text:
Reactivity of chlorobenzene and benzene in electrophilic substitutions
As we know, chlorobenzene is a deactivating substituent and benzene has a stable structure. Now I'm confused.
My question is — considering electrophilic aromatic substitution is chlorobenzene or benzene more reactive?

considering electrophilic aromatic substitution is chlorobenzene or benzene more reactive?

Let's expand the question to include all of the halobenzenes.
All of the halogens ($\ce{F, Cl, Br, I}$) are more electronegative than hydrogen.  Therefore they will inductively remove electron density from a benzene ring. However, you can also draw resonance structures for the various halobenzenes where the halogen substituent is donating electron density to the ortho and para positions of the benzene ring through resonance.
Therefore, in the case of the halobenzenes these two effects are working in opposite directions.  In order to determine which effect controls the reaction, we need some data.
The following table compares the relative rates of electrophilic aromatic nitration of the halobenzenes to the rate for benzene itself. If we assign benzene a relative rate of 1, we see that all of the halobenzenes react slower than benzene itself. This tells us that the inductive effect is stronger than the resonance effect in the halobenzene series.
\begin{array} \hline
 \text{Relative rates of aromatic electrophilic nitration} \\ \hline
\end{array}
\begin{array}{|c|c|c|c|} \hline
\ce{Ar-X} & \text{Relative rate}  \\
\hline
\ce{Ar-H} & 1.0 \\ \hline
\ce{Ar-F} & 0.11 \\ \hline
\ce{Ar-Cl} & 0.02 \\ \hline
\ce{Ar-Br} & 0.06  \\ \hline
\ce{Ar-I} & 0.13 \\ \hline
\end{array}
(data source: see p. 26 here)
However, when the halobenzenes do undergo electrophilic aromatic substitution, they react preferentially at the ortho and para positions, because, as explained above, the resonance effect donates electron density to the ortho and para positions.  Therefore, these two positions are the least deactivated and react preferentially.

then why iodobenzene is more reactive towards electrophilic substitution than fluorobenzene...? In one of my exam question the answer was fluorobenzene is more reactive than iodine counterpart and iodobenzene is more reactive than both chlorine and bromine counterpart

@maverick  The Table above shows that difference in relative rates of nitration for the 4 halobenzenes is less than factor of 10, a very small difference.  Further, the rate difference between fluoro- and iodobenzene is even smaller. So it wouldn't surprise me if in some reactions or under different conditions (temperature, solvent, etc.) fluorobenzene reacted a bit faster (or slower) than iodobenzene.  It would surprise me if fluorobenzene reacted >100 times faster than iodobenzene or vice versa.

$\ce{Cl}$ is a weak ring deactivating group. This means chlorobenzene is less reactive towards  electrophiles than benzene. But, due to the presence of non-bonding electrons, they can donate electron density through formation of $\pi$-bonds.
$\ce{Cl-}$ is still an ortho/para directing group.

Electrophilic substitution relies on the electrophile being attracted to the delocalised electrons in the benzene ring. But chlorine is more electronegative than carbon, and so draws electrons in the ring towards itself.
That makes the electron density around the ring rather less in chlorobenzene. It is less attractive for electrophiles, and so the reaction is slower.

Since chlorine is an electron withdrawing group, so it  withdraws electron from aromatic ring. As a result of
which the electron density of the aromatic ring decreases and deactivates the ring.
Chlorobenzene is less reactive than benzene towards electrophilic substitution reaction. As halogen atoms make the aromatic ring less reactive towards the upcoming electrophile, they are regarded as deactivating groups but they act as ortho para deactivators due to the development of partial negative charge on ortho and para postion. Hence chlorobenzene is ortho para deactivator and is less reactive than benzene.

That's precisely what all the above answers are saying.

As you have said the chlorobenzene is deactivating -This should be a bit of a foreshadowing.
The chloro substituent will function to withdraw electrons from the benzene ring and reduce its electron density, this will make it less reactive with electrophiles and reduce the reaction rate.

Chlorobenzene is less reactive than benzene towards electrophilic substitution reaction due to - I effect.As chlorine is electron withdrawing group it deactivates the benzene ring and reduces the electron density on benzene ring and hence make the aromatic ring less reactive towards the upcoming electrophile.

Chlorobenzene is less reactive than benzene towards electrophilic substitution reaction due to - I effect.As chlorine is electron withdrawing group it deactivates the benzene ring and reduces the electron density on benzene ring and hence make the aromatic ring less reactive towards the upcoming electrophile.

Trying to make it easy: electrophile means electron loving so in haloarenes the halogen group pulls electron from benzene. But electrophiles need electrons so benzene is more active.


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Why are chlorides, sulfates, and bicarbonates of Ca and Mg solely responsible for the hardness of water?
When soap is dissolved in water - irrespective of the anion - $\ce{Ca^2+}$ and $\ce{Mg^2+}$ form precipitates. Why are just chlorides, sulfates, and bicarbonates the only specie considered as "hardness" in water?

If I remember correctly, hardness is defined by the Mg and Ca content of water. Temporary hardness is the mineral content which can be removed by boiling and permanent is the total Mg and Ca content - this is what counts when the precipitate with the fatty acid is being formed.
Now typical potable water will only contain sulphide, bicarbonate, halide and perhaps nitrate anions. Other anions either don't occur naturally or they are harmful so they are not considered in the context of drinking water.

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Why do we use external pressure to find force exerted by gas on a piston in a closed vessel during its expansion?
I was reading a book in which i got confused due to this. A gas is in a vessel covered by a friction-less piston. Gas pressure is more than outer atmospheric pressure (which is constant) due to which it expands. Since it expands and move the piston outward, it does work on it which is calculated by force times displacement. In finding force we use Force = Pressure times Cross section area of piston. In finding this force to calculate work why are we using Pressure on piston by atmosphere and its weight and not pressure exerted by gas. Why? Can you also tell how the correct work done by gas can be calculated.
Thanks.

Very possibly a duplicate of http://chemistry.stackexchange.com/questions/37246/work-done-by-the-system/37250#37250

For a frictionless, massless piston, the force per unit area that the gas exerts on the inner face of the piston is always equal to the force per unit area applied on the outer face of the piston (since the piston is massless, ma = 0).  However, for an irreversible process, the force per unit area exerted by the gas at the interface is composed both of the "thermodynamic pressure"  that one could determine from the equation of state for the gas (e.g., the ideal gas law) applied locally at the interface,  plus "viscous stresses" in the gas that are determined by the rate at which the gas is deforming.  Furthermore, if the expansion or compression is rapid enough, the conditions within the cylinder are not even uniform spatially.  So figuring out the force per unit area that the gas exerts at the interface for irreversible expansion is problematic.  However, fortunately, if the force per unit area on the outside is specified by manually imposing a force per unit area (e.g., if the atmosphere is imposing the pressure outside), the work done by the gas on the surroundings can still be determined.  This is because the force per unit area exerted by the gas at the interface will adjust to always match the force per unit are on the outside (again, since the piston is massless and ma = 0).
In the case of a reversible process, the situation is much simpler, since here, viscous stresses are negligible, and the force that the gas exerts per unit area on the inner face of the piston can be determined by applying the ideal gas law (or real gas equation of state) to the spatially homogeneous gas in the cylinder.  In this case, however, the force imposed on the outside must be controlled very carefully, so that it differs only slightly from the average inside gas pressure during the entire expansion or compression (and the gas pressure within the cylinder remains essentially uniform spatially).

why are we using Pressure on piston by atmosphere and its weight and not pressure exerted by gas.

Because the gas is expanding against the atmosphere. Put yourself in the shoes of the gas. Imagine that you're inside that container, trying to push the piston outwards. What are you pushing it against? The external pressure, right? The work done by the gas is, essentially, how much effort you are putting into pushing out the piston.

Can you also tell how the correct work done by gas can be calculated.

Since you're reading about work done, whatever book you are reading should really have a formula telling you that the work done by the gas is:
$$w = \int_1^2\!\mathrm{đ}w = \int_1^2\!p_{ex}\,\mathrm{d}V$$
where 1 and 2 are the initial and final states respectively.

+1 except for that it is recommended to use a different đ symbol for inexact differentials.

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How do you know when the first compound is completely distilled and how do you know when the next compound starts
Did a distillation between two solvents. Would the answer have to do with temperature?

You could improve this question a lot by asking it properly, that might actually get you even better answers.

There is no such point when the first solvent is completely distilled and then the next one starts. From the very beginning, both solvents boil together, though the vapor initially is relatively enriched with the lower-boiling one. The cutoff point is set somewhat arbitrarily. And yes, the answer surely has something to do with temperature, because temperature is the only parameter we can control.
Like NotNicolaou said, if the boiling points of your solvents are reasonably far away, you may separate the mixture almost completely by collecting the fraction that distills near the boiling point of the low-boiling one. Otherwise, you might not be able to get pure compounds, unless you use an industrial rectification column.

The answer would depend upon how different the boiling points of the two solvents were.
If the boiling points were completely different, say one component boiling at 40 °C and the other at 250 °C, then heating the mixture to 40 °C would remove all of the component boiling at that temperature, leaving the 250 °C boiling component behind. This the principle of a rotary evaporator, with the high boiling component often being the product of a chemical reaction.
If the boiling points are similar, say 70 °C and 80 °C, then it can be harder to get complete separation. Often in these cases, three 'fractions' are obtained: clean 70 °C material, clean 80 °C material, and a small amount of a mixture which distills off between the two. In this case, a fractionating column (such as a vigreux column) can be used, to get better separation. The boiling point of the distillate can always be measured, as an internal thermometer is often used and as such its possible to see whether the boiling point of the distillate is cleanly 70 °C, cleanly 80 °C or some in-between number (indicating a mixture is being distilled).
Its also worth noting that many solvents form azeotropes with water, and in these cases (such as benzene and water) it is not possible to cleanly distill the two components apart despite the different boiling points, since the mixture has a defined boiling point which is neither that of benzene nor that of water.


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Weighing reaction mixture
At school we do an experiment with baking soda and vinegar (in a cup) both sealed in a ziplock bag. The bag is masses on a triple beam balance (example 32.7 g). Bag is tilted to allow the vinegar and baking soda to mix.
The students note the signs of chemical reaction and mass the bag again. There is always a 0.3 g – 0.5 g loss of mass. How is this explained? Cheap ziplock bags? Is some of the mass from the endothermic reaction converted?

Think about the reaction:
$$\ce{NaCHO3(s) + CH3COOH(l) \to CO2(g) + H2O(g) + Na+(aq) + CH3COO- (aq)}$$
The resulting $\ce{CO2}$ will make the bag expand, therefore increasing the volume and the subsequent buoyant force. This force opposes the force of gravity, making it seem like the mass of the matter inside the bag has decreased.


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Optimal CFM for kerosene to burn?
I am going to be making a kerosene forge, and I need a refresher on calculus-based chemistry.
Given a certain distribution of alkanes, what is the amount of air (21% O$_2$) needed (in cubic feet per minute [CFM]) at 21C to burn kerosene at a rate of X mL/s?
The distribution of alkanes is unknown but from Wikipedia I've got that the molecules have no more than 16 and no less than 6 carbon atoms. The density is $\mathrm{0.795~g~cm^3}$. If any more info or help is needed, I'll be glad to pitch in where I can.

This is something of a standard reaction stochiometry question, although the situation is a bit more involved than the ones you normally see in textbooks.
The first step is to work out what the chemical reaction we're talking about actually is. One complication is that "kerosene" doesn't name a single molecular species, but is instead a mix of hydrocarbons. But the hydrocarbons are similar enough that we can probably approximate the mixture as a "typical" hydrocarbon of roughly the right type and mass. Wikipedia approximates kerosene as dodecane, and even gives a combustion reaction (saving us the effort of balancing one):
$$\ce{2 C12H26(l) + 37 O2(g) → 24 CO2(g) + 26 H2O(g)}$$
Now that we have stoichiometry, we just need to convert the quantities of reactants involved to numbers of molecules (moles). We're dealing with rates of consumption for both reactant, so let's just run the numbers for a fixed period of time, say 1 minute.
We have an amount of kerosene we want to burn ($\ce{X\; mL*s^{-1} * 60s}$) and a conversion from volume to mass (the density), and all we need to get the "moles of kerosene" would be the molecular weight of kerosene. Since we're approximating kerosene as dodecane, use the number for dodecane ($170.34\;  \ce{g*mol^{−1}}$).
Using the knowledge from the balanced chemical reaction equation that 2 moles of dodecane require 37 moles of oxygen to combust completely, we can determine how many moles of oxygen we need to burn the given quantity of kerosene.
Now we need to convert moles of oxygen to a volume of air. If we assume oxygen and air is an ideal gas (a safe assumption under normal conditions), we can use the ideal gas law to interconvert moles ($n$) into volume ($V$): $PV = nRT$, where $P$ is the pressure, $T$ is the temperature, and $R$ is the ideal gas constant (about $8.314\;\ce{L*kPa*K^{−1}*mol^{−1}}$). Or, if we're dealing with standard temperature and pressure (about 1 atmosphere at sea level and about room temperature), then 1 mol of ideal gas takes up about $22.4\;\ce{L}$. So we can find out what volume of oxygen we need.
But we're not dealing with pure oxygen, so we have to divide by 0.21 to get the volume of air which would contain that volume of oxygen. (This works because the 21% is a volumetric percentage.) The final little bit is to convert the number of liters of air used per minute into cubic feet used per minute.
So there's really no calculus involved - it's just lining up the conversion factors and multiplying through. The tricky bit is probably figuring out the chemical reaction formula you want to use (e.g. the assumption of dodecane). It also helps to know that things are easiest if you do most of the calculations in SI (metric) units, and only convert to non-SI at the end.
One final caveat is to keep in mind that we assumed that kerosene was well represented as dodecane. If your kerosene is substantially different from dodecane, the reaction will have to be adjusted. We also likely assumed that the input air was at standard temperature and pressure, and was dry (the 21% value is for dry air). Warmer/cooler air, air at altitude, and humid air will all change the calculation. Also keep in mind that the calculation is for combustion under ideal conditions - incomplete mixing and non-ideal combustion issues might mean you need more air in order to ensure complete combustion.

After the forge heats above 1000C, will incomplete mixing be a factor? That's well above the autoigniton point of kerosene.

@tuskiomi You can only autoignite if there is oxygen to react with. If you have incomplete mixing, you'll by definition have some air-poor pockets where there is insufficient oxygen to react ... but all that's getting more into "chemical engineering" territory, rather than plain "chemistry". My main point was that in practice there will be more to building such a forge than the straight stoichiometry considerations might suggest.


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Change in spin angular momentum during transition in one-electron atoms
I have a really basic qustion about spin and electronic transitions.
In one-electron systems (like H or $\ce{He+}$), will the electron change its spin during an electronic transition? Let's say the electron de-excites from 5p UP to 3s, will the spin be UP or DOWN in the 3s orbital?
From crystal-field theory I remember that a d-d transition where the electron had to switch to opposite spin was spin-forbidden (although it may still happen due to certain vibrational states). Is this still true for one-electron systems?

In general, transitions in which the "direction" of spin changes are forbidden. More formally, the spin conservation rule (also referred to as the Wigner rule) states that

For both radiative and radiationless transitions,
  transitions between terms of the same multiplicity are spin-allowed,
  while transitions between terms of different multiplicity are
  spin-forbidden.

But spin-orbit coupling relaxes this rule, so in general, lines corresponding to spin-"forbidden" transition are just much weaker and not totally absent from the spectrum. And the bigger the atomic number, the stronger is the spin-orbit coupling, and consequently, the bigger are the intensities of spin-"forbidden" transitions.

$
\newcommand{\linop}[1]{\hat{#1}}
\newcommand{\ket}[1]{\left|\,{#1}\right\rangle}
\newcommand{\abs}[1]{\left|\,{#1}\right|}
$
With respect to your question on addition of spin angular momentum you raised in comment section: you formulated it incorrectly which confused me, so my comments (now deleted) were wrong. Thus, I decided to write a short answer here.
So, for angular momentum $J$ the simultaneous eigenstates of $\linop{J}^{2}$ and $\linop{J}_{z}$ operators are given as follows,
$$
\begin{align}
        \linop{J}^{2} \ket{j(j+1),m} &= j(j+1) \hbar^{2} \ket{j(j+1),m} \, , \\
        \linop{J}_{z} \ket{j(j+1),m} &= m \hbar \ket{j(j+1),m} \, ,
\end{align}
$$
where $j$ could take non-negative integer and half-integer values,
$$
    j = 0, 1/2, 1, 3/2, \cdots \, ,
$$
and $m$ ranges from $-j$ to $j$ in integer steps
$$
    m = -j, -j+1, \cdots, j-1, j \cdots \, .
$$
When there are two sources of angular momentum (like two electrons with their spins) we first we have to make a choice when specifying the state of the system: either we use the uncoupled picture $\ket{j_{1}, m_{1}, j_{2}, m_{2}}$, which leaves the total angular momentum unspecified, or we use the coupled picture $\ket{j_{1}, j_{2}, j, m}$, which leaves the individual components unspecified.
The very language OP uses (i.e. talking about multiplicities of states) suggest that we choose the second picture for further discussion. In the coupled picture the question arises as to the permissible values of total angular momentum quantum numbers $j$ and $m$. It is known from quantum theory of angular momentum that the allowed values of $m$ follow immediately from the relation $m = m_{1} + m_{2}$, while the situation with the allowed values of $j$ is more complicated: its permited values are given by
$$
    j = j_{1} + j_{2}, j_{1} + j_{2} - 1, \dotsc, \abs{j_{1} - j_{2}} \, .
$$
All this is true for any kind of angular momentum (orbital or spin) with the only one exception: with respect to half-integer values of $j$ we note that for orbital angular momentum, where the Born interpretation requires cyclic boundary conditions to be satisfied, only integer values are admissible.
Now for each and every electron we have a source of spin angular momentum characterised by quantum numbers $j = 1/2$ and $m = \pm 1/2$. Thus, in accordance with what is written above for a system of two electrons we have 4 possible state which in the coupled picture are written as follows: $\ket{1/2, 1/2, 1, -1}$, $\ket{1/2, 1/2, 1, 0}$, $\ket{1/2, 1/2, 1, 1}$ and $\ket{1/2, 1/2, 0, 0}$. The first three states are three components of a triplet, the last one is the only component of a singlet.
Note, however, that projection quantum numbers for individual components ($m_1$ and $m_2$) are not specified in the coupled picture, so while it might be tempting to think that for $\ket{1/2, 1/2, 1, -1}$ state they are $m_1 = m_2 = -1/2$ this is wrong. And that what was essentially wrong with the question raised by OP in the comments section: he used the coupled picture terminology ("multiplicities") and at the same time specified the values of projection quantum numbers for individual components ("two electrons both have spin of -1/2"). This could not be dome simultaneously.

Thank you. Follow-up question: What happens when two valence electrons in different orbitals both have spin of -1/2? According to the Clebsch-Gordon series we should get a total spin angular momentum of -1 and 0, leading to multiplicities of -1 and 1. Is that correct? Or should I take the absolute value of the sum of the two electron spins?

As a minor point, if you are looking for more information, phys. chem. and  spectroscopy  textbooks usually call the two coupling schemes described here as $j j$ and Russell -Saunders coupling


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Does combustion occur more easily at higher oxygen concentrations or just higher oxygen partial pressure?
I saw a trailer for The Martian and started thinking about space flight.  I know that spacecraft and spacesuits often operate at low pressure with high oxygen content (near 100% $\ce{O2}$ at about 20 kPa), which is harmless to humans and keeps the mass of the space vehicle lower.  Humans do just fine with an oxygen partial pressure of 20-30 kPa as long as other gases present are in safe amounts.
But an issue I hadn't considered is the effect on combustion.  A lot of people mistakenly believe that the Apollo 1 fire was due to using pure oxygen in the test, but that's only part of the issue, and astronauts routinely use nearly pure oxygen at low pressure even today.
So my question is, is flammability affected by the total concentration of oxygen in the atmosphere, just its partial pressure, or some combination of the two?
As a specific example, what affect would a 30 kPa almost pure oxygen atmosphere have on flammability?

When you're dealing with questions like "how well does this burn", you are intrinsically dealing with the reaction rate of the combustion reaction.
When looking at chemical reaction rates, they are typically proportional to the "chemical activity" of the species reacting. (There's complications due to the stoichiometry of the reaction, but those corrections are confined solely due to the nature (formula) of the reaction, rather than any particular concentration/pressure conditions.) The definition of chemical activity can get complex in certain cases, but for most normal reaction conditions, the "chemical activity" of a reactant is proportional to its molar concentration - or in other words, how many molecules of the reactant are in a given volume.
For solution reactions, we typically use the concentration directly, but for gaseous reactants, we can take advantage of the ideal gas law and realize that at a fixed temperature, the concentration of a substance is proportional to it's partial pressure. Which means the chemical activity of a particular gaseous reactant will be proportional to its partial pressure.
So in determining the reaction rates of such things as combustion, it's the partial pressures of oxygen and other reacting gases that matter, rather than the total pressure. Absent any side reactions, or considerations such as thermal transfer, adding or removing an inert gas should not affect the rate of reactions.
(But note for combustion, which is highly exothermic, thermal transfer effects can't necessarily be neglected. The ability of the inert gas to either carry away excess heat - or act as a blanket to keep heat near the reaction point - could substantially change how quickly a fire burns in practice. But I'd expect it to be a complex relationship based on air circulation, convection considerations, etc.)

R. M.'s answer is incorrect. The flammability of an oxygen-nitrogen mixture is much more dependent on oxygen concentration than on oxygen partial pressure. See "Oxygen Partial Pressure and Oxygen Concentration Flammability: Can They Be Correlated?", Journal of ASTM International, 2016.

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text:
Why does p mean −log(x)?
$\mathrm{pH} = -\log[\ce{H+}]$ while $\mathrm{pOH} = -\log[\ce{OH-}]$
Why does $\mathrm p$ represent $-\log$ of something? Is it due to some historical reason or due to some scientific reason?

@TanMath Today, the symbol $\mathrm p$ is interpreted as an operator ($\mathrm px=-\lg x$). If you are interested in the etymology of this use, you might get better answers on http://hsm.stackexchange.com/.

@Loong can you give me evidence for your claim? I cant find anything about p being an operator (except for physics stuff)

@TanMath You can find this definition in the international standard ISO 80000-9 Quantities and units – Part 9: Physical chemistry and molecular physics as well as in the IUPAC Green Book.

@Loong is that online?

@TanMath The Green Book is available online: http://www.iupac.org/fileadmin/user_upload/publications/e-resources/ONLINE-IUPAC-GB3-2ndPrinting-Online-Sep2012.pdf

According to the international standard ISO 80000 Quantities and units – Part 9: Physical chemistry and molecular physics, the symbol $\mathrm p$ is interpreted as an operator $(\mathrm px=-\lg x)$. It is used in particular (but not only) in the definition of the quantity $\mathrm{pH}$:
$$\mathrm{pH}=\mathrm pa_{\ce{H+}}=-\lg\left( a_{\ce{H+}}\right)=-\lg\left(m_{\ce{H+}}\gamma_{\mathrm m,\ce{H+}}/m^\circ\right).$$
where $a_{\ce{H+}}$ is the activity of $\ce{H+}$ in solution and $\gamma_{\mathrm m,\ce{H+}}$ is the activity coefficient of $\ce{H+}$ on the molality basis at molality $m_{\ce{H+}}$. The standard molality $m^\circ$ is chosen to be equal to $1\ \mathrm{mol\cdot kg^{-1}}$.
The definition of $\mathrm p$ (and $\mathrm{pH}$) given by ISO is actually quoted from IUPAC Quantities, Units and Symbols in Physical Chemistry (Green Book).
In its current form, this definition can be traced back to a recommendation given in
  R. G. Bates and E. A. Guggenheim
  Report on the standardization of pH and related terminology
Pure Appl. Chem., 1960, Vol. 1, No. 1, pp. 163–168
http://dx.doi.org/10.1351/pac196001010163
which reads

There already exists international agreement that $\mathrm{pH}$ should be written and printed on line in roman type. We recommend that, with the unique exception of $\mathrm{pH}$, the operator $\mathrm p$ (printed in roman) should denote $-\log_{10}$.
  For example
  $\mathrm p m_\mathrm H$ means $-\log_{10}m_\mathrm H$
  (…)
  where $m$ denotes molality (…)

$[{\ce{H+}}] = 10^{-\text{pH}}$. In other words $[{\ce{H+}}]$ is $10$ to the power of $-\text{pH}$.  Actually, this definition was published in a German journal, but luckily power in the mathematical sense is written as Potenz.
Note that the correct definition of $\text{pH}$ is now $\text{pH} = -\lg a_{\ce{H+}}$ where $a_{\ce{H+}}$ is the activity of $\ce{H+}$


text:
How do water insoluble medications get absorbed in the blood stream?
If a chemical is ingested (supplement or drug) orally, which is insoluble in water, how can it be absorbed into the body?  Is there something inside a capsule that the compound is mixed with?

Think of this: fat is insoluble as well, yet you may eat fat, and it will get digested. How so?

I'm fat in general, so maybe I'm capable of absorbing anything. :-) Is there a chemical reaction in our stomach that turns fat into something else so we can absorb it. I do know a person with Colitis can absorb a lot more than usual, not to their benefit. Especially MSG, gluten, chemicals that cause chemical imbalances (from MacDonalds, Dunkin Donuts, Burger King, etc..), etc...

Long story short, we've been able to absorb fats way before any supplements or drugs came along (and way before we became humans, too). The digestive enzymes in the bile split the fats into a fine emulsion and work on them. Here my knowledge becomes somewhat patchy. Let someone with the better grasp write an answer.

+1, thanks Ivan!  I just found some good material after some reading.  Processing a chemical to be "water solubilized" seems to be how they do it.  According to some material I found, Vitamin E needs to have this done during manufacturing.  Whereas Vitiman B and C, are naturally water soluble.  Hopefully we get a response.

Regarding fats - a simple explanation is that the enzymes secreted into the small intestine break them down into smaller pieces, which are then absorbed. You can read Wikipedia for more info https://en.wikipedia.org/wiki/Fatty_acid_metabolism

About the vitamin E, are you referring to tocopheryl acetate? I am guessing it is more water soluble, but I can't be sure.

Yes, Exactly! ...

Thanks :P
I just checked its solubility, <0.1g per 100ml, in high school inorganic chemistry you'll be told it's insoluble, but as a fat-soluble vitamin(which means low daily intake requirement)  it seems well enough for supplementary purposes.

Well, firstly the capsule itself is not entirely insoluble. You can see this for yourself - if you drop one into water, it will start to dissolve very slowly and the water will turn cloudy. This process usually goes faster in the stomach, which has a very acidic environment, or the small intestine which is more alkaline. The capsule material is designed to dissolve when it's eaten, hence releasing the drug - if it didn't dissolve, the whole capsule would just pass through your digestive system and come out... at the other end.
On to the issue about the drug itself. In general designing a drug is not very simple at all, because it has to survive the very same enzymes that break down our food - if the enzymes break down the drug then it becomes useless. For example, if you tried to eat insulin it wouldn't do you any good - insulin is a protein and will be immediately digested by protease enzymes. If the drug survives those enzymes, it has to be able to diffuse through the lining of the small intestine in order for it to enter the bloodstream. The cell membrane, being a non-polar lipid bilayer, doesn't easily let polar or charged molecules diffuse across it. So normally drugs have to be somewhat lipophilic, which is a fancy way of saying non-polar. But at the same time, as you rightly note, they also have to be water-soluble. Most drugs try to strike a balance between the two.
One way that drugs are absorbed is by reversible protonation. Many drugs have amine functional groups with a $\text{p}K_{\text{aH}}$ of around 6-8. At the physiological pH of 7.4, an equilibrium will exist in which both the unprotonated and protonated forms exist in significant amounts. The unprotonated form can cross cell membranes more easily, and the protonated form is more soluble in water.
There are other drugs that don't diffuse across cell membranes very well, due to their large polarity. They may be absorbed via different means, such as active transport (in which the cell uses energy in order to let the molecule enter), or facilitated diffusion (which is the same as diffusion, except that a protein helps out). An example is levodopa; it is structurally similar to the amino acid phenylalanine and the transport protein for phenylalanine can't quite tell the difference. Being polar molecules, these drugs are already water-soluble.
In fact, that's only half the story, because different drugs have different targets. Some need to go to the brain and for this to happen, they have to cross the blood-brain barrier. Some need to stay in the intestines to kill off bacteria - so you have to make them really polar so that they can't diffuse into the bloodstream. Absorption is only a small part of a drug's journey in the human body. If you are interested in finding out more, look for a textbook on medicinal chemistry and read the section about pharmacokinetics - it should have all the info you want.

Not to forget, some if not most capsules are digestible, not just soluble. Many are made of protein or starch, which can be digested into amino acids and glucose, thereby releasing the drug within.

@busukxuan cool - didn't know that :)

Apart from acknowledging orthocresol’s fine answer, allow me to add that nothing is ever entirely insoluble in something else. There is always entropy to be gained by having just a few molecules of A diffuse into B, so tiny bits of (say) hexane will end up in water.
Hexane and water are extreme examples here. Most drugs are designed to be somewhat semi-polar as are most hallucinogenics (albeit not necessarily designed). They will dissolve in aquaeous solvents well enough to be taken up. One often teaches solubility as black-and-white, when in reality it should be millions of shades of grey.

Compounds like organic amines, which are basic and would be insoluble in water are very often turned into water soluble salts by treating them with acid yielding hydrochloride salts, sulfates, citrates and so on. Examples include ephedrine and methyl phenidate, both of which are marketed as hydrochloride salts in Europe.

OP's asking about things which are used despite being insoluble...

What is "OP"? ..


text:
Enthalpy of formation and bond energies of CH2(OH)2?
I was wondering if I could have the bond energy and enthalpy of formation in kJ/mol for $\ce{CH2(OH)2}$?
I honestly searched the Internet and my textbook and I couldn't find it.  I'm sure I could have missed it somewhere but nevertheless I need to calculate the energy produced by the reaction of $\ce{CH2O}$ and $\ce{H2O}$.
If someone could give me that info that would be exceedingly helpful.

Firstly, regarding the bond energy, you will have to specify exactly what bond you are talking about, you can't ask for the bond energy of a compound because compounds have many bonds. This doesn't apply for diatomics (for example, you can ask what the bond energy of $\ce{N2}$ is; although that is loose usage of terminology, the meaning is clear). But methanediol is obviously not a diatomic.
Secondly, from what I understand, you need to find the enthalpy change for the hydration of formaldehyde: $\ce{CH2O_{\;(aq)} + H2O_{\;(l)} -> CH2(OH)2_{\;(aq)}}$ and your approach was to find the enthalpies of formation of the individual compounds. That's a good idea and more often than not, it works because  enthalpies of formation are widely tabulated and easy to find. But it's often worthwhile just searching for the enthalpy change of the reaction you are interested in, especially if it is a relatively well-known or simple reaction (like this one). Anyway, this recent paper gives the value of $\Delta _{\text{r}}H$ for the reaction above as $-39.0 \text{ kJ mol}^{-1}$.


text:
How does valency change in a reaction?
There is a reaction given in my book:
$\ce{2FeSO_4 +Heat -> Fe_2O_3 + SO_2 + SO_3}$
I just noticed that on the left $\ce{Fe}$ has valency $2$, while it has $3$ on the right. How can it change valency?
Changing the number of electrons is easy, but this only changes the charge, not the valency. Does this mean that the number of protons changes?

Yes, valency can change, just like that. Some elements (in fact, most of them) can manifest different valencies; that's their nature. If you look again at this very reaction, you might notice that Fe is not the only element to change its valency here; S does the same. Reactions like this one are called redox reactions. There is a huge body of knowledge that explains what valencies (or rather, what oxidation states) an element can have, and which of them are more stable; the answer by @Raghav Gupta touches on that in more detail.
As to the number of protons (if you mean the protons in the element's nucleus), it has nothing to do with valency. Really, Fe contains 26 protons; is its valency 26? Absolutely not, it is 2 or 3. Also, the number of protons in an element can never be changed at all. If you change it, you'll get another element; also, that would be a nuclear reaction, a thing which is out of scope of chemistry textbooks.

The valency of an element is the number of hydrogen atoms that can combine with or replace (either directly or indirectly) one atom of the element. In other words, the valency of an element is the number of electrons an atom of the element uses to combine with atoms of other elements - it is the combining power of an atom of the element. In an atom, the valence electrons are the electrons that can be used in combining with other atoms - these are the electrons in the orbitals of the outermost shell (also called valence shell).
That means valency  is not dependent on number  of proton.
Iron (Fe) Atomic no 26, so 26 electrons to be filled in different orbital. So the electronic configuration is:
$\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6}$. This is the way of filling of electrons according to aufbau's principle. But the exact arrangement is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2}$
So Iron can lose two electrons from the outermost 4s subshell to show a valency of two. Then it can lose another electron from the 3d subshell (penultimate) two show a valency of three. When it shows the valency of two the electron configuration is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 3d^6}$.
When it shows the valency of three the electron configuration is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 3d^5}$
And the +3 valence state is stabler than the +2 valence state, because the d-orbital contains 5 electrons (half-filled). As you know fully filled and exactly half filled orbitals are more stable than the orbitals having different no of electrons, i.e. less than half filled and more than half filled but not fully filled.


text:
Inert in Cathode when Electrolysis
Why Platinum, Gold, Carbon are inert? In electrolysis, it's not react in anode because anode is doing oxidizing, but how about cathode? Are they still inert?

Well, yes, they more or less are. What would you expect?

I mean, why they are not react, cahtode doing reduction, while Gold could does reduction the most, but why Gold doesnt resact?

Why would it? Metallic gold is already a reduced form.


text:
Why does butanoic acid have a higher boiling point than isobutyric acid?
Why does butanoic acid have a higher boiling point then isobutyric acid, when they have similar molar mass and molecular structure?
Butanoic acid ($\ce{C4H8O2}$) has a molar mass of 88.11 g/mol, the boiling point is 163 °C, and intermolecular force of H-bond.

Isobutyric acid ($\ce{C4H8O2}$) has a molar mass of 88.11 g/mol, the boiling point is 155 °C, and intermolecular force of H-bond.

Same reason why n-Butane has a higher boiling point than Isobutane: More intermolecular van der Waals interaction.

In general, linear chains are more easily structured and packed than branched ones.  You need more energy to disturb the more relatively packed structure.This could explain why butanoic acid has a higher boiling point than isobutyric acid, and yet they are isomers.

Structuring and packing is usually an argument when talking about the melting point, right? How relevant would it be for liquid $\to$ gas?

There is higher energy needed to get away from liquid.

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


text:
How does crystal field stabilisation energy relate to the stability of a complex?
I know how the crystal field stabilization energy is calculated, but how does it reflect stability? CFSE measures the energy relative to the barycentre, not the initial energy of the $d$ orbitals. For example,a $d^5$ configuration in a weak ligand field would have zero CFSE, but that certainly does not mean that the complex is unstable.

You nailed it, and there is not much left to add.

@Jan So it doesn't relate to stability?

Lets start from the beginning
Firstly the metal orbitals (eg- like 3d) are degenerate,
now as soon as the ligand surrounds the metal this degeneracy between the orbitals is lost (continuing the same example - the energy of five 3d orbitals is no longer the same)
The orbitals of the subshell are repelled by the ligands depending on their orientation relative to that of ligands, and they split into few sets
Now, the orbitals are generally grouped into 2 sets (taking a very classic high school example of splitting in case of octahedrally surrounded central metal atom by ligands)
One is between the axes and other along the axes and so now we assume the degeneracy between these 2 sets of orbitals of a subshell,
And the important part is that the cfse that is calculated,is with respect to the energy of the whole original set of orbitals (ie wrt barycentre)
( Eg-when the whole 3d was degenerate and ligand field was strong to increase the energy of complete set of orbitals but not enough to split them, as the ligands come closer the splitting happens, that is the reason why in standard textbooks we generally have 3 diagram representation of this phenomenon, what i am talking is about the 2nd to 3rd diagram transition)
And now finally how is this splitting energy is related to stability
Lets say we measure the energies wrt barycentre , so now the orbitals of higher energy will have positive energy and of lower will have negetive energy, and because of the splitting most of the times this relative energy (whose fancy name is cfse) comes out to be negetive, that is the energy of the system has decreased because of splitting and hence more stable
so its not really wrong to call crystal field splitting energy as crystal field stabilisation energy
Though there also many cases where cfse comes out to be positive,but then again there are other interesting and intriguing explanations for them!


text:
How to calculate the equivalent weight of a metal in a metal sulphate?
The question is:

$\pu{2g}$ of a metal is present in$\pu{4.6g}$ of its metal sulphate. Find the equivalent weight of the metal.

I understand the concept that equivalent weight of an element is the molecular weight divided by the n-factor (in this case it's the valency), but I couldn't arrive at a conclusion.
Could you also help me understand how equivalent weight of a metal is related to the equivalent weight of the sulphate? In my textbook, it's written like this:
$$\textrm{eq. wt. of metal} = \textrm{eq. wt. of metal sulphate(2-)}$$
and then they carried on with the calculations.
PS I know the answer, I need to know how one should conceptually arrive at it!

I assume that if you are dealing in equivalent weights then you must be well associated with the concepts of equivalents.
Now,
$$ \textrm{equivalent} = \textrm{weight} / \textrm{eq. weight}$$
and since compounds combine in equal amounts of equivalents we have
$$\frac{\textrm{weight of metal}}{ \textrm{eq weight}} = \frac{\textrm{weight of sulphate}}{ \textrm{its eq weight}}$$
$$\rightarrow 2/x = 2.6/48$$
Hence $x= 39.92$.


text:
Must compounds formed between solids be a solid as well?
My teacher gave us a long list of compounds and asked whether they are solid, liquid, or gas at room temperature and normal atmospheric pressure. I only know the more common compounds (eg. carbon dioxide, ammonia) and elements but not the rest.
Can I assume that a compound is solid because its constituent elements are solids at normal temperatures and pressures as well? If so, does the same goes with liquids and gases? If so, what if its constituting elements are not of the same physical state?
If not, then how can I determine the answer? Or must I memorize the physical state of the compounds in order to answer the questions?

No, they don't have to. In short, you have to memorize. The physical state of a compound has nothing to do with the states of its elements.
Examples:
$$\ce{Hg~(liquid) + Br2~(liquid) -> HgBr2~(solid)}$$
$$\ce{C~(solid) + S~(solid) -> CS2~(liquid)}$$
Now, there are some common patterns (an ionic compound is most certainly a solid; a compound with small non-polar molecules is probably a gas), but that's another story.

Are there any exceptions for ionic compounds that are not solids?

Of course there are; look for ionic liquids.


text:
How to use the Nernst Equation
So I understand the Nernst Equation, but I don't have some info. (Please bear with me, I'm in secondary.)
I'm doing a Cu - Al cell. My E cell is negative (-1.33). Does this have an affect on the equation?
What exactly does E0 stand for? I understand how to find E cell, but do I square it by something? I can't find online how many electrons are swapped. How many electrons are swapped?
How do you factor in the pH? I've calculated the pH and the number of ions in it, but where do i add this in on the equation? Please lay it out how you would in the equation so I can see. (I have numerous pH's)
I'm stuck on R (gas constant, which is 8.31 (volt-coulomb)/(mol-K)). I can't figure out the volt-coulomb as i don't know the voltage. How can I calculate the coulomb? Then the volt-coulomb? And is Mol-K (Mole)(Kelvin)? If so, do I factor in the temperature in Kelvins twice?
For N (n = number of moles of electrons exchanged in the electrochemical reaction) do i just multiply the moles per electron?
Finally - What exactly am I dividing to calculate Q? (How do I calculate Q?)
Could you do one full equation as a template for me please? One pH level is 4.3
Please explain this all in basic jargon - or just normal English if you prefer. My vocabulary is limited. Thank you!

I can help more if you have an example of a question from past-year competition archives.

I have everything sorted except N- Could you please directly answer how many electrons are exchanged- I'm unsure if it's 6, 3, or 5 (adding them both together). Thank you!

Asking the faceless masses of the internet for help that you need very quickly can be a dangerous proposition, but let's see where this goes.
Here's the Nernst Equation:
$$E=E^\circ - \dfrac{RT}{nF}\ln{Q}$$

$E$ is the measured cell potential
$E^\circ$ is the standard cell potential, assuming 1.0 molar concentration, 1 atmosphere of pressure, and 25 degrees Celsius
$R$ is the ideal gas constant - $8.314\ \mathrm{\frac{V\cdot C}{K\cdot mol}}$
$T$ is the temperature
$n$ is the number of electrons transferred
$F$ is the Faraday constant $9.648\times 10^4\ \mathrm{\frac{C}{mol}}$
$\ln$ is the natural logarithm
$Q$ is the reaction quotient, which is related to the amount of reactants in solution.

Finding $E^\circ$

You look this value up. Actually you look up the values of $E^\circ$ for your half-reactions.
You say you are doing copper-aluminum, so your half-reactions are probably:
$$\ce{Cu^2+ + 2e- -> Cu}$$
$$\ce{Al^3+ + 3e- -> Al}$$
Here is a link to Wikipedia's Standard Electrode Potential Data Page
$$\begin{array}{r|c}
\mathrm{half-reaction} & E^\circ/\mathrm{V}\\ \hline
\ce{Cu^2+ + 2e- -> Cu} & +0.337 \\
\ce{Al^3+ + 3e- -> Al} & -1.662\\
\end{array}$$

How many electrons are exchanged?

You determine this number from your half reactions.
$$\begin{array}{r|c|l}
\mathrm{half-reaction} & E^\circ/\mathrm{V} & n\\ \hline
\ce{Cu^2+ + 2e- -> Cu} & +0.337 & 2 \\
\ce{Al^3+ + 3e- -> Al} & -1.662 & 3\\
\end{array}$$

How do you factor in pH?

You only factor in pH if $\ce{H+}$ or $\ce{OH-}$ appear in your reactions. In this case, neither does. However, high pH might shut down your reaction since both $\ce{Al(OH)3}$ and $\ce{Cu(OH)2}$ are low solubility.

How do I factor in the number of ions?

I suppose you mean the concentrations of $\ce{Cu^2+}$ and $\ce{Al^3+}$? These concentrations impact $Q$, the reaction quotient. For simple half reactions like yours, $Q=\frac{1}{\mathrm{ion\ concentration}}$

Units

The volt-coulomb and kelvin-mole units on $R$ are just what they are, units. You do not need to determine them; they are part of the value of $R$. These units are necessary to that all of the other units cancel leaving just volts as the unit of $\frac{RT}{nF}$

Putting it all together

You now have what you need to determine the half-reaction potentials at your nonstandard values. Just plug in what you have for $E^\circ$, $R$, $T$, $n$, $F$, and $Q$. Then you can calculate your cell potential.
Positive cell potential is spontaneous, so we want to plug in values for reduction (as written) and oxidation (reversed) to get the largest possible positive number. In this case (in which I use the standard potentials since i do not know your concentrations or temperatures):
$$E^\circ =\mathrm{ +0.337\ V -(-1.662\ V) = +1.999\ V}$$
Copper is reduced; aluminum is oxidized.
However, likely you are doing this experiment to calculate or verify something, likely either $E^\circ$ or ion concentration. If this is the case, leave that quantity as a variable and solve for it.
However, your $E$ is negative, which suggests electrolysis. Is this what you are doing?

Actually the value of $n$ in this case depends on how you combine the half-equations - so for the reaction $\ce{3 Cu + 2 Al^{3+} -> 3 Cu^{2+} + 2 Al}$ (or the reverse) you'd have $n = 6$, but for the reaction $\ce{3/2 Cu + Al^{3+} -> 3/2 Cu^{2+} + Al}$ you'd have $n = 3$. In the end the maths works out to be the same because of how the reaction quotient $Q$ is defined

I have everythng sorted except N- Could you please directly answer how many electrons are exchanged- I'm unsure if it's 6, 3, or 5 (adding them both together). Thank you!

@Kiwi - see orhocresol's comment. My explanation focuses on keeping the half-reactions separate for as long as possible. Then you do not need to worry about the number of electrons exchanged in the full reaction.

@Kiwi - refer to my comment above. In the first equation I wrote you have 3 copper atoms giving up 2 electrons each for a total of 6. These 6 electrons are accepted by 2 aluminium ions. Therefore $n = 6$. Note that $n$ is a stoichiometric coefficient, and although it's how I described it above, it's not a "number of electrons transferred". It's more like "number of moles of electrons transferred per mole of the reaction as written". By that I mean, the proper explanation should be: if you react 3 moles of Cu and 2 moles of Al$^{3+}$, 6 moles of electrons will be transferred, and hence $n = 6$.

@Kiwi the reason for the distinction is because obviously you can't have 1.5 copper atoms so the same explanation fails for the second equation I wrote.... but if you have 1.5 moles of copper then 3 moles of electrons are transferred. Ben's method of keeping the half-equations separate is also entirely valid and personally I think it is a more "proper" explanation as $n$ is an explicit stoichiometric coefficient in a chemical equation. However you do have to be able to keep track of which half-reaction has which value of $n$ as usually they will not be the same (as it is in this case).


text:
Calculate water hardness from grams of CaCO3
We have to figure out what the water hardness in mg/L or ppm is for a $\pu{20ml}$ solution of $\pu{0.400M}$ $\ce{CaCl2}$.
We learned that the formula for hardness is mg/L of calcium carbonate per liter. I started by calculating that there would be $\pu{0.801 g}$ of $\ce{CaCO3}$ precipitate if reacted with $\ce{Na2CO3}$.  I then converted $\pu{0.801 g}$ to mg and got $\pu{801 mg}$. I finally divides this by the original $\pu{20ml}$ or $\pu{0.02 L}$ but this gives $\pu{40050 mg/L}$ but the correct answer is $\pu{400 mg/L}$.
I feel like I might be making a mistake with units as my answer is 100 times greater than the correct answer but I don't see the problem.

Unless I'm missing something, I don't see the problem with your working either. Perhaps check whether you got the original concentration of $\ce{CaCl2}$ given in the problem right (the volume of the solution does not affect your final answer). The given answer may well be wrong, it's irritating but it does happen.
There is one thing I want to comment on though: never round off your answer too early. Since you obtained the intermediate answer of $801 \text{ mg}$, I am assuming you used a precise molar mass for $\ce{CaCO3}$ - I will take this to be $100.09 \text{ g mol}^{-1}$. (If you used the value $100$ you would have obtained $800 \text{ mg}$.) Your calculations should be:

Number of moles of $\ce{CaCO3}$ formed upon addition of excess $\ce{CO3^2-}$

$$\eta_{\ce{CaCO3}} = (0.400 \text{ mol dm}^{-3})(0.02 \text{ dm}^{3}) = 0.008 \text{ mol}$$

Mass of said $\ce{CaCO3}$

$$m_{\ce{CaCO3}} = (0.008 \text{ mol})(100.09 \text{ g mol}^{-1}) = 0.80072 \text{ g} = 800.72 \text{ mg}$$

Hardness of water in $\text{mg L}^{-1}$

$$[\ce{CaCO3}] = \frac{800.72 \text{ mg}}{0.02 \text{ L}} = 40036 \text{ mg L}^{-1}$$
This is your final answer and if you want to round to the appropriate number of significant figures, you should do so only at this point. Your problem is that you rounded too early (after step 2) and this means you obtained an imprecise final answer, $40050 \text{ mg L}^{-1}$.
Here, the appropriate number of sf is technically 3, since the original concentration of $\ce{CaCl2}$ is given to you in 3 sf. Since you obtained the imprecise answer $40050 \text{ mg L}^{-1}$, you would have rounded this to $40100$. But if you stuck to the precise value at step 2 and obtained the answer $40036 \text{ mg L}^{-1}$, you would round this to $40000$.

I understand now I missed something while reading the instructions.  Each sample had been concentrated by a factor of 100.  Thanks

Happened to me lots of times :) spend 30 mins on a simple question before realising I skipped one line of the question text.... Glad you found it!


text:
Calculate the amount of energy released when 500 cm³ of methane gas at STP reacts with excess air according to the equation:
$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} \qquad \Delta H = -890\ \mathrm{kJ\ mol^{−1}}$$
I am not totally sure how to work this out.
What I did, though, was:
$\ce{CH4}$ $M = 16\ \mathrm{g/mol}$
So, $500\ \mathrm{cm^3} = 31.25\ \mathrm{mol}$
$\ce{CH4}$ $H_\mathrm f = -890\ \mathrm{kJ/mol}$
-> For $31.25\ \mathrm{mol}$ we have $27\,812.5\ \mathrm{kJ}$
If I am wrong, can someone explain me why?

How you find the moles from volume? Did you assume $500\ \mathrm{cm^{3}}$ is $500\ \mathrm g$? This is wrong, because the density of $\ce{CH4}$ is not $1\ \mathrm{g/cm^{3}}$. You have to use another way to calculate moles of the gas. Think about ideal gas law or how much volume for STP gas.

Yes.. I just got it, I am supposed to use the volume of methane gas under standard conditions which is 22.4l, like any other gas... Thank you!


text:
Calculate the mass of butane gas that would be needed to heat 724 cm³ of water from an initial temperature of 7.44 °C to 50.7 °C
The thermochemical equation for the combustion of butane is:
$$\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(l)} , \quad\Delta H = -5748 \ \mathrm{kJ\cdot mol^{−1}}$$
So far, I've calculated the heat released by water as follows:
$$\begin{align}Q&= m \times c \times \Delta\vartheta\\
&= \rm 0.724\ g \times 4.18 \times 43.2 \\
&= \rm 130.7\ \frac{kJ}{mol}\end{align}$$
I don't know how to carry on from here.

Firstly, you need to understand what is happening. Butane gas is burnt which releases heat energy. This heat energy is absorbed by water. As the water absorbs this heat energy, it increases in temperature from $\mathrm{7.44}$ to $\mathrm{50.7}$ degrees Celsius. So in your in working out, the formula you have used is correct but you have written the 'heat released by water' which is incorrect. The water absorbs heat energy which is why it increases in temperature.
Secondly, always remember to write down the units when you are using equations. That way you won't make silly mistakes and always end up with the correct units. For example, in your working out you got that the heat energy absorbed was $\mathrm{130.7\ kJ\ mol^{-1}}$. It should be clear that this the wrong unit as energy is measured in joules, not joules per a mole.
Now, the first step in doing this question is to calculate the energy that is required to heat the water up to the desired temperature: $$q = mc\Delta T$$
$$= \mathrm{724\ g\times4.18\ J\ g^{-1}\ K^{-1}\times43.26\ K = 130918.6\ J = 130.9\ kJ}$$
Note how the final answer in joules which is the correct unit (also my answer differs slightly to yours because for $\Delta T$ I used $\mathrm{43.26}$ rather than $\mathrm{43.2}$)
So now we know that we need $\mathrm{130.9\ kJ}$ of energy to heat the water up. This means that the burning of butane must release $\mathrm{130.9\ kJ}$ of energy as $\Delta H = q$.
By looking at the equation that is given, it can be seen that burning 1 mole of butane releases $\mathrm{5748\ kJ}$ of energy (you can tell it releases energy as $\Delta H < 0$ ). So to calculate the number of moles of butane required: $$n_\text{butane} = \mathrm{\frac{130.9\ kJ}{5748\ kJ\ mol^{-1}} = 0.023\ mol}$$
Note you know that you have used the right equation as the units that we got for the final answer is in moles which is correct
Now that we know the number of moles of butane that is required, all we have to do is convert it into grams: $$m_\text{butane} = \mathrm{0.023\ mol\times58.12\ g\ mol^{-1} = 1.32\ g}$$
Therefore your final answer is: 1.32 g
Note your final answer should be to 3 significant figures as the data in the question is to 3 significant figures


text:
Molarity is dependent on temperature?
Okay I know that as the temperature rises the number of particles in a given volume decreases. The molar volume likewise increases. So the number of molecules per liter would decrease as the temperature rises meaning that a 1M solution is only 1M at a specific temperature.
But why is molarity dependent on temperature? I mean sure the density does decrease as the temperature rises but still there is a small chance that the molarity will stay the same.

First, there is very very little chance it will remain same. Density will surely change with temperature.
Secondly, even if there is a small chance that it will remain same, then also there is an even greater chance that it will change.
So we will have to look at the general case and conclude that molarity depends on temperature.

We have n = M * volume of solvent.
n = weight of compound / molecular weight.
Weight of compound / Volume * mol. Weight = M.
Density/ mol weight = M.
Now since molecular weight of a compound is constant . We have density directly proportional to the Molarity.
As you agree with the variation of density with temperature I am sure you agree to the change of Molarity also now.


text:
What is the justification for Ligand Close Packing Model?
The LCP model predicts that nonbonded atom to atom distances remain approximately the same when the central atoms are changed. What is the theoretical justification for this? thanks


text:
Fractional Distillation
Why does the long length of the fractionating column in fractional distillation enable more effective separation of two liquids in a mixture?
(basically what's the purpose of the fractionating column being long in relation to the effectiveness of the fractional distillation process)

It simply spreads the temperature change over a longer distance, enabling separation of distillates with boiling points close to each other. "Generally the component parts boil at less than 25 °C from each other under a pressure of one atmosphere."
Imaging reading a mercury thermometer one cm tall as compared with one 100 cm tall: it would be difficult to distinguish temperatures on the smaller one. In the fractionation column below, it helps keep the outputs separate.


text:
Solutions of several salts
What happens when you have a solution of a salt, and then try to dissolve another salt in that solution? I know about the common ion-effect etc, but how do you solve problems like this:
How much AgCl can you solve per L in a 3.8g/L NaCl solution?

To solve a problem like this, we need to know the solubility product constant $K_\mathrm{sp}$ for $\ce{AgCl}$, where at equilibrium,
$$K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]=1.8\times 10^{-10}\ \mathrm{M^2}$$
This relationship stipulates that under equilibrium conditions at all times, the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ must satisfy this relationship.
If the solution contains just $\ce{AgCl}$, we can determine the solubility. The concentrations of both ions must be equal based on the chemical formula.
$$\begin{array}{lcl}
x&=&[\ce{Ag+}]=[\ce{Cl-}]\\
x^2 &=& 1.8\times 10^{-10} \\
x &=& \sqrt{1.8\times 10^{-10}}=1.342...\times 10^{-5}
 \end{array}$$
We have $1.342\times 10^{-5}\ \mathrm{mol/L}=1.92\times 10^{-3}\ \mathrm{g/L}$ of $\ce{AgCl}$ in soltion.
However, if we already have some $\ce{Cl-}$ in solution, then things are a little different. Let's say we have $3.8\ \mathrm{g/L} = 6.5\times10^{-2}\ \mathrm{mol/L}\ \ce{NaCl}$ in solution, now we have a different concentration of $\ce{Cl-}$. When $\ce{AgCl}$ is added, some will dissolve, and the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ will both increase by some amount $x$.
We solve these problems using the ICE (Initial, Change, Equilibrium} approach:
$$\begin{array}{c|r|l}
\ & [\ce{Ag+}] &[\ce{Cl-}] \\ \hline
\mathrm{I} & 0 & 6.5\times 10^{-2} \\
\mathrm{C} & +\ x & +\ x \\
\mathrm{E} & x & 6.5\times 10^{-2} + x\\ \hline \end{array} \\
\ \\
K_{\mathrm{sp}}=1.8\times 10^{-10}=x(6.5\times 10^{-2} + x)\\
1.8\times 10^{-10}=x^2 +(6.5\times 10^{-2})x \\
0 = x^2 +(6.5\times 10^{-2})x - 1.8\times 10^{-10}\\
x = 2.77\times 10^{-9}, -0.065$$
One of the values for $x$ is nonsensical. The concentration of $\ce{Ag+}$ cannot be negative. Thus the amount of $\ce{AgCl}$ that dissolved in one liter was $9.77\times 10^{-9}\ \mathrm{mol}=1.38\times 10^{-6}\ \mathrm{g}$.
Note that since $x\ll 6.5\times 10^{-2}$, we could have simplified to $6.5\times 10^{-2}+x \approx 6.5\times 10^{-2}$ and then our maths would not have needed to invoke the quadratic theorem while getting the same answer within significant figures. This approximation is often valid. However, unless you are taking a test, you should always be able to Wolfram Alpha.


text:
Line of maximum density in liquids
Why is the thermal expansion coefficient zero along the line of maximum density in liquids? Why is $(∂^{2}V/∂T^{2})_{P} > 0$ along the same line?

'Cause it's maximum - that's how extrema are found analytically! I don't think it's a place for basic questions about derivatives.


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What does the "h" "k" and "(5.2.2.02,6)" mean in dibenzo(h,k)-4-oxatricyclo(5.2.2.02,6)undec-8,10- diene-3, 5-dione?
I am preparing to do an experiment that involves synthesizing dibenzo(h,k)-4-oxatricyclo(5.2.2.02,6)undec-8,10- diene-3, 5-dione via a Diels-Alder reaction.
Could someone please explain the nomenclature for this molecule? Specifically, what do "h", "k" and 5.2.2.02,6 mean?
Below is the reaction:

I'm not very good with organic chemistry but this got me thinking, and after checking this I think the [h,k] denotes which positions the oxatricyclo~ group is bonded to the "main" ring. Maybe this is clearer: http://i.imgur.com/Iudam83.png

@Molx Interesting, but unfortunately, it's not even wrong.

The systematic (but not necessarily preferred) name tricyclo[5.2.2.02,6]undecane describes the structure of octahydro-1H-4,7-ethanoindene (see this answer on how to name tricyclo compounds):

Expanding the name from tricyclo[5.2.2.02,6]undecane to 4-oxatricyclo[5.2.2.02,6]undec-8,10-diene-3,5-dione adds the functional groups, which leads to the structure of 3a,4,7,7a-tetrahydro-4,7-etheno-2-benzofuran-1,3-dione:

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), isomers of fused polycyclic ring systems are distinguished by lettering each peripheral side of the parent component using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1,2’, b for ‘2,3’, etc.
For example:

Accordingly, fusion of two benzene rings to the above-mentioned parent structure 4-oxatricyclo[5.2.2.02,6]undec-8,10-diene-3,5-dione at the relevant sides yields the name dibenzo[h,k]-4-oxatricyclo[5.2.2.02,6]undec-8,10-diene-3,5-dione.
(Note that this name does not necessarily correspond to the preferred IUPAC nomenclature and notation.)

Regarding Loong's correct answer, unsaturated fused polycyclic hydrocarbon azulene atom numbering1 (top left), and peripheral bond (edge) letter “numbering” for fusion nomenclature (top right).
Below is the skeleton of your product (without the two fused benzenes, before the dibenzo[…]… nomenclatoric operation), atom numbering2 (bottom left).
However, letter “numbering” of edges (bonds) is not used for polycyclic saturated hydrocarbon nomenclature, as the names are constructed in a different way3 (also there's no single continuous edge loop defined). But if someone would try to do that similarly for unsaturated polycyclic ring nomenclature, he/she would come with bond letter-ing, including those h,k (bottom right).

It's really strange that this very notable chemical seller uses this weird, wrongly constructed name as (the only) product name (webarchived) (or did you see it elsewhere?); see e.g. PubChem record for alternatives. (By the way, I think that a decent name, although maybe not preferred as well, is 3,4-(9,10-dihydroanthracene-9,10-diyl)oxolane-2,5-dione, maybe yet better with bridging nomenclature using prefixes like benzeno, furano, epipyrano, …: 11,15-dihydro-9,10-[3,4]furanoanthracene-12,14-dione (see e.g. examples in BB2013 P-25.4.4.1))

Note that the non-peripheral, fusion atom numbers are appended with Latin letters, but these are unrelated to edge (bond) letters.
Note the different numbering for saturated polycyclic hydrocarbons, starting at “fusion” (wrong term in this context), or bridges junction; also no letters added anywhere.
The dibenzo derivative would be named as an unsaturated pentacyclic hydrocarbon, pentacyclo[6.6.5.02,7.09,14.015,19]nonadeca‐2(7),3,5,9(14),10,12‐hexaene.


text:
Hill system with isotopologue
I'd like to use Hill notation (i.e. CH, then alphabetical), but I am not sure how to include isotopologues (i.e. carbon-13) in this notation.
For example, what would be the preferred form of ethylamine (with 13C)?

C13CH7N
CH713CN
12C13CH7N
12CH713CN
Other

I'm leaning toward using #2 but was trying to find the std practice, but I didn't see anything about isotopologues.  Thanks.

I can't point to a definitive authority, but in most of the papers I read, #5 other is used most often.  Specifically for ethylamine:
$\ce{^{13}C\,^{12}CH_7N}%edit$
I've always assumed the reason for this choice is to accentuate the presence of the isotopically unusual atoms.  I suppose a good test for my theory would be formulas with multiple, isotopically distinct atoms of an element such as iron, for which the most abundant isotope is not the lightest.
Hopefully other answers will be along shortly to point to something more authoritative, such as a IUPAC report, guideline, or document.

I've been using that formula notation, for the reason you say. To accentuate the presence of 13C. If nobody has a more authoritative source, I'll use that notation, calling it a modified Hill system if need be, with C, H, then alphabetical by element, with isotopologues listed in reverse by natural abundance (i.e. 14C13C12C). Thanks.

Sorry, just, what does this formula exactly mean? I'm not really experienced in working with isotopologues. Especially, what does the superscript before an atomic symbol mean?


text:
Best Method To Make Large Graphene Oxide Sheets?
There are so many methods to make graphene oxide now, I'm not exactly experienced enough to know which is ideal. I want to know what the best method is for making large graphene oxide membranes. The goal is to have a surface area of a square meter.
Graphene oxide is usually made by filtration so the two requirements I have are:

that it can be made by filtration over porous aluminum oxide (for stability in water as revealed here: bit.ly/1HExnui)...well any method really, as long as Al^3+ cations can be incorporated into the graphene oxide in some way
that the method is known for success with the helium permeation test (as in essentially only water can pass through the graphene oxide).

I've asked a few questions already here on stackexchange and it always feels strange to simply ...demand or extract information. Thank you all.


text:
LiOH base ionization reaction
What is the base ionization reaction for LiOH? Usually, the reaction is like this: $$Base + \ce{H2O->H}[Base] + \ce{OH-} $$
But $\ce{LiOH2+}$ makes no sense. So what could the reaction be?

What you call bases is in fact a subset thereof. Besides that, there is a whole class of inorganic bases, also known as alkali. These are the hydroxides (sometimes called basic salts) of alkaline metals. Their ionization is somewhat different:
$$\ce{NaOH->Na+ + OH-}$$


text:
Hybrization of H in HNO
There is a problem that states: Find the molecular geometry of each molecule and the hybridization of each atom in the molecule. The two molecules are HNO and HCN. I found HNO to be sp3 hybridized and HCN to be sp hybridized. However, it says that only N and O are hybridized in the first. While all three are hybridized in the second. Why is hydrogen listed as sp hybridized?

That is a very good question; I can only assume an error in your book. Note that hybridisation should always be discussed on a per-atom level, and note that both the $\ce{O}$ in $\ce{HNO}$ and the $\ce{N}$ in $\ce{HCN}$ can also be assigned various possible hybridisations. P.S: Welcome to chemistry.stackexchange.com. Take a [tour] and refer to the [help] for any questions about the site.

Hybridization refers to the orbitals that each atom is using to make bonds.  In $\ce{HNO}$, the $\ce{N}$ and $\ce{O}$ are $sp^2$ hybridized.  This means they form a sigma bond with an $sp^2$ orbital from each and the pi bond from the interaction of the non-hybridized $p$ orbital on each.  The $\ce{N}$ also has a $sp^2$ orbital overlap with the $1s$ of $\ce{H}$, which forms a sigma bond.  Each of the lone pairs on the $\ce{N}$ and $\ce{O}$ is in an $sp^2$ orbital.
In $\ce{HCN}$ the carbon is $sp$ hybridized, as is the $\ce{N}$.  $\ce{C}$ and $\ce{N}$ form a sigma bond with $sp$ overlap, and the pi bond from the interaction of both of their non-hybridized $p$ orbitals (to form a triple bond).  The lone pair of $\ce{N}$ is in an $sp$ orbital, and the $\ce{C-H}$ bond is from an $sp$ overlap with the $1s$ of $\ce{H}$.
None of these molecules have $sp^3$ hybridization.

Note that OP explicitly also asked why hydrogen is listed as hybridised. Care to expand on that?

Hydrogen only forms bonds with its 1s orbital.  It can't hybridize to form bonds  as its electron fills a 1 s orbital which is too different in energy from the 2s and 2p oritals to mix.  I believe either OP misinterpreted something regarding this statement or there was an error in the book.

You should edit your answer to include that ;)


text:
Bisulfate vs Bisulfite to neutralize off gassing in paint
I need to neutralize the off gassing in wall paint that contains one or more of the antimicrobial compounds of  isothiazolinone, chloromethylisothiazolinone, methylisothiazolinone, benzisothiazolinone, or octylisothiazolinone.
I applied and used sodium bisulfate rather than sodium bisulfite. Do I need to now rinse it off and find the sodium bisulfite or will the sodium bisulfate work?
Link to study using sodium bisulfite.

Welcome to Chemistry.SE! Take the [tour] to learn more about this site. In particular, no one here can (or will) give medical advice. Good luck!

I am not a doctor. In this case, my first suggestion would be to leave the area and hire someone to apply bisulfite and another coat of a different paint. The two compounds will have very different action.

I am just trying to find out if bisulfite and bisulfate are totally different things. I have the room sealed off and medical treatment has already been  administered. I found this site while searching for answers about what the difference between the two chemicals sodium bisulfite and sodium bisulfate are. I can simply wait 2 months to open up the room again, but could really use the space to work as it is my home office. Thank you for any help.

To closevoters: Am I misinterpreting the question? Since I don't see how this is medical advice. For now I removed the background.

Yes, bisulfate and bisulfite are two completely different things.

Related: Sodium Bisulfate vs Sodium Bisulfite

I feel the question @Loong found is sufficiently close to warrant a duplicate closing.

Just wanted to thank you all for the help. I sincerely appreciate your wisdom. A chemist, I am certainly not! I was able to find the sodium bisulfite at a home brewing/wine making store. I am going to try that to see if it will neutralize the off gassing chemicals in the paint. According to a scientific study in Sweden, it "should" work to neutralize the chloromethylisothiazolinone (an antimicrobial additive) in the paint. From now on, I will stick to organic milk paint, so as not to cause an allergic reaction to my husband!
Sincerely Cathy C.


text:
How to determine force constant of C-C bond in ethanol?
How would one determine the theoretical force constant of the C-C bond in ethanol?
We are asked to calculate the frequency of the C-C bond (theoretical), but to do this we need a force constant, which is not provided to us.
I cannot find a source of this information. The only way I see to calculate the force constant is from the frequency of the bond, which is what we are asked to calculate.
This professor has a reputation of omitting important details

Use the fundamental stretch function below to determine what is the frequency of the C-C bond and what is the wavenumber?
$$\bar{\nu}\mathrm{(cm^{-1})}=1303\sqrt{f\left(\frac{1}{m_1}+\frac{1}{m_2}\right)}$$
where $f$ is the force constant in N/cm, and $m_1$ and $m_2$ are the masses in atomic mass units (u). Compare your calculated values with the experimental value. Provide the reference.

Is this question from a textbook (or at least, are you using a textbook)? It seems unlikely that most intermediate or advanced texts in this field do not contain a table to "typical" force constants. You may also want to try a Google search. That first hit looks promising.

Also, try to refrain from copying in your question as an image. It makes searching more challenging.

Ok sorry Ben. I'm not really familiar with that notation used for writing out the mathematical formulas. I'll need to work on that.

I performed your exact search and looked at the first result already. The PDF costs around 37 euro and the information isn't mentioned in the abstract, which is free. :(

You can estimate the force constant (as I have done here) and calculate the frequency of the stretch or, as you request, go the other way. Regardless, you have 2 unknowns and thus the problem you present isn't solvable. Below, I work it from the assumed value of the force constant: to go the other way, you'll need to locate the absorption spectral line for the C-C stretch in ethanol and go from there. Either way, you'll need to make some assumptions regarding reduced mass.
From page 496 in Organic Chemistry by Brown, Iverson, Anslyn, and Foote, the formula
$$\tilde{\nu} = 4.12\;\sqrt{{k\over\mu}}$$
is given, where $\tilde{\nu}$ is the frequency in wavenumbers, $k$ is the force constant in units of dyne/cm, and $\mu$ is the reduced mass (in amu, or atomic mass units). They state that for a single bond, the force constant can be taken as $5\cdot 10^{5}$ dynes/cm (an approximation, and we do have to start somewhere). The 4.12 differs from the constant in your given equation due to unit conversion between dynes/cm and N/m.
Now, on to your problem. The reduced mass term is (generally)
$$\mu = {m_{1}m_{2}\over m_{1} + m_{2}}$$
which is for two masses. I posit that in ethanol, one carbon has a "mass" of 15 amu (that's the carbon with 3 hydrogens attached to it) and the other has a "mass" of 31 amu (the carbon plus 3 hydrogens and the oxygen). To go beyond this involves a significant amount of advanced work that is beyond the scope of your problem (but if you're interested, Molecular Vibrations by Wilson, Decius, and Cross is the place to start).
We calculate
$$\mu = {15\times 31\over 15+31} = 10.1$$
and substitute into the first equation
$$\tilde{\nu} = 4.12\;\sqrt{{5\cdot 10^{5}\over 10.1}} = 916\;\mathrm{cm}^{-1}$$
From some notes for a class at Caltech, it appears we can expect a carbon-carbon single bond to appear anywhere in the range of 800 to 1200 cm$^{-1}$.
The calculated number for the C-C stretch here (towards the lower range we'd expect) is due to our "bloated" reduced masses, and that makes intuitive sense (if we do the same calculation for ethane, with "lighter" reduced masses for each carbon, we get $\tilde{\nu} = 1063$ cm$^{-1}$).


text:
Bends in cis Isomers
It’s been said over and over in my biochemistry books that the cis-double bonds between the Cs in the hydrocarbon tails in triglycerides causes a kink. This kink disappears when the bond is isomerized into its trans conformation. What is it about the cis bond that causes the kink?

Do you know what a kink is? And what the shape of a cis double bond is?

Yes. But why is the cis double bond shaped like that?

Does @BenNorris's answer solve your problem, or are you asking about the reason for cis/trans isomerism and the shapes of the respective isomers?

"But why is the cis double bond shaped like that?" - The cis isomer, by definition, is the one in which the substituents on the double bond are on the same side. This leads to the kink that you see in the pictures that @BenNorris uploaded in his answer. Rotation of the C-C single bonds adjacent to the double bond is restricted, so once an alkene is in the cis configuration, it will stay as a cis alkene unless something helps it isomerise (in a biochemistry context, probably an enzyme, but you may not come across that yet so don't worry).

@orthocresol Why is there a kink when you have both Hs on the same side? Do the Hs just physically weigh down on the bend or what?

You'll have to go and read up about electron pair repulsion and the concept of hybridisation in organic chemistry. The bond angles of the sp2 carbon (i.e. the ones in the double bond) are 120 degrees. How much chemistry background do you have? I don't think you should be reading biochemistry without having taken some organic chemistry in the past.

The best way to demonstrate this kink is with pictures. Usually we have some motivation to try to represent the molecule on a single line. So for oleic acid (cis-octadec-9-en-1-oic acid), we often get pictures like the following:

Elaidic acid, the trans-isomer is also sometimes depicted unhelpfully to save space:

The second structure for elaidic acid is not as misleading as the corresponding structure for oleic acid. A better bond-line structure for oleic acid is:

Better yet, here are some space-filling structures from Wikipedia based on crystal structures of both molecules.
Oleic

Elaidic

And here's $\gamma$-linolenic acid, which has three cis alkenes in it:


text:
Oxalate ion dentiticity - why is it bidentate and not tetradentate?
Why is the oxalate ion a bidentate ligand? Why does it not act like a tetradentate ligand (having four oxygen atoms, with lone pairs)?

The oxalate anion cannot be tetradentate. While there are four oxygen atoms, the ion is planar and the oxygen atoms are pointing in different directions.

The best any metal can do is bond to two of them, although there are two bonding modes possible:

Perhaps you were envisioning some $\kappa^4$-bonding mode that is
"side-on". However, most of the nonbonding electrons (orbitals) are also in the plane.
This is not to say that oxalate cannot form four bonds to metal centers; it just cannot do so with only one metal center. Oxalate can act as a bridging ligand


text:
Reaction of CaO with C
$\ce C$ is a good reducing agent and hence it can remove oxygen from $\ce{CaO}$. But instead of forming $\ce{Ca}$ and $\ce{CO}$, the products are $\ce{CaC2}$ and $\ce{CO}$.
Why does this happen?

Why would it not? C is a nonmetal, and it is common for nonmetals to form compounds with metals. Oxygen gives oxides, nitrogen gives nitrides (if you succeed in making him react, which may be not quite trivial, but that's another story), chlorine gives chlorides, etc. Now, C is less eager to take extra electrons than N or O, so it would form ionic carbides only with active metals. Well, Ca is pretty active.


text:
Carbonyl oxygen as nucleophile
Aldehydes and ketones have electrophilic centres at their carbonyl C, since the carbonyl O polarises the bond and attracts electron density towards itself on account of its high electronegativity. But despite this electron density, the oxygen atom hardly ever acts as a nucleophilic centre. If at all, it is the alpha C, which in enols and enolates acts as a nucleophile. I only know of the Wittig reaction in which the O also reacts with the reagent (the P atom in this case). Is there any reason for the low reactivity of the oxygen? It is joined to C with a pi bond, which in any case is easy to break on attack by a reagent.

It does react, for example in the formation of enol ethers, and technically the protonation of a carbonyl group is also an example. It's a much harder nucleophile than the alpha carbon and therefore reacts with different things.

Is there any reason for the low reactivity of the oxygen? || Yes. It is in $sp^2$ hybridization state, meaning it binds its lone pairs extremely strong.  It seems, however, that it can react with strong electrophyles like proton. It also should act as one in synthesis of pyrilium salts, but I was unable to find a proposed mechanism. Also, technically the nucleophilic oxygen in Wittig synthesis is derived from carbonyl oxygen, but is not one at the moment of attack

@orthocresol Isn't protonation of the carbonyl an acid-base reaction though? (i.e. not a nucleophile-electrophile reaction)

@SasQ Those labels aren't mutually exclusive. H+ is an electrophile, and anything that gets protonated is acting as a nucleophile towards it.

Well, in that case we would have to treat the entire conjugate base as a leaving group that leaves the proton, wouldn't we?

@SasQ Yes. But there's nothing wrong with that, is there? [Heads up: You need to tag me if you want me to get a notification.]

The carbonyl oxygen can act as a nucleophile but it is strongly dependent on the conditions of the reaction.
Enolates can react as a nucleophile through either the $\alpha$-carbon or the oxygen.

(source)
Changing the electrophile
Alkylation is a common reaction of enolates and provides a good insight into the factors affecting the selectivity. If methyl iodide is used as the alkylating agent in a polar aprotic solvent then the reaction produces mostly the carbon alkylated product. Switching to methyl bromide or methyl chloride increases the amount of oxygen alkylated product and if methyl tosylate is used then the reaction produces mainly the oxygen alkylated product.
This thinking can be extended to other electrophiles. Many hard Lewis acids such as $\ce{R2BCl}$ and $\ce{R3SiCl}$ (credit to @Jan for pointing this out) will give almost exclusively oxygen substituted products.
Why does this affect the regioselectivity?
Traditionally, the ratio of carbon- to oxygen-substituted products is explained through "hard–soft acid–base (HSAB) theory". Using this model, the outcome depends on the balance of two factors, namely the orbital interactions and electrostatic interactions between the reactants.
As can be seen from the diagram above, the HOMO of the enolate has the greatest contribution from the $\alpha$-carbon and so reactions which are controlled by orbital energy factors take place through there. Conversely, the greatest negative charge is on the oxygen due to its greater electronegativity and so if the reaction is controlled by electrostatic factors then it will take place at the oxygen.
This allows us to explain the effect of changing the electrophile from $\ce{CH3I}$ (mainly C-substituted product) to $\ce{CH3OTs}$ (mainly O-substituted product). The increasing electronegativity of the substituent increases the positive charge on the methyl carbon, meaning that electrostatic interactions become more important in the case of $\ce{CH3OTs}$. Conversely, the $\ce{C-X}~\sigma^*$ orbital (the LUMO of the electrophile) will be lower in energy in methyl iodide than in methyl tosylate due to the poorer energy match between the $\ce{C}$ and $\ce{I}$ atomic orbitals compared to $\ce{C}$ and $\ce{O}$. This brings the LUMO of $\ce{CH3I}$ closer in orbital to the enolate HOMO, making the orbital interactions more important in that case.
Likewise, boron or silicon central atoms are small and are highly charged, and so have strong electrostatic interactions with the oxygen end of the enolate. These would be considered "hard acids", and the carbonyl oxygen a "hard base". By comparison, neutral carbon electrophiles are "soft acids" and so exhibit less affinity for the oxygen.
This explanation might suffice at an introductory level. However, in truth, the concept of HSAB theory doesn't hold up when generalised; it also ignores fundamental concepts such as whether the reaction is thermodynamically or kinetically controlled (which generally affect the reaction outcome). HSAB theory is therefore not actually the correct way to treat this issue of ambident reactivity (i.e., a species reacting at two or more different parts).
The failings of HSAB theory, as well as a more sophisticated and correct model based on Marcus theory, are described in detail by Mayr et al. in a review (Angew. Chem., Int. Ed. 2011, 50 (29), 6470–6505).
Changing the solvent
Apart from the electrophile, changing the solvent is one of the easiest ways to influence the outcome of the reaction, particularly in the case of alkylation. In polar aprotic solvents, the metal cation is strongly solvated but the enolate is weakly solvated due to the lack of acidic hydrogens on the solvent to form hydrogen bonds. In polar protic solvents, the oxygen end of the enolate is strongly solvated by hydrogen bonding which hinders reaction at the oxygen, strongly favouring C-alkylation, even with an alkyl tosylate.
Some more reading can be found here, here, here and here.

The commonplace reaction of a ketone with a Grignard reagent involves the oxygen acting as a nucleophile ... towards the magnesium.
The view from the other side
Yes, the negatively polarized carbon gets all the headlines as a nucleophile, as this is the function that ends up in the alcohol that we identify as the product after workup. The magnesium literally just comes out in the wash as a salt formed from the original halide in the Grignard reagent and the acid used in the workup.
Yet upon further review the magnesium, at the other end of that polar bond with carbon, is also there, and with its fairly high positive charge density it acts as an electrophile towards the carbonyl oxygen. The resulting addition of magnesium to oxygen contributes to the polarization of the carbonyl group and thus ultimately supports the nucleophilic attack on the carbonyl carbon:
$\ce{\color{blue}{C}H3\color{brown}{Mg}Br + CH3\color{brown}{C}(=\color{blue}{O})CH3 -> CH3-\color{brown}{C}(\color{blue}{O}\color{brown}{Mg}Br)(\color{blue}{C}H3)-CH3}$
(Blue = nucleophile, brown = electrophile)


text:
Does calcium carbonate remove calcium from our bones?
Is it true that milk contains calcium carbonate which removes calcium from our bones so we shouldn't drink it?

Your bones are made of calcium carbonate. Milk is a good source of calcium and is considered to be perfectly healthy to drink.

I object. Bones are made of calcium phosphate.

@IvanNeretin You're right! I have always believed that it is calcium carbonate but that seems like a myth.

You would need HF for that.

First of all, it doesn't contain calcium carbonate. Calcium carbonate is solid material that doesn't solve in water. Marble and limestone are calcium carbonate (https://en.wikipedia.org/wiki/Calcium_carbonate).
Even if it contained, it would be an excellent source of calcium. It is even used as dietary supplement.
https://ods.od.nih.gov/factsheets/Calcium-HealthProfessional/
To answer the implied question (can you drink milk?), @Curt Fs's reference is perfect answer.

No, milk is a perfectly viable source of dietary calcium.

authors: عبدالله إيتا


text:
Hydrogen phase transition
A sample of hydrogen at a pressure of 7.10×10-2 atm and a temperature of 26.4 K is compressed at constant temperature to a pressure of 18.5
Which of the following are true?
1. One or more phase changes will occur
2. The sample is initially a solid.
3. The liquid initially present will vaporize.
4. The final state of the substance is a liquid
The sample is initially a gas
MY ANSWER:
I feel that one or more phases will change and that the liquid present will vaporize as it surpasses the supercritical point.

possible duplicate of A sample of hydrogen at a pressure of 7.10×10-2 atm and a temperature of 26.4 K is compressed at constant temperature to a pressure of 18.5 atm

How is this any different from your previous question?

I accidentally double posted it after putting my answer in the body. Im fixing that now.

Say, do you think you could screenshot the phase diagram and upload it? It'll make it much easier to see. You can crop out the question since it's already been typed out.

Regarding the question itself, 1) you already said that the sample is initially a gas - so the "liquid initially present" can't be vapourised if there's no liquid initially present. 2) a note on terminology: the point on the phase diagram is called the critical point, and the substance is called a supercritical fluid. 3) A supercritical fluid only forms when both $T$ and $p$ exceed that of the critical point. Check the diagram on Wikipedia: https://en.wikipedia.org/wiki/Critical_point_(thermodynamics)#Liquid-vapor_critical_point


text:
Can water volume reduce with introduction of another chemical
Can a chemical be introduced to a volume of water which can actually reduce the volume of water in that container?

related http://chemistry.stackexchange.com/questions/34919/is-there-a-soluble-salt-or-compound-that-when-dissolved-in-water-results-in-a-s

Mithoron, http://chemistry.stackexchange.com/questions/23534/is-it-possible-to-have-volumesolution-volumesolvent-volumesolute is not a duplicate because this question is asking can Volume(solution) < Volume(solvent), i.e. there is no term for the solute volume.

Yes, at least if only the total volume should decrease:
Density of water: $\pu{1000 kg/m^3}$
Density of ethanol: $\pu{789 kg/m^3}$
Expected density of 52 Vol.% ethanol solution: $\pu{890 kg/m^3}$
True density of 52 Vol.% ethanol solution: $\pu{909 kg/m^3}$
All values are for the same conditions.
The volume of every ethanol-water solution is smaller than the sum of the original volumes of water and ethanol. For example if you add $\pu{52 mL}$ of ethanol to $\pu{48 mL}$ water you will get a solution with an volume of approximately $\pu{97.9 mL}$.
This effect is known as volume contraction and depends on the mixing ratio of both components in a nonlinear way. Its reason is the formation of additional bonding forces (e.g., hydrogen bridges) between the molecules, which results in a higher density.
Note: I don´t think that there is something that can increase density of water in such a way that the resulting total volume is actually smaller than the volume of water used.


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Calculate concentration gradients in FPLC
Our FPLC works with two solutions:

A: 20 mM Tris.HCl, 0.2 M NaCl, pH 7.8
B: 20 mM Tris.HCl, 1.6 M NaCl, pH 7.8

I'm interested in sample volume from 102 mL to 117 mL. The FPLC will have a linear concentration gradient: At 24 mL it will be 75% A and at 144 mL it will be 5% A (the rest is filled up with B).
For visualization, I made a graph (y = -7/1200*x + 0.89):

Now, I have a sample with the whole output between the two indicated points (102 mL and 117 mL, so I have 15 mL in my tube) and would like to calculate the NaCl concentration.

Percentage of both A and B would change 70% from 24 to 144mL, with A from 75% to 5% and B from 25% up to 95%. The change rate would be 70% during the 120mL period or $0.70/120$ per mL. At any given point $x$ mL, between 24 and 144mL, concentration of $\ce{NaCl}$ in $\ce{mol/L}$ is:  $$\ce{[NaCl]} = (0.75 - 0.7\times (x-24)/120)\times0.2 + (0.25 + 0.7\times (x-24)/120)\times1.6$$
For a 15mL solution between 102 and 117mL, the concentration should be the average of the two points when x = 102, and 117mL.

I try to find out the concentration over a certain time (namely between 102 to 117 mL), so in the end, I have 15 mL in a tube. I want to find the concentration of this 15 mL, not a certain infinitesimally small point along the gradient. I hope it is clear.

Sorry, didn't see it carefully. In that case, the concentration should be the average of the two points when $x$ = 102, and 117mL. Did I miss something else?

Yes, that's it, the result of 1.24826 seems right. If you edit it in your post, I'll accept it for closure.

In addition to Dejian15's answer, which is probably the simplest, I also figured out a solution for a general curve (also non-linear ones).
$$ \frac{\int^{117}_{102}(\frac{-7}{1200}x+0.89)*0.2+(1-(\frac{-7}{1200}x+0.89))*1.6~\mathrm{ d}x }{15} =1.24826$$


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What equipment to use to make liquid oxygen?
I'd like to do some experiments with liquid oxygen but as a first step I'll have to make some. Unfortunately I don't have any experience designing cryogenic experiments. I'm thinking I'll cool gaseous oxygen using liquid nitrogen as the coolant and then I want to collect the liquid oxygen condensate in a small dewar of some kind.
It needs to be high purity so I'll be using 99.9% purity "scientific grade" oxygen from a cylinder.
Does it make sense to use conventional condenser glassware like a dimroth condenser for instance? Alternatively what's the best equipment to use and what's the best way to rig a cryogenic setup like this?
Also what safety precautions should I take?

You can probably just condense air with liquid nitrogen in a dewar, but it is dangerous.

Ah, I forgot to mention that this has to be pure oxygen, not liquid air.

Fortunately for safety, oxygen boils at a lower temperature than nitrogen. Fractional distillation isn't 'hard', but complex at cryo temperatures.

I think that does not really make any difference, as I find it highly unlikely that you will condense much more then oxygen. Sure, there is the possibility to trap some carbon dioxide, but apart from that I  don't know what other gasses you have in your lab. Any way, you can add more details to your question by hitting the [edit] button.

@JonCuster That is incorrect. The boiling point of nitrogen is about -195 degree Celsius and that of oxygen is about 183 degree Celsius.

Mea culpa - I'll claim lack of oxygen...

@Martin Thanks. I need it to be quite pure so I'll be using high purity grade oxygen (99.9%).

You can use something as simple as a test tube immersed in liquid nitrogen (in a Dewar) that is charged with oxygen from tubing connected to your oxygen cylinder. But this will be open in the air. Otherwise any closed system I imagine it has to be specialised equipment since any oxygen evaporation can create high internal pressure in a closed system. (So a Schlenk tube would not be a good idea)

Simply put an open flask in the LN2 and it will fill with liquid oxygen from the air.
Or better: DON'T DO THAT!! What the heck do you want with it?!?
LO2 is highly dangerous. If it comes in contact with anything combustible, it will blow up vigorously. Together with lab solvents, it will likely explode. Make sure the flask or dewar is dead clean.
Chemistry is science, and science is about prediction. If you cannot predict, from knowledge, not hearsay, what your experiment will likely do, don't do it.

I really think this is the sort of question where if you have to ask then you don't understand the situation enough to do the experiment safely. Safety can't be an afterthought but must be planned into the experiment.

It has already been suggested to use liquid nitrogen to cool oxygen and that is probably your best bet.  You could research something called a cryocooler that can be used to condense air or pure oxygen but it is very costly and would not be worth it unless you plan on using a continuous supply liquid oxygen and/or nitrogen.  If you're just going to do few experiments, don't worry about the cryocooler idea.
But I will say this.  You can condense air from LN2 and end up with very pure LOX.  Immersing an empty test tube in a container of LN2 for a while, you will notice it slowly filling with a very very pale blue liquid.  But if you remove the tube from the LN2 and let it boil for a while, you'll notice it getting bluer and bluer.  Once it gets a little darker than sky blue, you can be sure you have nearly PURE liquid oxygen remaining.