text:
Toilet spray effect
When we want the bad smell in toilet to go away we use a toilet spray to clean the air.
But does it really cleans it, or just the elements in that spray have better effect on our smell sense. And we think that the bad smell went away while it is not, and the air contains a mix of bed and good smells?

Some toilet sprays can be classed as deodorizers (or air sanitizers in the US?), while others call themselves air fresheners. So it may depend on which brand you are using, but  generally the deodorizer spray is working a little harder than the air freshener.
Odor control products can work through a number of different methods:

Chemical transformation. Typically oxidization. This is why bleach is a good product for eliminating smell.
Adsorption or Absorption. Trapping airborne smelly molecules is a great way to keep them from your nostrils. Activated charcoal in-soles for your shoes are a great example of this. Zeolites are also used. An air purifier would be what you'd need to use in a bathroom to eliminate smells using this method. Absorption into other media can then also be used in conjunction with other methods (chemical)
Sanitizer or "biocide" agents. These kill odor-causing bacteria, fungi and other microorganisms. The main culprit for 'toilet smell' comes from the results of bacterial activity in your gut. Whilst they do reduce the quantity of airborne bacteria, they require a constant applcation, and so can't be classed as an antiseptic.
Masking. This what air fresheners do. Add a stronger, nicer smelling fragrance. Humans only have a few million olfactory receptors. If you keep them busy with volatile pleasant fragrances, malodorous molecules can't get a look in.
Sealing. Painting over or applying a sealant to a urine-soaked concrete floor or smoke-laden timber panel is often the only way to mask some smells.

Most air spray deodorizers (sanitizers) use a combination of the above methods. They almost all use some form of masking (except those that claim to be fragrance-free), as well as acting as a biocide and in some cases promoting chemical transformation as well.
There are growing concerns about the safety of using air sprays.
And just as a wonderful aside; one of the main causes of toilet odor is a molecule called skatole (3-methylindole), which also happens to be used in perfumery and also as a flavoring in ice cream!


text:
Phase diagram analysis
The figure is a portion of the titanium copper phase diagram for which only single-phase regions are labeled. Specify all temperature composition points at which eutectics, eutectoids, peritectics and congruent phase transformations occur.
I think i've found the eutectic and congruent point. (aprox 960 and 980 celsius) but how do I find the eutectoid and peritectic point?

Sorry for blurry image!

Assuming that the fields marked by x are solid and y is liquid, these are the points:

a: eutectic point. Cooling will turn the single liquid into two solids. The point is between the composition of the two solids involved.
b: eutectoid point. Cooling will turn the single solid into two solids.
c: peritectic point. Cooling from immediately to the left will turn the solid into a liquid and a solid. The point is outside the composition of the two solids involved.
d: congruent point: A solid will melt at this point to a liquid of the same composition.
e: incongruent melting points. A solid will melt to a liquid and a solid. Note that the left one is intimately associated with the peritectic point.

There are no peritectoid points in this diagram. A peritectoid point is similar to a peritectic point, but with all phases solid.

source: Stack Exchange

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text:
Finding reaction rate from pressure change
I conducted a lab in which different concentrations of $\ce{HCl}$ were used and $\ce{Na2CO3}$ was added. The pressure change was measured for a reaction with the different concentrations.
Now I am going to find the reaction rate for the reactions. I am thinking that using the pressure data, I can find the reaction rate by doing the following.
Consider:

${PV = nRT}$ ($R$ and $T$ are constant so they can be taken out of the equation)
${PV = n}$
$P = {\frac nV}$ (Knowing that $\text{c}={\frac nV}$)
${P} = \text{c}$

Seeing as pressure is directly related to concentration, then couldn't I just use the pressure data to find the rate and then relate that to the concentration change over time?
The alternative would be to use ${PV=nRT}$ for each second of the reaction and find the concentration change that way. However, I want to consider the possibility of doing it this way.
One problem I forecast would be the units: kPa/s vs. mols/L

You can't "take out" $R$ and $T$ from the ideal gas equation. If you do not know the value for a variable, then the most you can do is try to cancel it out in a ratio: $\frac{P_1V_1}{P_2V_2}=\frac{n_1RT}{n_2RT}=\frac{n_1}{n_2}$

What I'm trying to say is that during the experiment Temperature was constant .. I based my information on this http://www.chemguide.co.uk/physical/basicrates/pressure.html

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text:
Can anyone identify this uncaptioned diagram?

It's from the start of part 3 / chapter 19 in Lewin's Genes X (see this page in Google Books)...
Looks a bit like a Ramachandran plot(?) but the axes only go to 100 / 70ish, but I don't even know what molecule it is since it doesn't say and there's no image attribution page at the front/back.
Any sleuths out there?

You could of course always contact the creator of this image.

Also, sometimes books contain figure information in other locations other than the page the figure appears on.  This can appear in neighboring pages or sometimes in a section at the beginning or end of the book.

Why do you think that it is molecule? I guess it is ribosome, considering kind of one bigger one smaller parts aggregation.


text:
How to test oxalic acid in potatoes
How could you test that there's oxalic acid in potatoes.For example,

what chemical reacts to oxalic acid and then what will happen if it
  reacts?

If you want to determine one compound, i.e. oxalic acid, in a broad mixture of others (potato), you need to first separate them, using i.e. chromatography. It is almost impossible to have a reaction so specific, that it reacts with oxalic acid and with nothing else.

Why would you need such a test? Vegetables with a tongue-detectable amount of oxalic acid are edible (e.g. rhubarb). If for some reason you suspect that your potatoes were contaminated with oxalic acid, or that you have a cultivar which produces miniscule amounts of it, you can be sure that as long as it tastes like a potato, it's not harmful. It will have to taste about as sour as a lemon to start getting problematic, and I don't think anybody would eat such a sour potato.

If you want to determine the total content of oxalate you would need to extract it from the matrix with hot HCl 1M, for instance, since it will dissolve calcium oxalate crystals. Afterwords you can precipitate oxalate again at pH=4-5 with calcium nitrate, thus forming white crystals of calcium oxalate. You can redissolve them by adding sulfuric acid and identify oxalate by taking advantage of its reducing properties - if you add potassium permanganate it will be reduced from Mn +7 (purple colored) to Mn +2 (brown colour) and oxalate will be oxidized to carbon dioxide. But of course you start with a complex matrix, the easiest thing would be to use chromatography to separate and identify oxalate.


text:
What prompts sodium to give up an electron?
What triggers sodium to convert from elemental Na to Na+? I know that it wants to have a full valence shell and all that, but how does it just eject the electron out?

To asses the possibility of a chemical change, we need to look into the spontaneity as given by the thermodynamic Gibbs' Free energy change at constant T and P. For your particular case, the change occurs when an electron accepting ion is present to form the crystalline compound after the electron exchange, in solution where the hydration releases energy or in certain solutions such as ammonia, where solvation of free electrons also take part.
Case (1). The changes involved are $$\ce{Na_{(s)}->Na_{(g)}^+ +e^{-}}$$, $$\ce{M_{(g)} +e^{-}->M^{-}}$$ $$\ce{Na_{(g)}^+ + M_{(g)}^{-}->NaM_{(s)}}$$
Where $M$ is an electronegative atom.The ionization energy ($\Delta H_{I}$) and the sublimation energy ($\Delta H_S$) required in the first transformation need to be compensated for by the electron gain enthalpy($\Delta H_{eg}$) released in the second transformation and the lattice enthalpy($\Delta H_{F}\text{lattice}$) released in the final transformation. If it is compensated, the change will be spontaneous. Moreover, the enthalpy released (negative) must also compensate for the lost entropy since gaseous $M$ is converted to an ordered crystal. $\Delta G=\Delta H-T\Delta S$ tells you the compensation required for $\Delta G$ to be negative.
($|\Delta H_{eg}+\Delta H_F|-|\Delta H_I+\Delta H_S|>|T\Delta S|$)
Case (2) Elemental sodium in water might lose electrons to the hydrogen evolving hydrogen gas and forming dissolved sodium hydroxide.
$$\ce{Na_{(s)}->Na_{(aq)}^+ + e^{-}}$$
$$\ce{H2O +e^{-}->\frac 12H2 +OH-}$$
here, the relesed hydration energy compensates the required sublimation energy in the first step and the negative electrode potential in the second step. There is net positive entropy change which favours the reaction, making $\Delta G$ negative.  ($|\Delta H_{hyd}(\ce{OH^-,Na^+})|+|T\Delta S|=|\Delta G|$)
Case (3) very similar to case (2) but here the solvation is both of free electrons and that of metal ions by ammonia (liquid). The equations are similar except that now instead of the hydration enthalpy, solvation enthalpy will be released. the entropy change is also positive here, supporting the change.


text:
What is the group number or name for elements between group 3 and 4 (F-block) on the periodic table?
What is the group number or name for elements between group 3 and 4 (F-block) on the periodic table?  I noticed when I look at a periodic table of elements, the elements of the F-block between group 3 and 4 doesn’t have a group number.  Are they just refer as Lanthanide,and Actinide or Inner Transition Elements,  But why doesn't they have a group number, it least it appear that way.
Link to a wide version of Periodic table of Elements so you can see what I am talking about.
http://en.wikipedia.org/wiki/Template:Periodic_table_%28wide_large%29

I suppose IUPAC hasn't bothered with group numbers for the lanthanides/actinides quite yet because there would only be two elements for each group, and because there are few vertical similarities among the f-block elements. The entire row of lanthanides behaves very similarly due to the 4f orbitals being anomalously compact and not being available for bonding or ionization. The first half of the actinides is quite different to the lanthanides, as the 5f orbitals there can actually participate in chemical interactions, though the second half becomes more similar as the 5f orbitals are once again strongly trapped to the atom, this time by the very high effective nuclear charge.


text:
Why does electronegativity increase across a period?
What explains why electronegativity increases as you move across a period? Does it have something to do with the shielding effect of added electrons?

Good question. In reality electrons will be revolving around the nucleus in high speed. It is difficult to imagine, how electron will be added to electronegative element. I don't know whether some piling up of electrons would be there or not, when electron is added under such high speed. It is also to be noted that electron in high speed shows wave nature and not particle nature. As I am yet college student I haven't got convinced with those wave and particle nature.

I had a simple doubt that, even though atom will neutral, how could there be electric field, which is the reason for attraction of electrons of other elements by an electronegative atom. When I learnt coulomb's law, I got the answer. When the distance between two dipole charges is zero, there will be no electric field and no attraction or repulsion. As in atom we have protons and electrons separated by a certain distance, electric field will not be zero, so there will be attraction or repulsion. Though this was not the answer for your question, I thought it would be helpful for you.

The electronegativity is the tendency of an atom or a functional group to attract electrons (or electron density) towards itself.

This is because the attraction of bonding electrons by an atom increases with nuclear charge (Atomic Number) and decrease of atomic radius. Both these factors operate as we move to the right in period.

Does it have something to do with the shielding effect of added electrons?

The electron cloud in the inner orbits act as a shield and reduces the nuclear attraction to the outer orbits. Because of the shielding effect, the tendency of nuclear attraction reduces and thus electronegativity reduces.

ashu's answer is good, but it does not get to the real reason that electronegativity increases across periods. Electronegativity is dependent on other properties that are more intuitively dependent on position within a period.
The absolute (Mulliken) electronegativity $\chi$ (which correlates with the more familiar, but more abstract, Pauling electronegativity) of an atom is the arithmetic mean of its first ionization energy and its first electron affinity
$$\chi_A = (E_{i,A}+E_{ea,A})/2$$
These values behave in a predictable manner related to the effective nuclear charge idea ashu mentions. Consider the data from the second period (all values in kJ/mol):
    Ei      Ea     X
Li  520     60     290
Be  900     0      450
B   800     27     414
C   1087    122    605
N   1402    ~0     701
O   1314    141    728
F   1681    328    1005
Ne  2081    ~0     1041

Notice that $E_i$ increases and $E_{ea}$ also increases (though not as regularly) from lithium to neon. The electronegativities also increase.

Across the period, electronegativity increases. This is mainly due to the increase in effective nuclear charge. As the number of core electrons remain constant, the number of protons increases. The protons actually draw those electrons closer to the nucleus.

That was already said in earlier answers.

Along a period the effective nuclear charge increases thus decreasing the atomic radius. This favours the increase in electronegativity and therefore it increases along a period from left to right.

Electronegativity means tendency to attract electron,along the period atomic size decreases,atomic size decreases means there is more attraction of nucleus towards electron,more attraction of nucleus means there is more positive charge,and more positive charge means there is more tendency to attract electron.thus electronegativity increases.

Well the electronegativity increases across the period because the electrons are being added onto the same energy level,this increases the number of electrons of an atom,the increase of electrons of an atom also leads to increase in clear charge,in fact the nuclear charge increases more,this leads to attraction of electrons by the nuclear charge,hence electrons get attracted towards the nuclear charge hence electrons are not easily lost!

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text:
What pressure difference is needed across the length of a 15 cm vertical drinking straw in order to drink a liquid of density 1.0 g/cm³?
What pressure difference must be generated across the length of a $15\ \mathrm{cm}$ vertical drinking straw in order to drink a water-like liquid of density $1.0\ \mathrm{g\ cm^{−3}}$?
I am really unsure how to proceed on this one. I was thinking of using the ideal gas law for the air inside the straw initially, and then to increase the pressure difference so the volume of gas in the straw decreases. But this doesn't get me far at all. If any could help me with this one I would be very grateful!
P.S. This was probably clear but I am very new to chemistry, I am a mathematics student.

This is basic hydrostatics, so it's technically physics, not chemistry. The problem has nothing to do with ideal gasses, you solve it simply by a force balance. See here and here for the theory requirements needed to solve your problem. Otherwise, try posting on Physics.SE.

This is not thermodynamics. I thought mathematics students also study physics am I wrong ?

@Hexacoordinate-C Indeed in my mathematics study I did study quite some physics, but mostly in the areas of relativity theory, waves, optics and some assorted problems that give rise to (partial)-differential equations. The course I did on thermodynamics was just for fun, and this is one of the questions that the professor mentioned at some point. I then (naively) assumed that that must mean it lies within the realm of thermodynamics.

The formula is $p =dgh$
$d$ is density = $1000\ \mathrm{kg/m^3}$
$g$ is gravity = $9.80665\ \mathrm{m/s^2}$
$h$ is height = $0.15\ \mathrm m$
$$p = 1.0\ \mathrm{g\ cm^{-3}} \times 9.80665\ \mathrm{m\ s^{-2}} \times 15\ \mathrm{cm} = 1471\ \mathrm{N/m^2}$$
This is the pressure you must overcome (about 0.21335 psi) to drink your water.
See the note by @Nicolau Saker Neto for references.

This is ridiculous to use $g = 9.80665 m/s^2$ because the value of $g$ depends on where you are in earth. Using $9.81$ is enough. :-)


text:
What is the joke that is shown in this picture?
There is supposedly a joke shown in this picture relating to periodic trends.
Does anyone know what it is? And could they also explain where the elements on the eye exam came from?

To me it's just a joke about how instead of random letters they're using random element symbols, probably inspired in part by the fact that several elements are represented by a single letter. I fail to see anything related to periodic trends.

Not entirely random. The elements are arranged in the same order they are found in a periodic table, reading as you would normally. It merely represents scientists using the periodic table as an eye chart.

I find it bizarre that Mg was dropped.  I suspect the artist found this version of the periodic table aesthetically pleasing?

As commented by @long, the elements shown are not completely randomised. The elements in all the rows, except in the 3rd row ($\ce{Na, K, Ca}$), are arranged in same order as they are given in periodic table.
After studying them on periodic table for 15 minutes, it seemed to have no meaning. There are no periodic trends. $\ce{H, Li, Na, K, Rb}$ belongs to IA group, $\ce{Be, Ca, Sr}$ belongs to IIA group. And they are arranged in a random way, having no meaning.
The picture is clearly a joke showing that when scientist have eye check-up, they have to read elements instead of alphabets.

And that their left eye focuses on the left side of the periodic table.


text:
Calculating charge on amino acid from pKa
I have been given $\mathrm pK_\mathrm a$ values of an amino group, a carboxyl group and a side chain of cysteine. How can I find the ionic charge on it at different $\mathrm{pH}$ values?

For this, you can use the Henderson-Hasselbalch equation. Using the degree of dissociation, $\alpha$, this can be written as
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\alpha}{1-  \alpha}$$
Rewriting to solve for $\alpha$:
$$\alpha = \frac{1}{10^{\mathrm{p}K_\mathrm{a} - \mathrm{pH}} + 1}$$
As stated above, $\alpha$ is the degree of dissociation, meaning the degree at which $\ce{H+}$ is dissociated from the ionizable group. At $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, $\alpha = 0.5$, meaning that $50\%$ of the ionizable groups in question are deprotonated. If $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + 1$, about $90\%$ of all groups are deprotonated and if $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + 2$, about $99\%$ of all groups are deprotonated. (If you want to, take a look at the shape of the $\alpha$ distribution for $\ce{-SH}$ in cysteine at Wolfram Alpha.)
From a textbook I found the following $\mathrm{p}K_\mathrm{a}$ values for cysteine:
\begin{align}
 \mathrm{p}K_\mathrm{a}(\ce{-COOH})   &= 1.9\\
 \mathrm{p}K_\mathrm{a}(\ce{-NH_3+}) &= 8.35\\
 \mathrm{p}K_\mathrm{a}(\ce{-SH})     &= 10.5
\end{align}
From these values, $\alpha$ can be calculated for each ionizable group at the desired pH and this will give you the net charge of the amino acid. Upon deprotonation, the following changes in charge occur for the ionizable groups:
\begin{array}{lclcr}
 \ce{-COOH} &:& 0 &\rightarrow &-\\
 \ce{-NH_3+}&:& + &\rightarrow &0\\
 \ce{-SH}   &:& 0 &\rightarrow &-\\
\end{array}
As an example, let's calculate the charge of cysteine at pH 10. Using the Henderson-Hasselbalch in a spreadsheet yields the following $\alpha$ values:
\begin{array}{lcr}
 \ce{-COOH}  &:& 0.9999\\
 \ce{-NH_3+} &:& 0.9781\\
 \ce{-SH}    &:& 0.2403\\
\end{array}
This gives the following total charge for cysteine at pH 10:
$$(-1)\cdot0.9999+(+1)\cdot(1-0.9781)+(-1)\cdot0.2403=-1.218$$

How can I calculate pI from this ?

No problem! pI is the pH at which the molecular charge is zero. This can be calculated using the tools explained above. The easiest approach would probably be to find pI by calculating total charge at different pH values in a spreadsheet.

So do I have to do this by trial and error ? No other option ?

You could use the equations above to derive an analytical expression. What you need is an expression for each of the three alphas (as given above) and the "equation" used to calculate the total charge at the bottom of the answer with total charge set to zero. From this it should be possible to solve for pH, i.e. for pI. Personally, I think using a spreadsheet is easier.

ok so i would have to use variables and solve algebra ...oohooh...I undrstnd why u find spreadsheet easier :P

@oveoyas I suddenly feel I am not able to grasp the concept of Henderson-Hasselbalch and to derive degree of ionisation. I am trying to make organic buffers... I am at a loss. Do you have a resource that I can use to get a grasp of it?

I just wanted to show you how you could set up the equation if you don't know that :-). If you find an expression for the ionic charge of one side group as a function of pH, you're well on the way. You just do that for all the side groups, and add them together, and say that overall you want that to equal 0.
Then you can solve for pH in a program that does such a thing. You can also graph this, and find the isoelectric point (pI) by visual inspection. I've done both below, mostly for fun, since oveoyas has provided a sufficient answer :-)

Do mention the "program that solves pH".


text:
Red glow in the dark pigments
I want to do some spectrometry experiments with glow in the dark paints. The red paints I've read product details for all seem to require a uv light to charge. Are there any red glow in the dark paints that can be charged with visible light?
I was also wondering why most red phosphorescent pigments don't have the ability to glow as long as green pigments.

Do you mean fluorescent or phosphorescent?

Oops, I meant phosphorescent. Thanks for clarifying.

The wikipedia page on Phosphoresence has a red pigment. Calcium Sulfide. I can imagine that the lower energy (higher wavelength, lower energy) is more stable because the triplet is not as high in state energy as lower wavelength emitters. Because of the stability the time to emit is longer and the intensity lower.


text:
Is there any substance that's a 0-0-0 on the NFPA diamond?
Is there any substance that's a 0-0-0 on the NFPA diamond? Basically a perfectly safe chemical.

Since I'm closing this, I guess I should offer some kind of explanation. Essentially, it comes down to the fact that we do not need an exhaustive enumeration of every harmless chemical known to man. Furthermore, nobody can claim that there aren't enough answers already. Thus, I don't see any reason why this needs to be kept open.

Not sure how this questioned could be more focused.  It is a well-defined YES/NO question and the answer is clearly YES.  Voting to reopen.

Water!

You can find plenty of others if you perform this search on Wikipedia (but then, go check the MSDS for the individual compounds before doing anything serious with them).

Wow, that was really obvious. I feel dumb.

I think this does not hold for all "types" of water. Imagine what pure water could to the the osmotic pressure in your cells. I don't know for sure, but I wouldn't drink a liter of distilled/deionized water.

A liter in your stomach would probably be ok, a liter in your lungs would cause trouble.

Any source on that?

@jori a Litre of distilled water would be far less troublesome than 10 litres of standard tap water. The difference in osmotic pressure from being deionised is small and not a notable hazard.

Since water can actually put fires out, maybe it should get a -1 red rating.

@JuanPerez if even possible

Sodium chloride. Has a 0-0-0 rating with no special hazards.

Compressed nitrogen and oxygen gases are rated 0-0-0, but it must be understood:

Any relatively inert gas such as nitrogen could still suffocate you. Presumably the 0 rating is meant to interpret a base level of risk, not zero, as distinct from a higher level.

Oxygen, of course, is listed as an oxidizer on the bottom of the diamond (special hazards).

Also Calcium carbonate. Same thing as sodium chloride- 0-0-0 with no special hazards.


text:
If I add some KOH, where will the reaction shifts?
I just got a final test and one of the question is:
$\ce {Ni^2+}$ + 6$\ce{NH_3}$ $\ce {<=>}$ $\ce{[Ni(NH3)6]^2+}$
If we add some aqueous $\ce{KOH}$ after the reaction reach equilibrium, where will the reaction shifts?

What are your initial thoughts?

Think about complex stability, NH3 dissociation equilibrium and Ni hydroxide precipitation.

@jonsca i think NH3 is a base and KOH is a base, so it looks like add NH3 on reaction, so the reaction will shifts to right

I think @RauruFerro has the right idea.

The equilibrium will shift to the left, as $\ce{Ni^2+}$ is removed from the equilibrium by the precipitation of $\ce{Ni(OH)2}$:
$$\ce{Ni^2+ + 2 OH- \rightleftharpoons Ni(OH)2\downarrow}$$
The precipitate forms when $[\ce{Ni^2+}]$ and $[\ce{OH-}]$ are sufficiently high to exceed the solubility product of $\ce{Ni(OH)2}$ ($K_{sp}=5.48 \times 10^{-16}$ at 25 °C, source). The concentration of $\ce{Ni^2+}$ before addition of $\ce{KOH}$ depends on the stability constants of the ammin complexes (there are several species in equilibrium with each other, as the ligands are exchanged stepwise, and not all at once). Nickel hydroxide will dissolve again when $\ce{NH3}$ is added, as this shifts the equilibrium of ammin complex formation to the right.


text:
Isothermal vs. adiabatic compression of gas in terms of required energy
I am working on an instruction manual of sorts to be used with an introductory course in thermodynamics. As an example of problem solving, I attempt to answer the following question:

If you have little energy available, would you rather use an
  isothermal or an adiabatic process to compress a gas?

My analysis is as follows:
The work required to compress a gas from volume $V_0$ to volume $V$ is $$W=-\int_{V_0}^{V}PdV$$
For the isothermal compression, the ideal gas law, $P=\frac{nRT}{V}$, is used and inserted into the equation above:
$$W=-\int_{V_0}^{V}\frac{nRT}{V}dV=-nRT\int_{V_0}^{V}\frac{dV}{V}=nRT\ln\left(\frac{V_0}{V}\right)$$
For the adiabatic compression, $PV^\gamma=P_0V_0^\gamma \iff P=P_0\left(\frac{V_0}{V}\right)^\gamma$ is valid. Thus, the required work is $$W=-\int_{V_0}^{V}P_0\left(\frac{V_0}{V}\right)^\gamma dV=-\frac{P_0V_0^\gamma}{-\gamma+1}\left(V^{-\gamma+1}-V_0^{-\gamma+1}\right)=\frac{PV-P_0V_0}{\gamma-1}$$
       $$W=-\frac{P_0V_0-PV}{\gamma-1}$$
This result indicates that the magnitude of the required work depends on the properties of the gas in question. Thus, it may vary which of the two compression processes requires the lowest amount of energy.
Now, finally, to my question:
Intuitively I would assume that the isothermal work is usually lower than the adiabatic work, as compression lowers the volume of the system and therefore usually increases the temperature if heat exchange with the surroundings is not allowed. Isothermal compression requires heat transfer to the surroundings to maintain constant temperature, lowering the pressure of the system and thus lowering the resistance to compression compared to the adiabatic compression (where heat exchange is not allowed).
Is the isothermal work actually smaller than the adiabatic one in most cases or is my argument flawed?

A better way to express the adiabatic work is
$$W=-\frac{P_0V_0-PV}{\gamma-1}$$.

It should be noted that the expressions derived by the OP apply only to reversible processes, since s/he has assumed (without mentioning this explicitly) that $P_{ext} = P_{gas}$.  Since this is designed for a course, this should be made explicit.

To solve this, try to use what I call the "graphical apparatus". For an isothermal process:
$$
\begin{align}
PV&=\text{constant}\\
P\mathrm{d}V&=-V\mathrm{d}P\\
\frac{\mathrm{d}P}{\mathrm{d}V}&=-\frac{P}{V}\\
\end{align}$$
for adiabatic process:
$$
\begin{align}
PV^\gamma&=\text{constant}\\
\frac{\mathrm{d}P}{\mathrm{d}V}&=-\gamma\frac PV
\end{align}$$
Therefore, starting at the same point on a P-V graph, the curves for an adiabatic and isothermal processes will diverge and the adiabatic curve will have a steeper slope. For the same reduction in volume (the graph in the picture is for expansion, not for contraction. In case of contraction, the curves will be reversed, i.e. adiabatic curve will be above the isothermal curve, and will enclose greater area under it for the same reduction in pressure), more area will be enclosed by adiabatic, and since the area $\int P\mathrm{d}V$ gives the work required, isothermal work is smaller than adiabatic for the same reduction in volume.
Your argument is correct. To provide more mathematical support to it, you can observe the fact that it is both increase in temperature and reduction in volume which increases the pressure in adiabatic process and only reduction in volume increases pressure in isothermal process. The exponent of volume in adiabatic equation ($PV^\gamma=K$) is $\gamma>1$, as compared to 1 in the isothermal equation. Hence, pressure change is more sensitive to volume change and is larger in magnitude.

But the work is given as $-\int PdV$ not as $\int PdV$ .

Your final equation for the isothermal work should read (you made a sign error):
$$W=nRT\ln{\frac{V_0}{V}}$$
You can make the comparison with the adiabatic case much more easily if you re-express the adiabatic work as:
$$W=\frac{P_0V_0}{\gamma - 1}\left[\left(\frac{V_0}{V}\right)^{\gamma -1}-1\right]=\frac{nRT_0}{\gamma - 1}\left[\left(\frac{V_0}{V}\right)^{\gamma -1}-1\right]$$
But, for the isothermal case, $T=T_0$. So for the isothermal case,
$$W=nRT_0\ln{\frac{V_0}{V}}$$
Now you can see how much more easily it would be to compare the two amounts of work.

We can prove it wrong .
See my comment on the above answer .

@Koolman PdV is the work done by the system on the surroundings.  -PdV is the work done by the surroundings on the system.  Do you disagree with this?

Yeah , according to chemistry conventions Work = $-\int PdV$ which is work done by gas (system ).

https://i.stack.imgur.com/UtzRZ.jpg
Given on https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Thermodynamics/Path_Functions/Work

Your reference is incorrect.  I stand by what I said.  Don't believe everything you read on line.   Learn to think for yourself.

Hello Chester, I think there's some confusion here. Wrt the IUPAC 2007 document page 56: equation given is $\mathrm{d}U=\mathrm{d}Q+\mathrm{d}W$. This makes it clear that the work done by system is taken $\mathrm{d}W=-P\mathrm{d}V$ i.e. with the negative sign. The footnote also states: "$W>0$ indicates an increase in the energy of the system", which can only happen when we define $W=-\int P\mathrm{d}V$. This is also  what Koolman was trying to convey in his comment to Satwik's answer.

@GaurangTandon  The document may say that, but the analysis in the document is incorrect.  Can you spot the error?  The work done by the system on the surroundings is always dW=PdV and the work done by the surroundings on the system is always dW=-PdV, irrespective of  which form of the 1st law equation is used.

@GaurangTandon The error in their analysis is the following: In the example, the compressional work done by the surroundings on the system is obviously positive, and the pressure is obviously positive.  But,  their diagram, $\Delta h=(h_{i}-h_{f})$, rather than $(h_f-h_i)$.  So the work done by the surroundings on the system should really be $W=-Pa\Delta h=-Pa(h_f-h_i)=-P(V_f-V_i)=-P\Delta V$.  They just screwed up the math.

@Koolman  See my comment to Gaurang Tandon detailing the error made in your cited reference.

@ChesterMiller Thanks for your reply! Except that when you write "In the example" and "but their diagram", I am unable to see what diagram you're referring to. The document I linked you to has no diagram on page 56. Which page are you referring to?

@Gaurang Tandon I was referring to this link posted by Koolman:  https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Thermodynamics/Path_Functions/Work  He has been arguing that the work done by the gas on its surroundings is -PdV, and the work done by the surroundings on the gas is +PdV, and I have been arguing the opposite.  My previous post points out the error in the analysis contained in the link he cited.

@GaurangTandon Each and every textbook on thermodynamics correctly states that the work done by the gas on its surroundings is PdV, and the work done by the surroundings on the gas is -PdV.  See Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics or Denbigh, Principles of Chemical Equilibrium

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


text:
What is the reason behind the characteristic colours of elements?
I've been trying to find out about this for some time now and when I searched online, I couldn't get any proper reasons.
My question is: What is the reason for the characteristic colours of elements? For eg: Why does Sodium have a characteristic colour of silvery-white?
Is there any solid scientific reason behind this?

Colour is a property of the light reflected or transmitted by any substance. Depending on the substance's characteristics (energy levels of constituent levels, bonding, intermolecular forces), there are certain available energy levels (molecular orbitals, bands etc) for the electrons. Whenever any wide spectrum energy is available to a compound (considering only the visible spectrum), then depending on the various energy gaps between the available to the electrons, different wavelengths will be selectively absorbed and re-emitted, but now in any random direction, therefore the reflected and the transmitted light will be partially deficient n these wavelengths. The absorption and re-emission is very characteristic of all substances, and hence, the wavelengths absorbed and the resultant spectrum (transmitted or reflected) will be characteristic. Depending on this spectrum, various colours can be seen for different compounds.
Sometimes, the energy gap between the electron energy level is very high and therefore the light in the visible spectrum cannot excite the electron. No wavelength will be absorbed and it will act as a transparent glass. If say, the gap corresponds to a energy gap of $E=h(4000A)$, then the blue light will be absorbed and the remaining light will be white-blue=yellow. This is the general way in which colors are produced. The energy gaps can be anything, the band gaps in metals, molecular orbital energy difference in covalent compounds, orbital energy difference in isolated elements, or simply the energy difference between the various vibrational modes in a gas/solid.


text:
Selectivity in aldol condensation between pivaldehyde and acetone
In a crossed aldol condensation (condensation between aldehyde and a ketone), the primary products are such, that the carbonyl group of the ketone remains intact, and the $\alpha$-hydrogen of ketone participates, i.e. gets removed by the base to form the resonance stabilised nucleophilic enolate which later attacks the aldehyde. These are formed together with the self-condensation products, but the reverse products (aldehyde-derived enolate + ketone) are rarely formed.
I had a look at the mechanism, and also how a base catalyses condensation by favouring enolate formation.

Why is the ketone involved in enolate formation and not aldehyde in
  crossed/mixed aldol-condensation reactions?

Your example might not be very well chosen since in the given case, the aldehyde doesn't contain any $\alpha$-hydrogens, so it can't be enolized.

@Philipp I did not have my scanner yesterday. The only example I could post was what I found on the net, which was unfortunately this.

Ketones have two alkyl groups attached to the carbonyl moiety, which are electron-donating by hyperconjugation. That is, the overlap of $\ce{C-C}$ and $\ce{C-H}$ $\sigma$-bonds with the $\pi$-type molecular orbitals of the carbonyl increases the electron density around the carbon of the carbonyl group, thus making it less electrophilic.
Alternatively, I think it's also equally (if not more) valid to consider the reactivity from the perspective of the frontier orbitals. Alkyl substituents tend to slightly lower the energy of the HOMO of the carbonyl (by comparison to hydrogens), but concomitantly they raise the energy of the $\pi^{*}$ LUMO of the carbonyl, into which the electrons of the enolate would flow, resulting in lower electrophilicity (since the energy of the tetrahedral intermediate and transition state resulting from attack by the enolate are going to be correspondingly higher).


text:
What is the edge of a diamond like?
The chemical structure of a diamond is defined as an endless lattice in which each carbon atom is covalently bonded to four other carbon atoms situated at the four ends of a tetrahedron.  But of course they aren't endless; diamonds have edges and faces.  So what happens when the lattice reaches the edge and there aren't any other carbon atoms to bond to?  What is the "surface layer" of a diamond composed of?

This is a nice question. Although I do not speak with authority on diamonds, most crystal surfaces have several imperfections which mitigate the hanging valencies that would otherwise appear for the atoms (constituents in general) at the surface.

Impurities. This is the primary method which helps satisfy the unfulfilled valencies at the surface. Many metallic, non-metallic or compound impurities are usually incorporated throughout the crystal, but are usually high at the surface. By bonding with these excess impurities, the surface carbon atoms can fulfill their valencies. An extremely pure crystal surfaces (lots of "hanging" valencies) will quickly adsorb most materials that it comes in contact with. Chemisorption or physisorption depends on the nature of the crystal, metals prefer physisorption while non-metals have at least unimolecular chemisorption usually.
Another way to mitigate surface valencies is distortion. The surface layer might not follow the exact space-filling Bravais lattice structure and may form short range distortions in structure to try to make the unfulfilled valencies satisfied. There might be kinks, depressions, steps etc which at least help in completing the valencies of a few atoms (at the boundary of distortion, say, a "step" structure).
If none of these operate, there is an inherent surface energy or surface tension between any two phases which quantifies the desire of the surface atoms to fulfill their hanging valencies to resemble the atoms in bulk which are stably attracted. In case of metallic and ionic crystals, this is comparatively low, but may be high in a covalent crystal like diamond.

In all cases, the surface layer of constituents (actually first few layers) are different from the atoms in bulk in energy, stability, forces acting (unbalanced on the surface and balanced in the bulk) and their structure/orientation.

Nice answer too. So "what it looks like" depends on context of what is important. If we are trying to electroplate (1) is very important. If you want to polish the stone, like for diamond jewelry, then (2) is very important. If you wanted to use the material as a catalyst for a chemical reaction then (3) is very important.

At the surface of a diamond, the carbon atoms are generally bound to various other atoms (or groups of atoms) with lower valence.
In principle, pretty much anything that could act as a substituent group in organic chemistry could also be bonded to the surface of diamond, but in practice, you're most likely to see simple groups with low reactivity, mostly either plain hydrogen or various oxygen-containing groups (I suspect mainly hydroxyls and/or ketones).
According to Wikipedia (which itself cites Harlow 1998, The nature of diamonds):

"Naturally occurring diamonds have a surface with less than a half monolayer coverage of oxygen, the balance being hydrogen."

Although the diamond lattice itself is very stable and inert, these surface groups can participate in chemical reactions, and indeed, it's possible to alter the composition of the surface layer by reacting it with appropriate chemicals.  For an example, see this random paper I found on Google.


text:
Ozonolysis conditions for benzene
What are the precise conditions under which benzene undergoes ozonolysis? All anyone can tell me is that it undergoes ozonolysis under "vigorous" conditions. But, what vigorous conditions?

What are the precise conditions under which Benzene undergoes ozonolysis?

In Über das Ozobenzol, an early German publication on the reaction of benzene with ozone, Ber. Dtsch. Chem. Ges., 1904, 337, 3431-3433 (DOI)
, the authors write

Leitet man in solches Benzol 1–2 Stunden lang bei einer Temperatur von 5–10° einen 5-procentigen Ozonstrom, so scheidet sich eine gelatinöse Masse aus, welche stark opalisirt. […]
Lässt man das unangegriffene Benzol verdunsten, so bildet es eine weisse, amorphe Masse von furchtbar explosiven Eigenschaften. Selbst beim Uebergiessen mit warmem Wasser detonirte diese Substanz heftig.
Uebergiesst man aber das gelatinöse Product mit eiskaltem Wasser,
so entsteht eine krystallinische Modification, die wohl möglich noch
heftiger als die amorphe bei der geringsten Berührung, ähnlich wie
Jodstickstoff, explodirt.

To sum this up:

The authors prepared the primary triozonide of benzene by passing a stream of ozone though it at a temperature of 5–10 °C.

Once the remaining benzene evaporated at room temperature, a white, amorphous mass with frightfully explosive properties was obtained.

When warm water was poured over it, this mass detonated.

When cold water was poured over the amorphous, jelly-like mass, a crystalline modification was obtained, which detonated on the slightest touch, similar to to nitrogen triiodide.

In the 1950s, a group of Dutch researchers published a series of articles about their kinetic experiments on the ozonization reactions.
The second article in the series, Recl. Trav. Chim. Pays-Bas, 1951, 70, 1005–1019, (DOI) notes that the ozonisation of benzene was performed in chloroform at a temperature of 242 K (–31 °C).

In the late 1960s, ozonization of benzene again gained some interest as a possible source for glyoxal in industrial synthesis.
In Glyoxal from Ozonolysis of Benzene, published in Ind. Eng. Chem. Prod. Res. Dev., 1969, 8, 89–92
DOI), the authors describe the ozonolysis of benzene in a mixture of acetic acid and nitromethane at temperatures between –20 °C and 15 °C.

Summary

All anyone can tell me is that it undergoes ozonolysis under "vigorous" conditions.

Apparently, your sources never bothered to check the literature! While the ozonide itself may detonate on touch, claims on the "vigorous" conditions necessary for its synthesis seem to be fully unsubstantiated by the literature!

To be sure, the ozonolysis of benzene(C6H6) [As given in the book "GRB Organic Chemistry for JEE (Mains and Advanced)" written by O.P. Tandon ] will be carried out at low temperature(196 to 200 Kelvin). Benzene when treated with ozone in prescence of an inert solvent like chloroform(CHCl3) or CCl4 or CH2Cl2, benzene triozonide is produced which on reduction with zinc dust and water yields 3 moes of glyoxal.

Ozonolysis is the cleavage of an alkene or alkyne with ozone to form organic compounds in which the multiple carbon–carbon bond has been replaced by a double bond to oxygen.

The ozonolysis of benzene forms Glyoxal through the formation of benzene Triozonide as an intermediate.

The question is about the conditions for the reaction, not the mechanism. Although this mechanism is correct, it does not answer the question.


text:
Making ice from carbonated water
Sorry if this is not the place for asking this question, but I'm going to do a real life project and don't know where else to ask.
I'm going to prepare and sell snow cones next summer. For making snow cones, I was told by ice cone makers that ice should come from carbonated water.
As in my country is quite difficult to find a proper ice factory that could produce ice with clean water, I'm planning to make my own ice for snow cones.
I was thinking to buy bottles of carbonated water, pour them in many plastic bags and freeze those bags in my refrigirator. In that way I could get carbonated ice.
My question is:

Do you think that pouring carbonated water in plastic bags for producing ice would be a good idea? Or could I get adverse effects like water loosing sodium ?

If the carbonated water comes in plastic bottles you can freeze it directly in the bottles.

If I do that, I would have a problem taking out all the ice form the bottle.

Only if you don't want to cut the bottle. It would help also if you cool the bottles before pouring into bags. It's worth trying.

Could you define "good idea"?

There should be no issue with putting carbonated water in plastic bags and freezing them (considering they're clean and sterile, of course). I imagine the benefit of freezing carbonated water instead of regular clean water has to do with the fact that there is much more dissolved gas in carbonated water. As the water freezes, the gas is expelled from the ice (only the water molecules will solidify), creating many pockets inside the solid ice which would weaken its structural integrity. This would likely make it much easier to shave or break into small pieces for a snow cone. I've never frozen carbonated water myself though. You should be a bit careful because as the carbonated water freezes, the carbon dioxide it releases will create pressure in a closed container, so you may end up with torn plastic bags, ruptured bottles, or a damaged freezer in the worst case.


text:
Corrosion of Aluminum by Baking soda
An aqueous solution of sodium hydrogen carbonate corrodes aluminum foil quite readily. How do you explain this?

Sodium aluminate is produced. The reaction is slower than the usual $NaOH + Al$ and liberates $CO_2$.

Is hydrogen produced also, like with NaOH? I guess not, but I want to make sure.

Nope, water instead.


text:
Isoelectric point pH, mV, for a metal oxide nanopowder
I'm trying to make a solution of $\ce{TiO2}$ that will coat stuff in as dense a monolayer as possible.  To do so, I'm using nano sized $\ce{TiO}$ particles and trying to get them to get as close to each other as possible.  From what I understand, the way to get these little guys to get close to each other is to get the solution to its isoelectric point thereby eliminating any repulsion caused by charge.
So: what pH should I set my water and $\ce{TiO2}$ powder at to attain my isoelectric point?  I ask because finding such a number has proven to be elusive.  I've found pH numbers from 3.8 to 9 to attain isolectricity for such a solution.  Does anyone have any experience with the stuff such that a true pH to attain isoelectricity revealed?
In addition, the meter to which I have access for pH also measures mV.  Is there a way to relate mV to the isoelectric point?  Is it as obvious as it seems i.e. keeping adding an acid until my little meter says that mV equals zero?  Or is there something more going on (there always is)?  Are this mV value and the isoelectric point even related?
Thanks for any help!


text:
Why is Potassium in French and English not called Kalium?
The symbol from the periodic table for Potassium is "K". This fits my native language (dutch), where Potassium is called "Kalium". The same question applies to Sodium (Na), which again in Dutch is called "Natrium". A quick look at Wikipedia showed that Potassium is mainly used as a word in Roman based languages (French, Spanish) and to my surprise in English. What is the history behind this?

You can read this article for more information about the history  of nomenclature of potassium.

@NilayGhosh that looks like it would answer the question to me - mind if I write it up properly?

@tardigrade no problem.

(with thanks to @NilayGhosh, who provided the key source for much of this answer)
Short answer: Potassium was called Potassium (and Sodium Sodium) by Humphry Davy, who first isolated both; it was then renamed Kalium in Germanic countries because of a previous naming proposal by Klaproth, who was the first to show that potassium carbonate and sodium carbonate were different.
Until at least the Middle Ages no distinction was drawn between potash (potassium carbonate) and soda (sodium carbonate). Martin Heinrich Klaproth first distinguished them in 1797, and suggested the name 'kali' for the first and 'natron' for the second.
Metallic potassium and sodium were then isolated for the first time in 1807 by Sir Humphry Davy, by electrolysis of potassium hydroxide (KOH) and sodium hydroxide (NaOH) respectively. He named his discoveries Potassium and Sodium in the Bakerian lecture at the Royal Society on the 19th of November the same year:

Potasium and Sodium are the names by which I have ventured to call the
  two new substances

(source)
(During the lecture he also apparently produced a nugget of metallic potassium and tossed it into a flask of water, "where the lump skittered around the surface of the water before exploding in lavender flames".)
This paper was translated into German by Ludwig Wilhelm Gilbert (supposedly in Annalen der Physik, although I can't find the original paper in any of the volumes from  1809). According to the article on vanderkrogt.net Gilbert replaced Potassium with Kalium and Sodium with Natronium "in agreement with German nomenclature" (i.e. following Klaproth's suggested names). From this point on, generally speaking, English- and French-speaking countries followed Davy's naming system, while the Germanic countries followed that of Klaproth/Gilbert, as can be seen from the sidebar lists here (potassium) and here (sodium).
Regarding the atomic symbols mismatch: the international system of atomic symbols was devised by the Swedish chemist Jacob Berzelius, a rival of Davy's, who followed the naming convention used by Germanic countries. Although actually, when first published (1813, in Thomas Thomson's Annals of Philosophy), Berzelius did follow Davy's names and used the symbols Po and So. Within a year he had switched to K/Na. So since the first publication of Berzelius' system actually followed Davy and used Po/So, the real surprise is that we ended up using K/Na.

It derive from the word "potash" (pot ashes - ashes of vegetable).
Sodium hydroxide, which was obtained from the ashes of marine plants, are confused over time with potassium hydroxide, from the ashes of land plants and both were called "alqili", an Arabic word meaning "ashes of plants."
Later, both substances were differed between "vegetable alkali" for land plants and "mineral alkali" for marine plants. Later, they received the English names "potash" and "soda".
As a symbol for potassium is take the first letter of the Latin word "kalium".

While this answer does (somewhat) explain the sources of the names "Potassium" and "Kalium", it does not explain why English ended up adopting the former name, while most other languages settled on the latter.

According to the Dutch wikipedia, the word Potassium is based on the Dutch word "potas". So it's actually derived from your (our) own language. The explanation is the same as the last contributor already explained. It's kind of weird that we ended up using the word Kalium instead of Potassium though. But well, we did.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this, like the other post, does not answer the question why potassium was adopted in English.


text:
Wasn't Rutherford's assumption of gold foil to be made of single layer wrong?
My teacher taught about Rutherford's gold foil experiment today. Sir said that, Rutherford used gold foil of thickness $10^{-7}$ m and concluded atom to be of size $10^{-10}$ m. Sir also said that Rutherford assumed foil to be consisting of only single layer of gold atoms.
What I thought was, if Rutherford used gold foil of $10^{-7}$ thickness, which is greater than the size of atom [$10^{-10}$] which he had concluded. So, gold foil should be made of more than single layer of atoms. But this would be against his assumption. If we assume gold foil to be consisting of more than one layer of gold atoms. Then we will be in need to know the structure of gold, i.e to know whether it has atoms in ABAB.. or AAAA.. or ABCABC... type of arrangement. My book says that gold has ABCABC... or ccp arrangement, it mean that there is no vertical allignment of atoms, thus there will be no vertical allignment of nuclei. It makes significant changes to Rutherford's predictions with respect to deviations. Won't it?

You were confused about the result of the experiment. It did not measure the size of the atom, but the size of the nuclei. Nucleus is the area of concentrated positive charge in the core of the atom, and is only a very small fraction the size of the atom itself.
And the result, $10^{-15}\ \mathrm{m}$, was the size of the nuclei, while the size of gold atoms has the order of magnitude of ångströms ($1\ \mathrm{\mathring{A}}=10^{-10}\ \mathrm{m}$).
The measurement comes from the analysis deflection angle of alpha particles when hitting the nuclei. The one-layer-of-atom assumption leaded to Rutherford to use a mathematical model for one single collision to analyze the angular distribution. Because the nucleus is so small, even if there were a small number of layers, most alpha particles will still only have one collision and the result will be correct.
However, the alpha particle was passing through some hundreds of atoms and many particles actually were deflected multiple times. The calculation would be much more complicated. Even if he knew the number of layers of atoms and tried to do it the right way, he might not be able to arrive at the correct number. Of course people back then would not be able to know that, but nevertheless, the existence of nucleus is still evident and Rutherford did almost get the order of magnitude right.


text:
Are these two reactions equivalent?
I'm supposed to balance a redox equation. I arrive at
$$
\ce{H2O} + \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} \rightarrow \ce{NH3} + 2\ce{S} + 3\ce{OH-}
$$
The "official" solution is
$$
4 \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} \rightarrow 4\ce{NH3} + 2\ce{S} + 2\ce{H2O}
$$
There is no information about if this is occuring in basic or acidic medium. But IMHO in the first equation the charges and all atoms are balanced, so are these equations equivalent?

Let's start with the question: How do we get from your answer to the official answer?  We can transform your answer into the official answer by first adding 3 ammonium ions to each side:
$\begin{aligned}\ce{H2O} + \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow \ce{NH3} + 2\ce{S} + 3\ce{OH-} \\
+ 3 \ce{NH3}+3\ce{H^+} & \rightarrow + 3 \ce{NH3}+3\ce{H^+}\\
\ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow 4\ce{NH3} +3\ce{H^+} + 2\ce{S} + 3\ce{OH-}
\end{aligned}$
Then noting that the $\ce{H+}$ and $\ce{OH-}$ will 'neutralize' one another on the product side, producing $\ce{3H2O}$, one of which will 'cancel' the water molecule on the reagent side:
$\begin{aligned}
\ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow 4\ce{NH3} + 2\ce{S} + 3\ce{H2O} \\
-\ce{H2O} & \rightarrow - \ce{H2O}\\
4\ce{NH4+}+2\ce{S^2-}+\ce{O2} & \rightarrow 4\ce{NH3}+2\ce{S}+2\ce{H2O}
\end{aligned}$
Producing what is the 'official' result.  Now we can ask: are the two answers equivalent?  Because $\ce{NH4+}$ is a weak acid in water, it will dissociate and we can write your answer as:
$\ce{H2O + NH3 + H+ +2S^2- + O2} \rightarrow \ce{NH3 + 2S + 3OH-}$
Your answer is a balanced redox reaction somewhere in between acidic and basic conditions, which is a rather unusual formulation.  I'm not going to go so far as to say, this can't happen, but it is unlikely, and certainly not how we typically describe balanced redox reactions.


text:
How much electricity could be generated by a person's stomach acid?
Motorola has a device that is powered by stomach acid. Given that a battery consumes the electrolyte as it generates a charge, how much power could be generated using a device such as this? Let's assume the electrodes are large enough that the electrolyte is the limiting factor in this reaction. Would the body just generate more acid to compensate? Would there be any physiological side-effects from using such a battery?

Sounds like one for Randall from xkcd

The acid isn't the relevant consumable, the electrode is.

@sehe I was thinking the same thing, although this one may not have enough physic for him.

@Aesin In my understanding both the electrodes and the acid are consumed. Let's assume we have large enough electrodes so that the acid is the limiting factor.

Well a battery works on the principle of dissolving one of the electrodes with the electrolyte. This means that you would have a solution of the metal and electrolyte in your stomach. If consumed in large quantities or in quick succession, then you most probably would notice some physiological effects, however, I would assume they'd be very minor given the mass of the tablet in question. Other than under those circumstances, I would expect the physiological effect to be non-existant. Given this has been approved by the FDA, you can be pretty confident that it's safe.

Thanks, but I was wondering more about the case where the electrodes are large enough that the stomach acid is the limiting factor in the chemical equation, not the tiny microchip-sized battery that Moto is making.

The electricity generate depends mainly from the electrodes. This is the same misconception of lemons battery. In fact the energy of every battery comes from the metals and not from the acid use as electrolyte, as you can't generate energy from lemon juice but from the different potential between copper and zinc coin so you can't generate energy from a stomach acid but with it. However $HCl$ is not used normally for battery so I think you could not produce a big amount of energy even if you  ingest very big electrodes.

I know that the energy doesn't come from the acid. However it stands to reason that if you have only a tiny amount of acid relative to the size of the electrodes you won't have a very good battery.

@Mr.ShinyandNew安宇 yes you are right so It will depend on what you have eat at breakfast and how you are digesting it! Nice strange question!


text:
Concentrations at equililibrium

At $25 ^\circ C$, $K=0.090$ for the reaction
  $$\ce{H2O (g) + Cl2O (g) -> 2HOCl (g)}$$
  calculate the concentrations of all species at equilibrium for the case: $\pu{1.0 g} \ce{H2O}$ and $\pu{2.0 g} \ce{Cl2O}$ are mixed in a $\pu{1.0 L}$ flask.

I don't know where I go wrong in my calculations:
We have $\pu{0.0555 M}$ $\ce{H2O}$ and $\pu{0.023 M}$ $\ce{Cl2O}$. So at equilibrium,
\begin{align}
  \ce{[H2O]} &= 0.0555 - x,\\
  \ce{[Cl2O]} &= 0.023 - x\\
  \ce{[HOCl]} &= 2x\\
\therefore K &= \frac{[\ce{HOCl}]^2}{[\ce{H2O}][\ce{Cl2O}]}
\end{align}
After some moving about I get to
$$-3.91x^2 - 7.1\times10^{-3}x + 1.15\times10^{-4} = 0$$
Where I end up getting $x = -8.74\times10^{-3}$ or $x=8.4\times10^{-3}$.
And this is where it all goes wrong, doing some quick calculations, I see that it does not match what the book says the answer is. So I'm doing something wrong, but I don't know what it is or where it's at.

I suggest you to use this schematic:
\begin{array}{l*{5}{c}}\hline
  \text{Equation} &
    \ce{H2O} & \ce{Cl2O} & \ce{<=> 2 HClO}\\[2ex]\hline
  \text{Initial concentration} /\pu{M} &
    \pu{0.055 mol//L} & \pu{0.023 mol//L} & 0  \\[2ex]
  \text{Equilibrium variation} /\pu{M} &
    - x & -x & +2x \\[2ex]
  \text{Equilibrium concentration} /\pu{M} &
    \pu{0.055 M} - x & \pu{0.023 M} - x & 2x \\[2ex] \hline
\end{array}
At this point you have to substitute into the equilibrium equation you mentioned:
\begin{align}
  0.090 &= \frac{(2x)^{2}}{(0.055 - x)(0.023 - x)}\\
  0.090 &= \frac{4x^{2}}{0.001265 - 0.055x - 0.023x + x^2}\\
  4x^2  &= 0.090 (0.001265 - 0.055x - 0.023x + x^2)\\
  4x^2  &= 0.090 (0.001265 - 0.078x + x^2)\\
  4x^2  &= 0.000114 - 0.007x + 0.09x^2\\
  0     &=3.91x^2 + 0.007x - 0.000114\\
\end{align}
So the positive result is:
$$ x_2 = \frac{-0.007 + \sqrt{4.9\cdot10^{-5} + 0.00178}}{7.8}\approx 0.0046$$
An then equilibrium concentration are for $\ce{[H2O]} = 0.055 - 0.0046 = 0.050$, $\ce{[Cl2O]} = 0.018$ and $\ce{[HClO]} = 2x = 0.092$. This are my results.


text:
What's the difference between enthalpy of combustion and heat of combustion?
Wikipedia has two different pages for enthalpy of combustion and heat of combustion. Their definition is the same and they use the same symbol $\Delta H_c^\circ $
It says that enthalpies of combustion are always negative as these reactions are exothermic. The heat of combustion tables list positive values for all substances.
Why do these values differ in sign?

$|\Delta H_c^\circ|$ =$\text{heat of combustion}$

Looking at Wikipedia for the definitions:
The standard enthalpy of combustion is the enthalpy change when one mole of a reactant completely burns in excess oxygen under standard thermodynamic conditions (although experimental values are usually obtained under different conditions and subsequently adjusted).
The heat of combustion $(\Delta H_c^\circ)$ is the energy released as heat when a compound undergoes complete combustion with oxygen under standard conditions. The chemical reaction is typically a hydrocarbon reacting with oxygen to form carbon dioxide, water and heat.
On first inspection, the two seem the same, however, there is a very slight different use in terminology and wording.
1.) What are standard conditions?
Referring back to Wikipedia:
Standard conditions for temperature and pressure are standard sets of conditions for experimental measurements established to allow comparisons to be made between different sets of data.
In chemistry, IUPAC established standard temperature and pressure (informally abbreviated as STP) as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 100 kPa (14.504 psi, 0.987 atm, 1 bar)
As you know, the standard enthalpy of combustion can be referred to as as $\Delta H ^{\circ} _{\mathrm{total}}$ whose units are in kJ. The molar enthalpy of combustion is how much energy is released per kJ/mol. How is the standard enthalpy of combustion found?
Using the individual enthalpies of formation, and Hess's law, we can calculate the standard enthalpy of combustion, or $\Delta H ^{\circ} _{\mathrm{total}}$.
$\Delta H_{reaction}^\ominus = \sum \Delta H_{\mathrm f \,(products)}^{\ominus} - \sum \Delta H_{\mathrm f \,(reactants)}^{\ominus}$
The standard enthalpies of formation, or the negative values are located here https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table) and as well, generally in your textbook in an appendix along with the standard entropy of formation, and Gibbs free energy of formation, as all three are state functions.
2.) Why do the signs differ in the values from your textbook compared to the table in Wikipedia?
For the heat of combustion, it is generally expressed in units of higher heating value, lower heating value, and gross heating value. As the article states in Wikipedia, the heating value (or energy value or calorific value) of a substance, usually a fuel or food (see food energy), is the amount of heat released during the combustion of a specified amount of it.
In summary, the sign difference that I see is due to different terminology, they are specifically referring to units of HHV, LHV, and GHV and expressing those in the table below. Although, heat of combustion may also be calculated as the difference between the heat of formation $\big( \Delta H_f^\circ \big)$ of the products and reactants just as the standard enthalpy of combustion, it seems to me that the heat of combustion is specialized in a way to be applied to fuels, and as such, different from the standard enthalpy of combustion that we are accustomed to.
Perhaps someone could clarify a little more on this?

The lower heating value is the measured, or empirical heat of combustion. It is lower because some of the internal energy (enthalpy) that is released as heat goes into the production of water in vapor form.
Since standard enthalpy values require compounds in their standard states (water in liquid form), the higher heating value must be calculated. This is done by adjusting for the heat of vaporization of water for the given mass. Typically this done by adding the adjustment quantity to the measured value.


text:
Standards for drawing chemical molecules
I'll preface this by saying I haven't had a chemistry class in 10 years and molecular structures are not my thing. But I'm submitting a revision to a journal paper where I need to include figures of molecules and I need to make sure I get it right so I don't get nit-picked apart for it.
Is there a standard/guide/list of rules for how to create chemical molecule figures like the one below?

I'm looking for a comprehensive list of rules but as an example of things I would need to know:

Should I include the carbon atoms in the ring drawing?
At what angle should the entire molecule be drawn (I've seen rigid-body rotated versions of the same molecule before)?

Rather than answering those specific questions, I'd like to know if anybody (IUPAC maybe?) has a standard for drawing these figures.

It's also worth noting that most chemistry journals publish their own instructions for authors including guidelines for figures and chemical structures.

As it happens, there are a set of quite involved documents from the "IUPAC Chemical Nomenclature and Structure Representation Division".
There's a list here, but to avoid later link-rot, they are currently:

Jonathan Brecher Graphical representation of stereochemical configuration (IUPAC Recommendations 2006) 2006, Vol. 78, Issue 10, pp. 1897-1970
Richard M. Hartshorn, Evamarie Hey-Hawkins, René Kalio and G. Jeffery Leigh: Representation of configuration in coordination polyhedra and the extension of current methodology to coordination numbers greater than six (IUPAC Technical Report)
2007, Vol. 79, Issue 10, pp. 1779-1799
W. Mormann and K.-H. Hellwich: Structure-based nomenclature for cyclic organic macromolecules (IUPAC Recommendations 2008) 2008, Vol. 80, Issue 2, pp. 201-232
Jonathan Brecher: Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) 2008, Vol. 80, Issue 2, pp. 277-410
Andrey Yerin, Edward S. Wilks, Gerard P. Moss and Akira Harada: Nomenclature for rotaxanes and pseudorotaxanes (IUPAC Recommendations 2008) 2008, Vol. 80, Issue 9, pp. 2041-2068

I found "Graphical representation standards for chemical structure diagrams" to be exactly what I was looking for and the most helpful of the bunch. But man is it dry reading!


text:
Have there been no advances in the determination of effective nuclear charges since Clementi and Raimondi in the 60s?
Effective nuclear charge is a very important concept in chemistry, and is the basis for the qualitative explanation of many observed chemical and physical properties, including several periodic trends. Hence it seems to me to be of significant importance that these values are accurately determined (not only for neutral atoms, but for monoatomic ions too). Slater's rules for the calculation of effective nuclear charges, from the 1930s, are in practically any university-level general or inorganic chemistry book, though they are heavily outdated and inaccurate, being mostly of interest due to their simplicity and their historical value. Many books go on to mention updated values for atomic screening constants, calculated by Clementi and Raimondi in 1963 and 1967. This updated table of effective nuclear charges is significantly more interesting, but not used as often probably because all the values for each orbital in each atom must be looked up individually.
However, it's been 50 years since Clementi and Raimondi's first publication on the topic, and I have seen no clear reference to a more recent improved calculation. Surely it is not because the values they found are not without flaws. For example:

Screening constants have only been calculated up to $Z = 86$, leaving out a portion of the table, in particular the actinides which have interesting electronic properties
No relativistic terms whatsoever were considered in the calculation of the screening constants, which is deleterious especially for atoms beyond about $Z \simeq 70$
The calculations are not completely reliable as they are, having not converged well especially some lanthanides, creating odd kinks in screening capacities for electrons in the $4f$ subshell
The calculations converged to inaccurate electron distributions in several atoms, such as $[\ce{Kr}]4d^85s^2$ for $\ce{Pd}$ instead of the actual $[\ce{Kr}]4d^{10}$

By now we should be able to do their calculations again with much more accuracy and much more speed, both due to the development of physical and mathematical theory behind the calculations, and due to the stupendous increase in affordable computational capacity. So why has there been no improvement to effective nuclear charge calculations? Has no one bothered? Or have there been several attempts, none of which have gained wide acceptance since then (sort of like what happens with electronegativity)?

I would say the most important reason why no development for effective nuclear charges is that the concept of effective nuclear charges has fallen to disuse in the recent years.  I can think of two important reasons:

Effective nuclear charges is intended to be used to explain electronic structure of isolated atoms.  To this day some atoms can still pose a challenge, but those are the atoms that experience heavy relativistic effects and would need other special treatments than shielding effect along.  However, for the most part, chemists are not interested in atoms anymore.
Its theoretical framework is too simple.  Chemist need their results to be extremely accurate.  A 0.1 eV change can easily be a game changer in an organic reaction.  That, unfortunately, cannot be achieved by a model that has only one adjustable constant.

However, the idea still lives on.  For example, pseudo-potential or Effective Core Potentials can be considered as more general and modern version of effective nuclear charge concept.  Instead of simply changing the charge but keeping the Coulomb potential unchanged, ECP provides the entire effective potential of core electrons. Valence electrons and sometimes even sub-valence electrons has to be excluded from such treatment to be able to produce results with sufficient accuracy. These pseudo-potentials are actively developed and widely used today, can provide chemically relevant accuracy in complex molecular systems and may qualify as "improvement to effective nuclear charge", although they are evolved into different species already.

Analytical Dirac-Hartree-Fock-Slater screening function for atoms (Z=1–92) Phys. Rev. A 36, 467
seems to address the deficiencies mentioned in the question.
Effective nuclear charge is expressed as a function of radius for elements through uranium and relativistic calculations are used.
There is also A new set of relativistic screening constants for the screened hydrogenic model High Energy Density Physics vol. 7 page 169-179.

It always amazes me how good you are at pulling up sources! I'll look into those articles, hopefully with them I can analyse periodic trends more accurately.

As you say, effectively nuclear charges are an interesting tool to provide simple handwaving-type explanations of chemical and physical properties. The reality is more complex, and effective nuclear charges are not a useful tool for heavy elements, where more advanced quantum chemical methods are used. In short: the values of effective nuclear charges we have are good enough for what we want to do with them.

I can sort of understand that. I'm sure no one is going to rely entirely on effective nuclear charge to put forth an argument (at least not one subtle enough that it is sensitive to details in a quantitative analysis). However, you could say the very same about electronegativity, yet there is a decent amount of research into efforts to redefine it. Why would that get singled out?


text:
Quantifying Ascorbic Acid in Soft Drink
I have access to a GC/MS, FID, and a spectrophotometer. My first thought was to derivatize with trimethylsilylating agent, then run it by GC. Unfortunately the water is problematic. Heating it is likely to destroy the ascorbic. I don't have access to a manifold freeze-drier. I do have access to a rotovap, but I don't think that works well for removing water. I do have access to a freezer, but no way to vacuum the sample down to 0.006 atmospheres. Titration is out of the question because I don't have a clue what else is in the sample.

Is there something wrong with just running the sample through the GC/FID, and then with added internal standard of ascorbic acid? HPLC would probably be better, though. See e.g. http://dx.doi.org/10.1021/jf071710s

@ssavec - You cannot inject aqueous samples containing ionic species into a GC unless you want to clog your system with nonvolatile detritus. Ascorbic acid is very soluble in water, so doing an extraction first is out.

@BenNorris thanks for info, I just thought that almost everything is reasonably volatile under GC conditions. Obviously wrong. I intuitively expected the HPLC to be much better for this problem and now I also know why.

Titration is in the question because ascorbic acid is likely the principal, if not only, reducing agent.

Using iodine

For example, [this site] describes a titration using the iodine-starch redox indicator (the presence of iodine and starch together creates a blue/black color).
The active oxidant is $\ce{I3-}$, formed from $\ce{I2}$ and $\ce{I-}$:
$$\ce{I2 + I- <=> I3-}$$
The redox reaction then is:
$$\ce{C6H8O6 + I3- ->C6H6O6 + 3I- + 2H+}$$
Add a small amount of a saturated solution of soluble starch to your soft drink sample and then titrate against a known solution that is equimolar in $\ce{NaI}$ or \$\ce{KI}$ and $\ce{I2}$. When the dark color persists in your soft drink sample, you have your end point.
This site provides a full procedure (PDF).

Using $\ce{KMnO4}$

Potassium permanganate is its own indicator, since $\ce{KMnO4}$ solutions are purple and the reduced version is colorless (technically an insoluble brown solid).
If your soft drink is dark-colored, you may want to dilute it so that you can see the color change.
The redox equation is:
$$\ce{2MnO4- +6H+ 5C6H8O6-> 2Mn^{2+} + 8H2O}$$

Spectrophotometrically

Ascorbic acid absorbs UV light (but then so does other things in your soft drink, like the preservative sodium benzoate). However, you could use ascorbic acid's reducing power to turn on an absorbance in the visible. Here is one example. This paper may be behind a paywall, but the abstract contains the essential idea. In this paper, the authors describe a system using copper(II) and phenanthroline(phen). Ascorbic acid reduces copper(II) to (copper(I), which creates a coordination complex with phenanthroline that has $\lambda_{\text{max}}=460 \ \text{nm}$. The authors claim a detection limit of 40 $\mu\text{g/L}$, which is pretty good (40 parts per million).
$$\ce{2Cu^{2+} + C6H8O6 -> 2Cu+ + C6H6O6 +2H+}$$
$$\ce{Cu+ + 3phen ->Cu(phen)3+}$$

Another colorimetric assay

The Folin-Ciocalteu assay is another nice colorimetric assay. Reducing agents react with the phosphomolybdate and phosphotungstate compounds in the reagent mixture and produce a blue color. This method needs to be calibrated against some ascorbic acid standards because the redox chemistry is complex and not fully understood. The absorbance max of the reduced FC reagent is in the 670ish range, if I remember correctly.


text:
Calculate partial pressure of gases at equilibrium
I have the reaction $\ce{CO_{(g)} + Cl2_{(g)} <=> COCl2_{(g)}}$ where $K=4.5\times10^9$ at $100\ \mathrm{^\circ C}$
The first question was to find $K_p$ which I know is $K_p=K(RT)^{\Delta n}$ where $R=0.08206$, $T=273+100=373$ and $n=1-2=-1$. Put in the numbers and get $K_p\approx1.47\times10^8$.
Now comes the second question and this is where I get stuck:

Equal moles of $\ce{CO}$ and $\ce{Cl2}$ are reacted at $100\ \mathrm{^\circ C}$. If the total pressure at equilibrium is 5.0 atm calculate the equilibrium partial pressures of the gases.

I have tried several solutions, where I either end up with a wrong answer or hit a dead end. I just don't understand how to solve this.

The equations you need are
$$K_p=\frac{p_{\ce{COCl2}}}{p_{\ce{CO}} p_{\ce{Cl2}}}=1.47\cdot10^8$$
and
$$p_\text{tot}=p_{\ce{COCl2}}+p_{\ce{CO}}+p_{\ce{Cl2}}=5.0~\text{atm}$$
Additionally, $n_{\ce{CO}}=n_{\ce{Cl2}}\implies p_{\ce{CO}}=p_{\ce{Cl2}}$, making the system solvable.


text:
Why distinguish between polarity and hydrophobicity?
I have yet to see an example of a molecule that's both polar and hydrophobic or that's both non-polar an hydrophilic. If no such molecule exists, why use the terms "hydrophilic" and "hydrophobic" at all if we can just refer to them as being polar or non-polar? If there are such molecules, what are some examples and why do they exhibit non-polarity and hydrophilicty, or polarity and hydrophobicity?

Hydrophilicity/hydrophobicity/lipophilicity/lipophobicity and polarity/non-polarity are terms that overlap to a great extent in simple substances, but which are not strictly equal. The former terms can have a much more involved explanation depending on several factors, even if polarity is usually the largest by far.
Philicity and phobicity are determined by contact angles between substances. It is possible to vary how a same substance gets wetted by exploiting surface patterning, which fiddles with the surface tension between the substances in contact.
Another interesting example I can think of is that some large perfluorocarbons are non-polar, but they are also lipophobic. It is possible to mix water, oil and a perfluorinated solvent, shake, and then watch as three clearly distinct phases separate.

So if I understand you right, a molecule's polarity can be a contributing explanation for its observed hydrophobicity, but saying the reverse (hydrophobicity explaining polarity) wouldn't make much sense. I'll have to check out those wikipedia pages as well.


text:
X-ray fluorescence: where do expelled electrons go?
I'm studying the XRF method cause I'm about to use a WDXRF spectrometer at college so I'm getting all the information about the physics and chemistry behind it.
I'm just curious about one thing, as you know when a beam of x-ray hits the electron of an inner shell if the energy is just about lower than the electron binding energy, that electron is expelled and the hole is filled with an electron from an outer shell and a photon (fluorescence) is emitted with the same energy as the difference between the shells.
What about the electron that has been expelled from the atom, where does it go and what happens to it?

After being used as a target, most metal blocks are positively charged. Hence atleast a few electrons must be leaving the target block.

OK they leave, but where do they go, they must be attracted t something else right or do they become free electrons?

Once they are out of the target, the potential which is accelerating the projectile electrons also starts accelerating these ejected electrons and they all become indistinguishable.

There are a number of techniques that use excitation of inner shell electrons. Auger spectroscopy, and ESCA for example. But XRF is a "bulk" analysis method, not surface such as Auger. The excitation energy from the initial x-ray will be keV whereas the empty orbitals in the atoms of the sample will be a few eV. The excited electron rips trough the sample creating various ions until its energy is reduced enough so that the emitted electron can drop into some empty low energy orbital. (In general it would a "different" electron that drops into the vacancy and creates the fluorescent x-ray.)
In general the x-ray can penetrated deeper into the sample than an excited electron can escape. So the bulk of the excited electrons stay within the sample.


text:
The proteins "eluted" or "were eluted"?
In describing proteins, I've written "the proteins detached from the [Ni-IDA] matrix and were eluted from the column"
Should I use "were" or can something simply "elute" ?

eˈlute, v.
trans. To wash out, cleanse. Now spec.  [after German eluieren, in same sense] to wash (adsorbed matter) away from the substance that has adsorbed it.

It's transitive, so you elute something, rather than something eluting. Your original "were eluted" is correct.

God sorry I didn't even think of checking transitive/intransitive :| Not quite with it today, cheers Aesin


text:
Why does water cause a jet flame when poured over a bunch of tealights?
A neat party trick (typically for outdoors) is to build a pyramid of tealights, let them burn for a while and then pour a bit of water over it and see a large jet flame shoot out (see e.g. this youtube video).
I was always told that this is caused by the water spreading the liquid wax out allow more of it to burn at the same time, but I am not completely satisfied with that answer because, in my opinion, that cannot explain the large jet flame. Can someone explain why this jet flame occurs? Is it the wax vapour that catches fire, which can form at 370 $^o\ce{C}$, and is this the reason you need to build a close packed pyramid? But what is the role of the water then?

The wax doesn't need to become gaseous to cause the column of flame (if that's what you were thinking). Water, when in contact with the liquid wax at a couple of hundred degrees above its boiling point, will flash-boil in a very chaotic manner as if it were "exploding" ("transition boiling" in this graph). The water closest to the wax very quickly vaporises and tries to escape, but is temporarily blocked by the layer of liquid water above it, so the system ends up being vigorously churned to make way for the vapour.
The problem is that the turmoil will make tiny liquid wax droplets airborne, as if wax were being shot out of an atomizer, creating a wax aerosol. Suddenly, the same amount of wax has its surface area increased tremendously and is exposed to much more oxygen. Even though the wax in the aerosol quickly loses heat and may become solid again in a fraction of a second, if even a small region of the wax aerosol reaches ignition temperature, the flame quickly propagates as a huge amount of heat is dumped at once in the vicinity and triggers the ignition of all the airborne wax.
This trick is effectively the same as what happens if you try to put out a grease fire with water.

I think it's a mixture of both.
You can actually get the same sort of flame just using a pyramid of tealights, but it takes a while longer. I think the role of the water is get the process started, throwing around some wax vapour that can then ignite and self-sustain.


text:
In nuclear chemistry, how does a neutron split to form a proton and an electron?
I'm studying radioisotopes at the moment and balancing nuclear reactions isn't making sense in that more matter is coming out of the equation in negative β⁻ decay equations:
$$\ce{_6^{14}C -> _7^{14}N + e- + \bar{\nu_e}}$$
Notice how the original element has decayed into a new element with an unchanged mass number but an atomic number that has increased by one.
In positive β⁺ decay equations, it makes sense:
$$\ce{_{12}^{23}Mg -> _{11}^{23}Na + e+ + \nu_e}$$
How can you create something with the same mass but with another proton?

$n \rightarrow p + e^- + \bar{\nu}_e$

One neutron has changed into a proton, that's what has happened. Our convention in chemistry is to identify nuclear species by the proton number (or the "atomic number") and that is why a new nuclide with an increased atomic number is formed. Both the $\beta^+$ and $\beta^-$ decay follow identical paths: in $\beta^-$ decay, a neutron changes into a proton (thus giving no change in mass number and increase in atomic number), whereas in $\beta^+$ decay, a proton decays into a neutron giving no change in the mass number and a decrease in the atomic number.

What you are missing is that the atomic mass has not changed — mass is conserved — the top number on the isotope notation 14 for both carbon and nitrogen in first case (in fact, it goes down very slightly, but they rounded it up). The increasing atomic number (bottom number) just means a neutron became a proton.

You wrote the equations correctly. With beta decay a neutron is converted into a proton, and a beta particle is emitted from the nucleus. With positron decay a proton is converted into a neutron and a positron is emitted from the nucleus. In both types of decay the mass number (neutrons + protons) is constant.
As a side note a positron will be short lived. It will annihilate with an electron producing a pair of 511 keV gamma rays emitted in opposite directions.


text:
Why should colorimetric titration of a weak acid with a weak base not be done in laboratory?
Why should colorimetric titration of weak acid with a weak base (or vice versa) not be done in laboratory?
How can I draw a pH graph for this?

Here is a paper with titration curves of weak acids with weak bases.

When a weak acid reacts with a weak base, the equivalence point solution will be basic if the base is stronger and acidic if the acid is stronger. If both are of equal strength, then the equivalence pH will be neutral. However, weak acids are not often titrated against weak bases because the colour change shown with the indicator is often quick, and therefore very difficult for the observer to see the change of colour.

Source: Wikipedia
Drawing the titration curve is hard, because you generate a buffer system which buffers against acids and bases at the same time. Around the equivalence point the buffer system works equally well against acids and bases (assuming their $\text{p}K_\text{a}$ values are the same), resulting in a slow rise of the pH. You don't get the usual "jump" in the curve at the equivalence point, which makes it hard to detect the actual equivalence point (your indicator should fulfill $\text{p}K_\text{a}^\text{Ind} = \text{pH}_\text{equiv.}$, which is highly unlikely).
For the reasons given above I disagree with the last part of the Wikipedia quote, but having never done the experiment myself I cannot be sure.


text:
How can zinc metal still be present after the reaction?

A $\pu{15.8 g}$ strip of zinc metal was placed in $\pu{100 ml}$ of silver nitrate. When the reaction was complete, the strip of zinc had a mass of $\pu{13.1 g}$. What was the concentration of the silver nitrate solution?

I am confused on writing the equation for this problem, because doesn't single displacement occur? So, how can zinc metal still be there?

There is an excess of zinc, so some is left over. Look at the problem again with that in mind.

This is a flawed problem. In these types of redox displacements, the silver will plate onto the zinc strip.

As we can clearly see, the silver nitrate is the limiting reactant in this problem as there are still 13.1 grams of zinc left that have not reacted.
That means that $\pu{15.8 g}$ $\ce{Zn}$ metal - $\pu{13.1 g}$ $\ce{Zn}$ metal = $\pu{2.7 g}$ $\ce{Zn}$ metal has reacted with all of the $\ce{100 mL}$ silver nitrate solution with $\pu{13.1 g}$ excess zinc metal.
We are however given $\pu{100 mL}$ of silver nitrate, and we are interested to find out how much product is formed. Looking online, the density of silver nitrate is $\pu{5.35 g/cm3}$ and keeping in mind that $\pu{1 cm3} = \pu{1 mL}$. Note that since the silver nitrate is in solution, the density of the silver nitrate in solution is unknown, but for references sake, we will refer to the density of the solid silver nitrate which is the 5.35 grams per cubic centimeter.
Knowing that
$$\rho = m/V,$$
we can find the mass by
$$\pu{5.35 g/cm^3} = \frac{m}{\pu{100 mL} \times \pu{1 cm^3/mL}},$$
therefore our mass of silver nitrate (theoretically) is
$$m = \pu{535 g}.$$
In looking at the reactivity series, zinc is higher than silver, so single displacement does occur just as you predicted and therefore zinc "should not" appear as it then performs a redox reaction with the nitrate in silver nitrate.
See Wikipedia for the reactivity series of the metals.
If we write the chemical equation for this reaction we can see that
$$\ce{AgNO3(aq) + Zn(s) -> ZnNO3(aq) + Ag(s)},$$
and the balanced equation is:
$$\ce{2AgNO3(aq) + Zn(s) -> Zn(NO3)2(aq) + Ag(s)}.$$
Let's quantitatively calculate what would happen if we reacted each of our reactants.
Assuming that zinc is the limiting reactant and we have silver nitrate in excess:
$$
\pu{15.8 g}\ \ce{Zn} \times \frac{\pu{1 mol}\ \ce{Zn}}{\pu{65.38 g}\ \ce{Zn}}
\times \frac{\pu{1 mol}\ \ce{Zn}}{\pu{1 mol}\ \ce{Ag}}
\times \frac{\pu{1 mol}\ \ce{Ag}}{\pu{107.9 g}\ \ce{Ag}}
= \pu{0.002240 g}\ \ce{Ag}
$$
Assuming that silver nitrate is the limiting reactant and we have zinc in excess:
$$
\pu{535 g}\ \ce{AgNO3}
\times \frac{\pu{1 mol}\ \ce{AgNO3}}{\pu{169.9 g}\ \ce{AgNO3}}
\times \frac{\pu{1 mol}\ \ce{Ag}}{\pu{2 mol}\ \ce{AgNO3}}
\times \frac{\pu{107.9 g}\ \ce{Ag}}{\pu{1 mol}\ \ce{Ag}}
= \pu{169.9 g}\ \ce{Ag}
$$
In both cases, we expect to see aqueous zinc nitrate to form, and solid silver to precipitate. In looking at my example, all of the zinc is consumed in the reaction so we expect to not see any zinc metal left in the first reaction since that is the limiting reactant, because there is not enough zinc metal to react with the silver nitrate. In the second reaction, we see that silver nitrate is in excess because we have a much greater yield for silver, however, this reaction is "limited" due to not having enough zinc metal.
However, in your case, 13.1 grams of zinc metal are still remaining. Therefore, zinc must be the reactant in excess and thus the mass of silver nitrate is all consumed in the reaction. Therefore although you expected the zinc metal to bind to the nitrate in the redox reaction, it does! However, we have too much zinc metal and not enough silver nitrate, therefore we still see some zinc metal left along with our usual products of zinc nitrate and silver. I just did a basic stoichiometric limiting reactant problem so you can quantitatively see how this type of reaction works. In reality, I do not know the density of the silver nitrate solution, and thus the mass of the silver nitrate, which is much lower in the context of your problem.

To me it doesn't make any sense to find the mass of 100 mL of solid silver nitrate. This answer is overly complicated for a problem, which is somewhat simple...

@Martin-マーチン Absolutely the "answer" does not make sense! The 2.7g of Zn that was oxidized is 0.0413 moles, so 0.0826 moles of silver ions were reduced to metal and fell off the zinc. Hence the original silver nitrate concentration was 0.826 M (= 0.0826 moles/100 mL).


text:
De-/protonation of R-OH group by acids and bases
Let $\ce{R-OH}$ be a group with a $pK_a$ of 16. Are the following two statements correct?

In principle, the group can protonate another acid with $pK_a > 16$, and so (put differently), any "acid" with $pK_a > 16$ can be used to "deprotonate" the group
If I had a table with bases were only $pK_b$ values were given, in order to select these bases from the table which are able to deprotonate the group, I would need to choose the bases with $pK_b < -2$ because the $pK_a$ values of their conjugate acids (i.e. $pK_{a\text{H}}$) will be 16 or higher.

I don't think this is sufficient for a full answer, so I'll leave a comment. Re: your first statement, in order for the acid-base reaction to be product-favored, it is necessary that the pKa of the conj. acid be greater than the pKa of the acid on the reactant side. As for your second point, the relationship pKa + pKb = 14 is only valid for aqueous solutions under standard conditions. I don't think it's valid for all conj. acid/base pairs in other solvents, but I could be mistaken.

I would hardly call a substance with pKa > 16 an acid, but the first statement seems right in principle. Remember that acid-base reactions are equilibria, so if your pKa is say, 16.2, only a slight deprotonation will occur.

@GregE. That's what I'm saying, isn't it? pKa(R-OH)=16 and so it can protonate another acid AH with pKa(AH) > 16. Your comment about my second statement is right, pKa + pKb = 14 is only valid in H2O (because of the ionization product limit).

@chipbuster Sure, it's not an acid in the classical meaning, but technically speaking, it acts as an acid w.r.t. the other substance.

@TMOTTM, yes, I wasn't disagreeing with your first statement, sorry if I was unclear

@GregE. Sure.. About the second point, let's assume I had written pKa(R-OH)=12 (so below the pKa of water). In order to deprotonate the group, from the table then I would have to select the bases with pKb <=2. Do you agree?

@TMOTTM, yes, I believe that's correct.

Your first statement is correct. The basicity increases with $pK_a$, so a base with $pK_a > 16$ will be able to deprotonate your alcohol.
Your second statement is also in principle correct; however, the value of the sum $pK_a+pK_b$ is solvent-dependent, and 14 is the value for water at standard conditions. $pK_a$ and $pK_b$ values of acids and bases depend on the solvent as well. For example, sodium amide is a strong base in liquid ammonia, but not so in water due to the leveling effect of the solvent. Dissolving sodium amide in water will give a solution of sodium hydroxide, which is not basic enough ($pK_a=13$) to deprotonate the alcohol.
Furthermore, the amount of deprotonated alcohol will depend on the difference between $pK_a$ values of alcohol and base, since the acid-base reaction is an equilibrium. For example, a base with $pK_a=33$ will deprotonate your alcohol to a greater extent than one with a $pK_a$ of 16.5 .