out. At least using sunlight meant I was already outdoors and therefore a little safer.

I would think that in addition to diluting the volatile gases, shaking would dilute the thermal energy, i.e., lower the temperature.

Here's a really neat example: https://www.youtube.com/watch?v=D1fH6Vuajeo


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What are the methods used to measure the concentrations of these anions?
Chromate and dichromate anions has the following equilibrium in acidic solution:
$$\ce{2CrO4^2- + 2H3O+ <=> Cr2O7^2- + 3H2O}$$
How can one experimentally measure the concentration of each ion at a certain equilibrium?

pH indicators come to mind.  More elaborate techniques could include conductivity measurements, NMR measurements, density measurements, etc.

Surely the simplest way is UV-vis spectroscopy, unless the peaks overlap significantly.

certainly don't try spectrophotometry, I did that in an experiment and my results were atrocious, and this is the main reason why I am asking. The difference in peak wavelengths was about 5nm

O17 NMR seems to be a thing and should work.  Cr53... possibly two, but unclear. However at elevated temperature chromate/bichromate conversion may become too fast for distinguishable peaks to exist.


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Would sodium salicylate dissolve in HCl?
I've been searching around the web and the consensus seems to be sodium salicylate dissolves poorly in HCl because it would form salicylic acid. However, I don't understand why forming salicylic acid product (which is poorly soluble in water) would make sodium salicylate (which is the reactant) dissolve poorly in HCl. If HCl doesn't contribute to making it more soluble, wouldn't the polarity of the water molecule dissolve it anyway?

HCl is in almost all cases already aqueous for common chemistry, mostly dissociated into $\ce{H^+} + \ce{Cl^-}$ because it is a strong acid, which means water is the primary solvent, not HCl. Pure HCl under ambient conditions is actually a gas (thus certainly not a pure liquid solvent).
That leaves the compound in question insoluble in the water after being protonated by $\ce{H^+}$ to form the insoluble acid molecule.
The process I mentioned in the comment is illustrated here:

The HCl is aqueous. But if that's the case, why would sodium salicylate be poorly soluble in HCl just because it forms an acid when there are so many water molecules around?

The thing rendered insoluble is not the sodium salt itself; rather the post-reacted salicylic acid. It is an induced solubility, as it were

So HCl protonates the salt, turning it into salicylic acid, which is insoluble in water, thus its insolubility in HCl?

That's right -- the new compound doesn't revert to the sodium salt after accepting the proton. That means you can actually reduce the concentration of insoluble salicylic acid by adding base, e.g. NaOH, and tipping equilibrium

Following from resonance, why would the salt takes up the proton? I thought compounds already in resonance state wants to stay that way. Is it because of the trend of strong acid/base -> weak acid/base?

The principle here is that minimization of molecular energy wins every time. Resonance is a stabilizing, energy minimizing effect, but it does not render every molecule with resonance unreactive. Accepting a proton in this case actually renders a molecule with lower energy than the negatively charged ion. Many such molecules react with acids and other compounds, and that's a big chunk of organic chemistry.

So it's the solubility of the product that's considered, even though it's different from the reactant in this case?

Yes, effectively. You're just making a compound that's insoluble in water (salicylic acid), from one that isn't (sodium salicylate), using something else in the water (HCl), through ionic combination. It's a precipitation reaction much like what you've probably learned before, just now in the class of organic salts, not metallic salts.


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If you run an electrolytic cell long enough, does the conductivity decrease?
I've never seen this discussed so I wanted to ask here.  Consider the electrolysis of water with inert electrodes and electrolytes that won't be oxidized or reduced, such as sodium sulfate.  Since the migration of electrolytes in an electrolytic cell is responsible for the electrical current, does the gradual sequestering of electrolytes at the electrodes decrease the electrolyte concentration in the middle (between the electrodes), and cause a decrease in conductivity?
A couple thoughts occur to me:

the $\ce{H+}$ and $\ce{OH-}$ will diffuse from the electrodes into the cell water.  As the $\ce{H+}$ and $\ce{OH-}$ diffuse back into bulk they could carry the electrolytes with them.
the battery could be disconnected momentarily to mix the electrolyte solution.

If anything, gradual buildup of $\ce H^+$ and $\ce{OH}^{-}$ in their corresponding areas should increase the conductivity.

...does the gradual sequestering of electrolytes at the electrodes decrease the electrolyte concentration in the middle (between the electrodes), and cause a decrease in conductivity?

In a word, no.  But, it is actually very complicated.  The electrolyte in most simple lab electrolytic cells for water is not very viscous.  This means that convective transport dominates over diffusion, except very very close to the electrode surfaces, i.e. at the boundary layer, which is hundreds of nanometers to maybe a few hundred micrometers thick, depending on conditions.  Outside the boundary layer, i.e. in the vast majority of the space "between the electrodes", no, the concentration of the electrolyte will not decrease.  In fact it may build up to higher concentrations, since water electrolysis involves loss of water, leaving more electrolyte per unit of solvent.
Inside the boundary layer, a phenomenon that does distort the bulk concentration of electrolyte can happen.  This phenomenon is called concentration polarization.  For water electrolysis, this means that the cathode (which generates hydrogen gas and consumes aqueous protons) would be depleted in protons, i.e. would be more alkaline than the bulk solution.  The anode's boundary layer, which generates protons from water while evolving oxygen) would have an excess of protons. The cathode boundary layer would have excess sodium cation and the anode excess sulfate anion to maintain electrical neutrality.
The formation of the boundary layer is very fast, taking less than a second, not gradual.  But the boundary layers are very small.
So in the end, you had it right with your first idea:

the $\ce{H+}$ and $\ce{OH-}$ will diffuse from the electrodes into the cell water. As the $\ce{H+}$ and $\ce{OH-}$ diffuse back into bulk they could carry the electrolytes with them.


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Radiating chemical suggestion?
Given a flat and thin object laying on a table, such as a small piece of silver paper or a small piece of cardboard, I want to find a way to detect which face of it is facing up. The solution should be able to detect the faces even when the object is placed underneath other object (so that camera/vision based approaches are not feasible).
What I am thinking is to find a radiating chemical using which we can paint on one side of the object and paint the other side of the paper piece with a chemical or substance that blocks the radiation, we should be able to tell if the object is up or down.
My question is: Is there such substance/chemical and associated detector on the market that we can purchase and use?

I can't see much sense in this. You wanna radioactive credit card?

Credit card or a piece of paper is just an example to make it easier to understand. I am planing to use this on robotics.

Your rewrite of the question, to provide your actual use case, improves it dramatically.


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Is Iodoform test restricted for only methyl ketones?
Do only methyl ketones give iodoform test? Where can a keto methyl group be present in?

All compounds with $CH_3CO- $ or $CH_3CH(OH)-$ group answer iodoform test.


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What factors influence the resultant oxidation state of Fe in 3Ca(OH)2+2FeCL3?

$\ce{3Ca(OH)2 + 2FeCl3 -> 2Fe(OH)3 + 3CaCl2}$

What factors influence the resulting oxidation state in the iron hydroxide molecule in this reaction? Since iron's oxidation state ranges from +2 to +6, and there exists iron hydroxide (II), how do we know what the oxidation state will be in this reaction?
Would this equation yield $\ce{Fe(OH)2}$ under some conditions?
I googled and found that in order to obtain $\ce{Fe(OH)2}$ we react $\ce{FeCl2}$ and $\ce{KOH}$ with no air present:

$\ce{FeCl2 + 2KOH -> 2KCl + Fe(OH)2}$

Would my initial equation yield $\ce{Fe(OH)2}$ in the absence of air? If yes, why?

The oxidation state of $\ce{Fe}$ before the reaction is +3.
$%edit$

First of all, figure out what oxidation state do you have initially (that is, before the reaction). Then see whether it can change. Is redox reaction possible? Are there any strong reducing agents?

I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here. Additionally, you can visit this chatroom for more assistance.

@bon - thank you! I'll add that to my bookmarks.

@IvanNeretin - thank you! I'll read up on redox reactions. Ferrum's pre-reaction oxidation state is +3, of course. Fe seems to be the most potent reducing agent here, since it's able to "donate" up to 6 electrons, according to my copy of the Periodic Table.

Strength of a reducing agent is determined not by the number of electrons it may donate, but by its "willingness" to do so, which in case of $\ce{Fe+^3\to Fe+^6}$ is quite low. But that's irrelevant. See, you have $\ce{Fe+^3}$ and you want $\ce{Fe+^2}$; how's that possible? Some other element must reduce iron. What element that might be, if any?


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Determine the oxidation number of sulfur in SF6
I've just started learning about redox chemistry and oxidation numbers. I can work out most of the oxidation number questions, but I don't know how to determine the oxidation number of sulfur in $\ce{SF6}$. This is not an ionic compound, and doesn't contain $\ce{H}$ or $\ce{O}$, so it seems as if I have no starting point.
If you could, please answer the question and provide a simple explanation.

In the case of $\ce{SF6}$, sulfur would have the oxidation number of +6 because the charge being applied to the fluorine is +6. Similarly fluorine would consequently have an oxidation number of -1 since $$6x + 6 = 0 \Rightarrow x = -1$$
(the right-hand side is equal to zero since that happens to be the net charge on the overall chemical formula).

Although this is the correct answer, it doesn't really explain why fluorine is negative and sulphur positive. Why couldn't sulphur be -6 and fluorine +1?

This has to do with electronegativity, as the more electronegitive an atom is the more likely it is to gain electrons and in our case Flourine is a lot more electronegitive then Sulfur. Thus, Fluorine is more likely to gain electrons from Sulfur and retain a -1 charge.

Oxidation number doesn't really have much to do with gaining or losing electrons. The sulphur doesn't actually have a +6 charge. Oxidation numbers are just a formalism to help with balancing reactions.

That maybe true, however electronegativity helps predict weather the oxidization number will be positive or negative irrespective of weather that ion's charge actually retains that magnitude.

In case of covalent compound oxidation state is determined by cleavage of bonds. During cleavage of a  bond between two different assaign +1 charge to less electronegative atom and -1 to more electronegative atom for single bond. Similarly +2 and -2 for double bond and so on. F is most electronegative atom hence it always show -1 oxidation state because it form only single bond.

Being the most electronegative element in periodic table (placed in right-upper corner) fluorine always has negative oxidation state in compounds besides $\ce{F2}$ and due to the fact the fluorine is a halogen (group $17$)  the only negative oxidation state is $-1$.
So fluorine in $\ce{SF6}$ has an oxidation state $-1$. Then we can calculate o. s. for sulfur:
    $x + 6×(-1) = 0$
    $x = +6$

So sulfur in this compound has an o. s. of $+6.$


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Reduction at the cathode
I'm learning electrolysis and we learned that the positive ions go to the cathode which is where reduction takes place.
However, I got confused because when you learn about ionic bonding, the positive ions are the ones that lose electrons. And we learned that reduction is the GAIN of electrons. These two things clash. Can somebody please explain this to me?

Positively charged ions wouldn't usually lose more electrons, they've already lost some.
The positively charge ions you have are called cations. These go to the cathode, which is negatively charged, as opposites attract. So, positive ions go to an area of negative charge and gain electrons, i.e. reduction takes place.
Now how can we negate any confusion between cathode and anode (it's very easy to get mixed up)?
Let's examine the word cathode a little more closely. Essentially there are two parts, 'cath' and 'ode'. The key part here is the ending of the word, which is 'ode'. The Greek root of this means, according to the Oxford English Dictionary, way. So as long as you can remember that 'cath' usually refers to a positive and 'an' to a negative, you can put the two together. Cathode meaning 'positive-way', as in positive things go this way.


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Does HF react with ethanol?
HF is needed to be dissolved in ethanol in many cases, e.g. in electrochemical etching for fabrication of porous material or in sol-gel synthesis of transition metal oxides being doped with F. Does HF only dissolve in ethanol or can it also react with it to form some fluorocarbon compounds under ambient conditions? If it reacts with ethanol, it will introduce complications in my sol-gel synthesis (ethanol used as solvent and HF as dopant), which I need to take care of.

HF can't be converted to R-F because HF is a weak acid so it does not react with alcohols.

I don't think HF will react with ethanol under normal conditions. Nucleophilic substitution could be a possibility, but $\ce{F-}$ is worse nucleophile than $\ce{OH-}$, so this won't happen.

If your worried about HF forming fluoroethane, I wouldn't be. HF is for whatever reasons a terrible nucleophile in the liquid phase but a great nucleophile in the gas phase. You should not observe any appreciable reaction, especially when dilute.


text:
What decides whether a reaction releases light or heat
I understand the reason that a chemical reaction would create photons (physics is more my strong point), but why would it create heat instead? My only guess is that the light released somehow causes the molecules around it to gain kinetic energy through some process and thus create heat, but I don't know how this would occur. What decides whether a given reaction produces photons or heat?
I'm a chemistry noob so try to explain using noob-friendly terms. Also, I don't know what tags suit so please put appropriate tags if my tags aren't.

The amount of energy released by a reaction is what most often creates light. Burning, the energetic reaction usually caused by oxidation of something, creates light because the amount of heat released is large enough to heat the reactants or products to temperatures where the radiation emitted contains a lot of visible light (coals in a fire glow red because they are hot enough that black-body radiation has a significant component in the visible).
Most exothermic chemical reactions don't generate this amount of heat and, although they release heat, you don't see it. Hand warming pads, for example, often contain supercooled sodium acetate which can be made to crystallise by popping a small metal insert. The exothermic reaction (crystallisation) releases heat to warm your hands, but not enough to see visible light.
A very small class of chemical reactions produce light directly as the reaction leaves molecules in an excited state which decays with the emission of visible light. Glow sticks are an example of this.
Overall, though, most reactions produce heat because the products have lower energy than the reactants and the released energy has to go somewhere. And it usually just ends up as heat. No photons are usually involved as the energy passes directly into the vibrational and translational energy of the products. So the excess energy in the reactants' bonds ends up as vibrational or translational energy in the products.

Let me put this into a physics friendly form.
When you say heat, heat is not actually an entity of absolute definition like a photon. heat is transference of energy between two bodies. so when you say when heat and when photon is actually like saying, when energy, when photon.
it doesn't make sense, since photon has energy.
I assume you mean when it does transfer energy in form of light and when it doesn't.
Every chemical reaction has an enthalpy based on how different is the products from reactants. when the reactants has more energy than products, the reaction results heat to transfer into environment (exothermic) when the vice versa, heat transfers from environment to reaction (endothermic) which makes our energy balance solely right. in exothermic case, heat gets absorbed by compounds in environment. if the temperature rises from critical limit that makes electrons in atoms emit light, then photons emit and when not, we just have raise in temperature.
I hope I was simplistic enough.


text:
What determines the speed of charging of a battery?
I was told on physics.stackexchange that this question belongs to Chemistry, what do you think?

Recently we have seen the surge of superfast mobile phone charging. Just an example: http://www.bbc.com/news/technology-30708945
What underlying physical principle determines the speed of charging of a battery?

There are many factors limiting speed of charging. Here are just a few:

Run-away overheating -- some battery chemistries, such Li-ion, may even self-ignite.
Spalling of electrodes -- the physical stress of rapid charging can buckle electrodes and cause a composite electrode to break apart.
Side reactions -- to increase the charging current, higher voltage is required. This over-voltage causes undesired side reactions, such as lithium plating (and out-gassing in Ni-MH and lead-acid batteries).

To make cells charge more rapidly, each of these potential issues must be analyzed and both the cell and the charging system must be designed to overcome the problem. In some cases, there have been spectacular failures, even when millions have been spent on design.

In addition to the limits mentioned by DrMoishe Pippik there is also the intercalation rate that determines the charge rate.

As you may know, what determines speed of a chemical reaction is its kinetics.
I didn't read the BBC article, but charging a battery follows a simple reaction :
A + B(+) → A(+) + B
and the rate of this reaction is determined by:
R = K [A][B(+)]
as you can see it depends on K and concentrations.
the thing we can change to make the reaction speed really high is K.
K is determined by three things. first the nature of reaction and second the catalyst we use and we can change both and third the temperature.
so to make a super hell fast recharging battery you need to find the right material for reaction and right catalyst and right temperature.
I hope this was the answer you were looking for.

The geometry of the involved materials is probably also an important factor in this case.

Yea, of course, it's part of the nature of reaction.thanks for mentioning


text:
How can rocks absorb CO2?
In the beginning of the earth, there was at a certain point so much $\ce{CO2}$ in the air that most of it was absorbed by rock. The consequence of that was the snowball earth, because the atmosphere was empty; all the heat of the earth couldn't be more absorbed by $\ce{CO2}$, so it was getting cold. But how can rock ($\ce{SiO2}$?) absorb $\ce{CO2}$?

$\ce{SiO2}$ can't. Some other rocks probably can.

That is something of a simplified view of snowball (or slushball) earth events but the silicate weathering is right. Silicate rocks are weathered by carbonic acid. An example reaction might be something like this: $\ce{Mg2SiO4 + 4 CO2 + 4 H2O -> 2 Mg^2+ + 4 HCO3- + H4SiO4}$ You might consider asking this question on earthscience.SE instead.

Definitely a question for earth science stack exchange

rock ($\ce{SiO2}$)?

Rock is not strictly $\ce{SiO2}$. Rocks are composed mostly of silicates: minerals such as feldspars $\ce{CaAl2Si2O8}$ or $\ce{(Na,K)AlSi3O8}$, olivine $\ce{(Mg,Fe)2SiO4}$, pyroxene $\ce{CaMgSi2O6}$, etc.
There are two ways in which rocks can absorb carbonates, one being a more direct form of the other.

Weathering of rocks by carbonic acid. $\ce{CO2}$ reacts with $\ce{H2O}$ to form acid and bicarbonate: $\ce{CO2 + H2O -> H+ + HCO3^-}$. This acid would then trigger: $\ce{4KAlSi3O8 + 4H+ + 2H2O -> 4K+ Al4Si4O10(OH)8 + 8SiO2}$. This is the reaction of orthoclase (a type of feldspar, common in granites) to form kaolinite (a clay, aka dust or mud). All of the resulting cations ($\ce{Ca2+, Mg2+, K+, Na+}$, etc. and the bicarbonate end up washing to the oceans, where they precipitate carbonates: solid rocks composed of mostly calcite ($\ce{CaCO3}$) and dolomite ($\ce{CaMgCO3}$), thus overall removing  $\ce{CO2}$ from the atmosphere.
Directly reacting of $\ce{CO2}$ with solid rock. All of the above basically happens in-situ, directly transforming solid silicate rocks to solid carbonate rock. When this happens, minerals such as magnesite ($\ce{MgCO3}$) or siderite ($\ce{FeCO3}$) also occur. This is actually one of the way to do industrial carbon sequestration: instead of putting $\ce{CO2}$ in the atmosphere, for example in power stations, you inject it down to the rock to lock it up in solid minerals.


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How long will it take for a measurable change in mass (corrosion) in an iron copper corrosion cell to happen?
I am planning to do an experiment, in which I will test the affect of salt concentration on corrosion rate. Would there be a relationship? How long would it take for rust to form, that could be scrapped off and corrosion rate measured?

In order to get any measurements, you would probably need to leave the coupon in the solution for about a week or so.
To see the difference due to the salt concentration differences, I would recommend going even longer, probably at least 2-3 weeks.


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net charge nucleobases at alkaline conditions
While doing anion exchange chromatography of a short oligonucleotide, I have noticed that at $\rm pH=12$, the oligo retains longer on the column than the same oligo- at $\rm pH=8$. I thought the reason for this could be the change in the net charge of nucleobases.
Could anyone suggest what is the net charge of RNA nucleobases (adenine, guanine, cytosine, uracil) at $\rm pH=12$ (denaturing conditions)?

The conjugate acids of the adenine and guanine have pKa's between 9-10. Guanine also has a nitrogen with a pKa of 12.3, which would be mostly protonated at pH 12. The pKa for the phosphoric acid proton of a nucleotide is significantly lower than 7.
Therefore, in a pH 8 buffer, the basic nitrogens of each adenine and guanine are protonated, while the phosphate is deprotonated. This gives an overall net charge closer to zero than you would see at pH 12.
In a pH 12 buffer, where the pH is above the pKa of both the phosphoric acid protons and the nitrogen conjugate acid protons (except the last one on each guanine), all of these positions would be deprotonated. Therefore, the formal charge on the nitrogens would be zero (again, except the most basic nitrogen on guanine), but the phosphates would still have a charge of -1 each. This would lead to a net charge of significantly higher charge magnitude because there would be very few positive charges to balance out the negative charges.

At pH 12, the net charge should be approximately: $$(n_N +n_u) - (n_c + n_g).$$
Approximate net charge at pH 8: $$n_N - (n_a + n_c + 2n_g).$$
nN= number of nucleotides
nu= number of uracils/thymines
nc= number of cytosines
ng= number of guanosines
na= number of adenosines

Thank you for your explanation! Do you think do Uracil and Cytosine influence the net charge of the oligo? I have found that Uracil and Cytosine may get a negative charge on oxygen: "Uracil is a weak acid. The first site of ionization of uracil is not known. The negative charge is placed on the oxygen anion and produces a pKa of less than or equal to 12. The basic pKa = -3.4, while the acidic pKa = 9.389." (from wikipedia)

What do you think, in case I run the anion exchange chromatography at ph12.3, will Uracil have 1 or two negative charges and cytosine 1 negative charge on oxygen?

Actually, I hadn't thought about the acidity of uracil/thymine. I had only thought about them in terms of basicity. Also, looking again, cytosine does have a basic nitrogen (pKa = 12.2). Uracil's imide nitrogen has a pKa of 9.39, so it would actually be deprotonated in the process. Thymine's imide has a pKa of 9.7. I don't have a chance at the moment, but I will update those equations to show those changes over lunch. In short, cytosines will have a positive charge in both cases (though it will be partially deprotonated at 12), uracil/thymine will go from neutral to negative charge.

@alec_djinn Okay, so it's updated. Realize these are approximate because you're kind of near the pKa for some of them. The higher the pH, the more negative it will be, and I have no idea as to what conditions cause things start to react (e.g. deamination of cytosine to uracil, for instance). At pH 12.3, the imide on uracil is going to be deprotonated, but the amide won't, so it would have be -1. Cytosine, however, would be approximately half deprotonated. The protonated species would be +1, while the deprotonated would be neutral. Neutral cytosine doesn't have any especially acidic protons.


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Number of resonating structures (hypothetical)
To my knowledge, resonance happens where more than one lewis stucture can represent the compound with no shift in sigma bonds
so, suppose in a simple conjugated system even like 1-3 pentadiene , two forms are well known, i want to know if a third form where each atom has completely filled or completely empty p orbitals should also be taken into consideration, as technically, there is nothing wrong with it, I accept that it's "contribution" will be negligible, but it should be counted in all POSSIBLE resonating structures
https://commons.wikimedia.org/wiki/File:Naphthalene_resonance_structure.svg
similar argument here, a lot more can be created if we allow charge separation of pi bond
EDIT: The question is whether there is a convention of some sort which tells us which ones to count which ones not (if we cant tell the magnitude of difference in stability of the structure)

These resonance structures are valid but they usually provide a very small contribution to the overall structure.

I wrote that :)

So what is your question? What you have written is correct.

in a question where it asks us to tell the total number of resonating structure, theoretically, not only the ones in majority, so is there a convention which limits certain kind of structure to not be included?

also, http://chemistry.stackexchange.com/questions/15665/ozone-resonance-structure-differing-opinions?rq=1  here, how a sigma bond can be broken in resonating structures? i thought that it was a necessary condition taht only pi electrons move

For the reasons you outline, it really isn't a useful question to ask for the total number of resonance structures. Including some other restriction (e.g., how many fully octet resonance structures exist?), would improve the question. Even better, the question could be phrased as "Draw three important resonance structures of the compound below."

@jerepierre agreed,  one more thing, doesn't the definition of resonance only allowes us to move pi electrons? but then how http://chemistry.stackexchange.com/questions/15665/ozone-resonance-structure-differing-opinions?rq=1

No. Resonance structures can include the movement of sigma electrons, although these structures may not necessarily be significant contributors. There is no definition that says that only pi electrons can be moved in resonance contributors.

so only condition is position of nuclei, conservation of electrons and valid lewis dot structure ex-not pentavalent carbon etc

Resonances structures are a model of the quantum mechanical orbital mixing.  To my mind they are a complete farce.  They are also quite useful
A model is a simplification of a problem, usually based on ignoring negligible factors.  So this question really distills down to what is negligible and what isn't?
That depends on roughly two factors:

The conditions of reaction.

Are the conditions strongly ionic or acidic?  Then charge separation may become relevant.

The rigor of the problem.

Is this for a high school course? A Ph.D. thesis? A bar-room chat with your mates?
Unfortunately, like most of chemistry, there is not an absolute answer here.


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Which is the strongest base relative to ammonia?

Among these bases,

ammonia
methylamine
ethylamine
dimethylamine
trimethylamine
quinuclidine

1) Which is the strongest base relative to ammonia? Why?
2) Which conjugate acid would be the strongest and which will be the weakest? Why?

I'm going to assume I need to look at the relative basicity (which is in kcal/mol)

ammonia: 0
methylamine: 10.1
ethylamine: 13.2
dimethylamine: 16.6
trimethylamine: 21.1
quinuclidine: 28.9

Looking at these values I would want to guess that means quinuclidine is the strongest base; and that ammonia is the strongest conjugate acid and quinuclidine is the weakest conjugate acid.
Is this correct?
The more I think about it I think I should be looking at proton affinity because the largest proton affinity is going to be the strongest base, which is still quinuclidine. but I'm still unsure about the conjugate acid.

You've tagged this post as computational chemistry, does this mean you're looking for help with using computational methods to determine relative pKa etc?

i tagged computation chemistry just because that is the course i currently am in and didn't know what would be more appropriate to tag. this was off an experiment i did where we didn't need to find pKa so i'm assuming that means my professor just wants us to use proton affinity. do you happen to know the trend for proton affinity vs weak base. i still am uncertain if it's true that largest proton affinity means strongest base.

Let's have a look at some of the definitions in the gold book and see how they relate to your problem.

proton affinity
  The negative of the enthalpy change in the gas phase reaction (real or hypothetical) between a proton (more appropriately hydron) and the chemical species concerned, usually an electrically neutral species to give the conjugate acid of that species. Proton affinity is often, but unofficially, abbreviated as PA.

gas-phase basicity
  The negative of the Gibbs energy, $\Delta G_\mathrm{r}^\circ$ change associated with the reaction:
  $$\ce{B + H+ -> B+-H}$$
  in the gas phase. Also called absolute or intrinsic basicity.

I was unable to find an exact definition for relative basicity, so I am going to assume based on the following reasoning. First we define the basicity for ammonia, $Bf(\ce{NH3})$, and an arbitrary amine $Bf(\ce{NR3})$.
\begin{align}
\ce{NH3 + H+ &~-> NH4+}  -\Delta G_\mathrm{r}^\circ(\ce{NH4+}) &
Bf(\ce{NH3}) &= -\Delta G_\mathrm{r}^\circ(\ce{NH4+})\tag{1} \\
\ce{NR3 + H+ &~-> NR3H+}  -\Delta G_\mathrm{r}^\circ(\ce{NR3H+}) &
Bf(\ce{NR3}) &= -\Delta G_\mathrm{r}^\circ(\ce{NR3H+})\tag{2} \\
\end{align}
We can follow up and define the relative basicity of an arbitrary amine  to ammonia $Bf_\mathrm{rel}(\ce{NR3})$ as the difference of the independent basicities.
$$Bf_\mathrm{rel}(\ce{NR3}) = Bf(\ce{NR3}) - Bf(\ce{NH3})\tag{3}$$
With (3) we would actually describe the following reaction:
$$\require{cancel}
\begin{align}
\ce{NR3 + \cancel{H^+} + NH4+ &~-> NR3H+ + \cancel{H^+} + NH3}\\
\ce{NR3 + NH4+ &~-> NR3H+ + NH3}\\
\end{align}$$
So the relative basicity describes in this scenario how likely it is that the arbitrary amine is abstracting a proton from the conjugate base of ammonia, i.e. ammonium. The higher the relative basicity, the stronger the base.
The conjugate acid $\ce{NR3H+}$ is the strongest if its parent compound has the smallest basicity, meaning most likely to loose the proton. The weakest conjugate acid is obviously the strongest base.
But what have we actually compared? In principle we have discussed with this comparison how likely it is for the amines to bind a proton and we have chosen to do that in comparison with ammonia. We could chose to do the comparison to water and with a couple of transformations and simplifications we would arrive at a $\mathrm{p}K_\mathrm{a}$ which we would then compare. To make things easier, we could also directly compare the values of $\Delta_\mathrm{r}^\circ$ in (1) and (2); the lower the value, the higher the basicity.
If we now choose to compare the enthalpies instead of the Gibbs energies, then we would be ultimately comparing (gas-phase) proton affinities. We would be still looking at the same reactions, just a different characteristic property. We would still be observing the same trend.
TL;DR It does not matter if you look at proton affinities or relative basicity, you are still comparing the same effect with different numbers, that are all linked.


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Understanding Standard Hydrogen Electrode
I came across a question today.

A galvanic cell is composed of two hydrogen electrodes, one of which is a standard one. In which of the following solutions should the other electrode be immersed to get maximum e.m.f. ($K_a=10^{-5}$ for acetic acid and $k_a=10^{-3}$ for phosphoric acid)
(A) $0.1$ $M$ $\ce{HCl}$
(B) $0.1$ $M$ $\ce{CH3COOH}$
(C) $0.1$ $M$ $\ce{H3PO3}$
(D) $0.1$ $M$ $\ce{H2SO4}$

I thought "Well, seems like all I need to do is find which solution will give the most number of hydrogen ions because its hydrogen ions which will suck those electrons from the other half cell which will be the cause of current. And concentration is same for all. So there is no chance for a weak acid. So option (B) cannot be correct. All other seems like a strong acid. So it should be $\ce{H2SO4}$ as it gives $0.2$ $M$ of hydrogen ions for $0.1$ $M$."
But my thought was absolutely wrong. Why?

There are more hydrogen ions on the other side, anyway.

@IvanNeretin you mean on the standard half cell? But why is it?

Yes, that half. See, what is the standard hydrogen electrode? What is the concentration of $\ce H^+$ in it?

@IvanNeretin its unity. So do you mean $\log\ce{H+}-\log {\ce{SO4^2-}} = 10$ while if there was $\ce{CH3COO-}$ it would have been lot more...is that what you mean?

I don't see where that 10 comes from. Other than that, yes. The SHE is more acidic than any of our options. If we want to be as different from it as possible, we want to be less acidic.

@IvanNeretin thanks. You have helped me again. And that 10 came by putting the concentration of $\ce{SO4^2-}$ which is $0.1M$. I assumed full dissociation here. And after your help the answer posted by Elit is more clear now.

Even if we assume full dissociation (which may not be the case), there are a couple of issues here. First, we want to compare $\ce H^+$ with another $\ce H^+$ and not with $\ce{SO4^2-}$. Second, the $\log 0.1$ is nowhere near $\pm10$.

@IvanNeretin let us continue this discussion here

This is a concentration cell question where 2 electrodes of same stuff are dipped in their own ions and if there is non equal concentrations of anything a "potential" develops called electrode potential (to be understood in qualitative way)(kind of more concentration of reactant forces product formation)
as you can understand this would be a redox reaction where reduction would occur in one and oxidation in another, Now, by standard it means that all the components of the reaction (participants, i.e. reactants and products) are in standard state i.e. $1M, 1$ $bar$ etc, so, we know that concentration of $\ce{H+}$ in standard one is $1M$, we need to create biggest gap between them, as the concentration is really high and we cant compete with it, just reduce our own $\ce{H+}$ to increase their difference and force ours to undergo Oxidation to gas and standard undergoes reduction
Just for extra information,
$E=E_0-{RT\over nF}\ln Q$ and can be derived by $G=G_0+RT\ln Q,\;G=-nFE$ {$n$ is number of electrons being exchanged ex- in $\ce{2H+->H2},\;n=2$; $F$ is the charge on 1 mol electrons} and Standard Hydrogen is the standard bearer so, we assign its standard value as $0$, $E_0=0$.
EDIT : I can't comment due to reputation
, basically, assume complete dissociation and and concentration of $\ce{H+}$ in $\ce{HCl}$ would be $0.1M$ , $\ce{H2SO4}$'s second ionization is not complete like others but can be assumed for a good approximation, so concentration becomes $0.2 M$ , and basic thing to remember is
IF ONE SIDE  UNDERGOES OXIDATION THEN OTHER MUST UNDERGO REDUCTION
and if we strengthen either reaction, potential increases, concentration is only in a solvent, how can a solid would have active $\ce{H+}$ ions (everything is activity not concentration or partial pressure flat out, they are approximated, solids are assumed to have activity as unity) and log is defined as the power to $10$ that would give us that number


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What could happen when mixing bleach with other cleaning residue?
I have a wall covered in mold. I have used various cleaning wipes attempting to clean it, including a vinegar mixture. I now want to spray the wall with a bleach cleaner. I am concerned for fumes from what may be considered mixing. For how long is the previous cleaning considered active enough to mix with the current cleaning ingredients? In other words, if I cleaned with vinegar a week ago, has enough time elapsed to safely use the bleach?

Vinegar and bleach are not a problem. The problem mixing is ammonia and bleach which creates chlorine gas. On a wall bleach is definitely the product to kill mold.

Bleach reacts with acids to produce chlorine gas as well. I use this reaction in the lab to make chlorine (under controlled situations) for a particular reaction.

This answer is a bit hand-wavy because there are a lot of factors that we can't easily know in advance. You are unlikely to make very much in the way of toxic fumes from the small residues of vinegar left after it is dry. However if the wall is covered with mould, the room is probably poorly ventilated, so it is more reasonable to be somewhat concerned.
When mixing bleach with other cleaning products, the biggest concern is the reaction with ammonia, which produces extremely poisonous nitrogen chlorides. However chlorine bleach (hypochlorite ion) is a very reactive substance that will react with a wide range of things, including acids such as vinegar.
Acetic acid -- the "active ingredient" in vinegar -- is less volatile than water so it could easily remain a long time in a damp, poorly ventilated room. On the other hand, it will also slowly react with the wallpaper, the sheetrock, etc etc... Fortunately, since acetic acid warns us of food going bad, the human nose is very sensitive to it. If the wall doesn't smell vinegary, then there's not much left.
The main reaction product between acetic acid and chlorine bleach will be free chlorine gas plus smaller amounts of chlorine monoxide gas. This is a very simple reaction: hypochlorous acid is a very weak acid, so some hypochlorite ion is shifted back to acid form by the acetic acid (or any other acid); but hypochlorous acid is also very unstable, and decomposes to those materials plus chloride ion.
You will easily tell if this occurs: it will stink even more chlorine-y than the bleach, and probably make your eyes and the back of your throat sting. If that does happens, leave the room immediately and do not reoccupy it until it has had a long time to air out. Get in the shower to rinse your eyes and nose.
(A small fraction of the acetate will also be converted to chloroacetates. These are not volatile, but are extremely toxic. More of a concern for cleaning food-handling areas. In this case not a big deal unless you plan to eat your wallpaper.)
I would suggest that whatever you do, you first need to get good ventilation in there if you don't already have it. It is a common fallacy that "natural" things are harmless and evil "chemicals" will poison you. In fact, in a poorly ventilated space almost anything except fresh air is sub-optimal for your health. Apart from anything else, the mould spores can cause serious health issues. Furthermore, it is quite likely that ventilation is the root cause of the problem and you will not permanently fix the problem until that is addressed.
Once you have good ventilation, the quite small amounts of toxic gas that the bleach may produce from the quite small residue of acid, is not so concerning. Personally, I would probably use an oxygen (peroxide) bleach, but it isn't so much of a concern, and it doesn't fade the stains as much. You may not need bleach at all. Once you have had a while for the moisture level in the wall to drop, a lot of things will kill your mould relatively easily, although removing the unsightly blotches is another question.


text:
What is the melting point of glycerol in a vacuum?
Can anyone tell me what the melting point of glycerin would be in a vacuum?
From researching online (disclaimer: I'm definitely not a chemist) it seems the boiling point is $\approx \pu{170^\circ C}$ in a vacuum (vs. $\approx \pu{290^\circ C}$ at sea level), and the melting point at sea level is $\approx \pu{18^\circ C}$ so my best guess for a melting point is $\pu{-102^\circ C}$?

The melting point usually does not vary much with pressure. For glycerol it it even unclear in what direction. My best guess for the melting point of glycerol in vacuum is 18°C.

Is there any documentation for these types of industrial chemicals that would make this explicit? It would be useful to know exactly what the behaviour of the chemical is as the pressure drops to zero - would the manufacturers publish this information for example?

I don't think that someone would try to measure supposed small variations of the melting point of glycerol at low pressure. First of all it is probably not very interesting. Then one had to have glycerol of a purity that is normally not available. I believe that otherwise present impurities would have a much greater effect to the melting point than varying low pressure.
To get an idea of the size of the effect, see that the melting point of water is virtually constant between 10 mbar and 100 bar.
https://en.wikipedia.org/wiki/File:Phase_diagram_of_water_simplified.svg


text:
Alkaline Ester Hydrolysis Reaction
In an ester hydrolysis reaction using a dilute base what is the pH change before and after the reaction occurs. In this reaction I want to react methyl ethanoate with dilute sodium hydroxide. I want to know if the pH of the solution will change after I add the ester to the base.
If it doesn't, is there a reaction I can do with esters that will result in a pH change?

Assuming this is in water. The pH shouldn't change much because the pKa of alkoxides in water are pretty much the same as that of water itself.

Okay, thanks for replying. Is there some reaction that you can do with esters that will allow a change in pH?

The pH will definitely change... Write out the mechanism to see why.

Just any reaction where I can see a pH change that I can use with an ester. And how would I write out the mechanism? I am a beginner with organic chem, are there any websites you could recommend to be able to write out the mechanism?

Looking at the mechanism properly. At the start when it is only dilute NaOH, it is pH basic. But when I add the ester, it will react and form an alcohol and carboxylic acid, which is acidic. So wouldn't the pH of the final solution be lower than the start?

My bad. Jerepierre is correct. pH would definitely go down. But it wouldn't be a carboxylic acid. You would still end up on with pH>7.

Okay, thanks for that. Would you also be able to tell me where I can find an experimental setup for this type of reaction so I can actually do it? I am having a bit of ytrouble finding something online with actual measurements for things.

Alkaline hydrolysis of ester:

$\ce{RCOOR' + OH- -> RCOO- + R'OH}$

Almost all organic acids are weak acids:

$\ce{RCOO- + H2O <<=> RCOOH + OH-}$

Therefore, a lot of $\ce{OH-}$ is consumed during the esterification, with the product $\ce{RCOO-}$ generating little $\ce{OH-}$.
There is a net consumption of $\ce{OH-}$, making the solution more acidic, lowering the pH.