basic chemistry alkaline test for flavonoids!!! with reference please What is flavanoid physical and chemical properties, and how alkaline reagent test for flavonoids gives yellow coulour?

From this reference:

  Treatment of a flavonoid with a strong base will result in ionization of its hydroxyl groups with resulting large bathochromic shifts of both absorption maxima.


Isotopic mass to density I want to figure out/calculate the density ($\pu{g/cm3}$) of the various isotopes, here $\ce{^109Ag}$. I know it's isotopic mass is $\approx109$ and calculated the number of particles/gram. But I couldn't find out how to get the density.

From what I could find out over some mole maps I need to multiply the molar mass with a factor to get to the density but is that even possible or am I wrong?

This is possible, just use the density equation for a crystal lattice
$$ d = \frac{z\times M}{V_c \times N_\mathrm A}$$
Where $d$ is density; $z$ is the number of atoms per unit cell; $M$ ist he molar mass of the material (isotope in this case); $V_c$ is the volume of a unit cell; and $N_\mathrm A$f is avogadro's number.  Since silver is FCC, $z= 4$ ($\frac{1}{8}\times 8 + \frac{1}{2}\times 6 = 4$) and $V_c$ is equal to the lattice parameter squared ($a^3; a = \pu{0.409nm}$).

This method will give you a very close approximate value but is limited in accuracy to the experimentally observed density as it does not account for free space in the crystal due to vacancies in the lattice or voids at grain boundaries. For example, if you calculated the densities of iridium and osmium you find that iridium should be denser, but experimentally you will not observe this result because iridium has a sufficiently high concentration of vacancies that osmium is actually denser.


Preference of certain positions during chlorination of an alkene I am currently doing the questions:

Write equations for the reaction of chlorine with propene and 2-butene using structural formulas.

So I know that propene is $\ce{H3C-CH=CH2}$ and chlorine is $\ce{Cl2}$.
So:

From this I observe that naturally, the double bond is removed in the presence of an addition reaction and that the chlorine atom bonds to the $\ce{CH}$ pair and the other $\ce{Cl}$ atom does not bond to any of them but attaches itself at the end.

Why did the second $\ce{Cl}$ pair not attach itself to $\ce{CH2}$? Does this mean that atoms can only bond to $\ce{CH}$ pair and nothing more than that, e.g. not $\ce{CH2}$ or $\ce{CH3}$, etc.

Second question for the reaction of 2-butene with Chlorine. I know that 2-butene is $\ce{H3C-CH=CH-CH3}$

I noticed, like what happened above, the chlorine atom bonds to only the $\ce{CH}$ pair. Here however there are two $\ce{CH}$ pairs so the Chlorine doesn't need to attach itself at the end of the chain, unlike the first picture.

Are my observations correct? My textbook has no mention of this.

In your first example, propene plus chlorine, one of the Cl atoms bonded to a "CH" and the other, that you described as bonding to the end, bonded to a "CH2," so it's not just CH groups getting all of the chlorines.  Instead, it's the carbons at the ends of the double bond, however many hydrogens they might be attached to.

I think part of your misunderstanding lies in the way that you're writing/imagining the alkene.  The propene molecule has a shape, and the geometry of the molecule plays a major role in its reactions.  The double bonded section is more or less flat, and the CH3 is spread out in space.  The chlorine atoms attach to the ends of the double bond, wherever that may be.  The mechanism for this attachment is beyond the scope of your question, but if you're curious, you can find more information here.  At the very least, know that the chlorines do not attach to the carbons at the same time; there is an intermediate compound formed.

(If my drawing isn't clear, you can try building a model of the molecule.)

It would require the breaking of a C-H or C-C sigma bond in order for the chlorine to get to the CH3 carbon, and that's unlikely.  As I mentioned above, the C=C section is planar, so these carbons are open on the sides for attack from a halogen.  A bond still has to break for a new C-Cl bond to form, but in this case it is the weaker C-C pi bond that breaks.  With easy physical access and a lower energy barrier, the reaction goes forward.


Molar conductivity and concentration For strong electrolytes, according to Kohlrausch, molar conductivity decreases as the concentration increases.

( i don't know why I'm not allowed to attach a photo here in order to show you the diagram ), but however I'm trying to explain how the diagram looks.

In the diagram we have strong electrolytes like $\ce{HCl,\, H2SO4,\, KCl,\, LaCl3,\, Na2SO4}$ whose molar conductivity decreases as the concentration increases. But, what I don't understand is that in the diagram we have also $\ce{CH3COOH}$ which is a weak electrolyte, and his molar conductivity also is decreasing and even far more rapidly, as the concentration increases.

Can somebody explain this ?

Kohlrausch in 1874 he demonstrated that an electrolyte has a definite and constant amount of electrical resistance. From 1875 to 1879, he examined numerous salt solutions, acids and solutions of other materials. This resulted in the law of the independent migration of ions, that is, each type of migrating ion has a specific electrical resistance no matter what its original molecular combination may have been.So this means that the electrical resistance of the ions in the solution that are migrating, and conductivity is just the inverse of this.

CH3COOH which is a weak electrolyte, so does not dissociate completely. Since it's only the migrating ions that affect it, only the dissociated ions affect conductivity and specific electrical resistance no matter what its original molecular combination may have been.

At lower concentrations, since there are more water molecules for every CH3COOH molecule, more of these dissociate so there are more migrating ions meaning a larger conductivity. The molar conductivity is defined as the solution conductivity normalized by the total ionic concentration. So concentration increases without any significant increase in conductivity meaning the molar conductivity decreases rapidly.


Relation between First Law of Thermodynamics and Ideal Gas Law Thermodynamics has always been a tough thing for me. There are lots of assumptions in this subject (those assumptions, I know, are necessary, I know the science of thermodynamics is a very practical science).
First Law of Thermodynamics states mathematically:
$$\Delta U=Q+W$$
(with proper sign conventions must be used). This is just a law of conservation of energy and a very straightforward equation, but when we come to chemical thermodynamics this equation changes its form and becomes:
$$\Delta U=Q+p\,\Delta V$$
My intuition says as soon as pressure and volume comes in any equation it becomes specifically for gases. So, my first question is:

Why thermodynamical equations are just for gases?

Let's imagine an isothermal expansion of a gas (that simple piston and gas experiment) under a constant pressure, now work $W$ is
$$W=p\,\Delta V$$
but if use ideal gas Law equation i.e.
$$pV=nRT$$
$$p\,\Delta V = \Delta nRT + nR\,\Delta T\tag1$$

since the expansion is isothermal therefore $\Delta T = 0$ and I can think that during expansion no atom or molecule has been annihilated therefore $\Delta n = 0$, so after all we get
$$p\,\Delta V = 0$$
$$W=0$$

I want to know my mistakes in above consideration.

There is a question in my book:

  A swimmer coming out of a pool is covered with a film of water weighing $18\ mathrm g$. How much heat must be supplied to evaporate this water at $298\ \mathrm K$? Calculate the internal energy change of vaporization at $100\ \mathrm{^\circ C}$. $\Delta_\mathrm{vap}H^\circ = 40.66\ \mathrm{kJ\ mol^{-1}}$ for water at $373\ \mathrm K$

My book give its solution like this
$$\ce{H2O(l) -> H2O(g)}$$
Amount of substance of $18\ \mathrm g$ of $\ce{H2O(l)}$ is just $1\ \mathrm{mol}$. Since, $\Delta U=Q-p\,\Delta V$, therefore,
$$\Delta U=\Delta H-p\,\Delta V $$.
$$\Delta U=\Delta H-\Delta nRT$$
$$\Delta U=40.66 \times 10^3\ \mathrm{J\ mol^{-1}}-1\ \mathrm{mol}\times8.314\ \mathrm{J\ K^{-1}mol^{-1}}\times373\ \mathrm K$$
$$ \Delta U=37.56\ \mathrm{kJ\ mol^{-1}}$$

I have a lot of problems with this solution which goes directly to the foundations of science of thermodynamics. (I must say it's because of these books that science becomes a rotten subject, these books destroy the real essence of science).
How is $\Delta n=1\ \mathrm{mol}$?
Why temperature is taken as $373\ \mathrm K$ and not $298\ \mathrm K$, since the process starts at $298\ \mathrm K$ we should use it?
At $373\ \mathrm K$ the process becomes an isothermal one (latent heat) so $\Delta U$ ought to be zero, if we think the process of vaporization starts from $373\ \mathrm K$.

Any help will be much appreciated. Thank you.

Why thermodynamical equations are just for gases?

They are not. The equation $\Delta U = q + P\Delta V$  applies to any phase (gas, liquid, solid...) when only pV work is done. In the particular form of the equation you present, the pressure is in addition constant during the work.

Gases are (1) an easy way to introduce thermodynamics concepts because they allow for a simplified analysis and may exhibit dramatic behavior, and (2) they provide a link to understanding the behavior of other phases. They also happen to be inherently important for practical and historical reasons during the development in the science.

  I want to know my mistakes in above consideration.

There are no mistakes. Consider the ideal gas law $$V=\frac{nRT}{p}$$ If you assume that $p$, $n$ and $T$ are constant, then the dependent variable $V$ will also be constant.

What's probably causing the confusion is that the water is undergoing a phase change. We assume that the liquid does no work, that the change in volume is only due to the formation of vapor. In practice $\pu{\Delta n=+ 1 mol}$ for the gas, $\pu{\Delta n= - 1 mol}$ for the liquid, and $\pu{\pu{\Delta n= 0}}$ for all of the water. We ignore the work done when reducing the amount of liquid because it is small compared to that done when the gas is formed (the change in volume of gas is much greater).

  How is Δn=1 ?

See the answer to the previous question. You are converting $\pu{18 g}$ of liquid water into vapor. Since the molecular weight of water is $\pu{18 g/mol}$, you are converting $\pu{1 mol}$ of water.

  Why temperature is taken as 373 K and not 298 K, since the process starts at 298 K we should use it?

I agree, if the work is actually performed at a lower temperature than $\pu{373 K}$, so this is an estimate. It is assumed that water is "boiling off" the skin at $\pu{373 K}$ and $\pu{1 atm}$ vapor pressure (the boiling point of water at $\pu{1 atm}$ of pressure is $\pu{373 K}$). Maybe not an accurate portrayal of what is going on, but it gets you to practice the theory.

  I have lot of problems with this solution which goes directly to the foundations of science of thermodynamics. (I must say it's because of these books that science becomes a rotten subject, these books destroy the real essence of science).

The problems more generally are that (1) we have an intuition about the way the world works, based on everyday observations, and this intuition sometimes misinforms us; some of the effort of education is to develop a more accurate intuition; and (2) when teaching science, practice problems are sometimes abstract and don't reflect real life situations except approximately. For instance, when you dry yourself after swimming, you are far from a thermodynamic equilibrium, with air currents, sun heating your skin, and a low water vapor pressure all playing a potential role. Modeling all this is beyond the scope of an introductory course. Probably the most important approximations here (assuming a closed system free of mass flow) are that the enthalpy of vaporization is constant over a broad temperature range and/or (as you rightly pointed out) that the water evaporates at $373 K$.


How to interpret the Maxwell-Boltzmann distribution to find the activation energy? I understand activation energy to be the relative difference in energy between that of the activated complex/transition state (at some temperature) and the average energy of the reactants like shown here:

(Source: Wikimedia, User Bkell)

If this is correct, I have questions about graph like this one:

(Source: webchem.net (not available anymore, Internet Archive), retrieved via the Internet Archive)

Is the location marked activation energy supposed to be the energy level of the transition state provided the energy axis is absolute?
Can the Maxwell-Boltzmann distribution be done for heterogeneous systems (i.e., two chemical reactants)?
Unrelated to the graph, why does negative activation energy result in a decrease in reaction rate (per Arrhenius equation)?

The marked location corresponds to a level of kinetic energy in the reactants sufficient to result in a successful collision (energy wise, it says nothing about orientation).  The energy required for a successful collision is the gap in potential energy between the reactants and transition state.
A heterogeneous system is a system where the reactants are in different phases.  A common example is a gaseous reactant that collides with a solid catalyst.  In that case the Maxwell Boltzmann distribution can be applied to the gaseous reactant.
Can you clarify this?  What reactions have negative activation energies?  The reactions you'll commonly encounter have positive activation energies.


Is Km the concentration needed to keep the enzyme running at max speed? $K_m$ is the Michaelis constant, which is the concentration of substrate needed to achieve a rate of $\frac{V_{max}}{2}$. $V_{max}$ is the maximum number of molecules that can be reacted per second.

My lecturer has said that $K_m$ is the lowest concentration of substrate needed to ensure that the system runs at $V_{max}$.
I do not understand why. To me, I would have assumed that it would need to be at a concentration that would produce $V_{max}$ or something close to it.
I can understand that at a concentration that is too low, then the enzyme will make quick work of the backlog. However, I cannot see why exactly Km should be the point at which this happens. It all feels quite arbitrary to me.
Could someone explain, preferably intuitively, why this is true?

$K_m$ is the lowest concentration of substrate needed to ensure that the system runs at $V_{max}$?

No. $V_{max}$ is never reached, it is the limit of the rate when you increase the substrate concentration indefinitely. Less dramatically said, when the substrate concentration is about ten times higher than $K_m$, the rate will be about 90% of $V_{max}$ (and so on, so hundred times higher - about 99%).
In practice it is often difficult to get to hundred times $K_m$ because the substrate might not be that soluble or there is substrate inhibition at non-physiologically high substrate concentrations.

Could someone explain, preferably intuitively, why this is true?

As I read in a textbook, imagine a bicycle repair shop with 5 people working in it. When the number of bikes brought in doubles from what it is usual, they will repair more bikes. It does not matter, however, if you park a million or a billion bikes in front of their shop, they will be working at capacity in either case.
If you like math, you can also answer your question by looking at the relationship between substrate concentration and rate described by the Michaelis-Menten equation set up by  Leonor Michaelis and Maud Menten:
$$\mathrm{rate} = V_\mathrm{max} \frac{[S]}{K_m + [S]} = V_\mathrm{max} \frac{1}{1 + \frac{K_m}{[S]}}$$
When $[S] \gg K_m$, the fraction approaches 1 and the rate approaches $V_\mathrm{max}$.

My lecturer has said

That is what you heard. If you think the lecturer misspoke or you misheard, it is a good idea to check a textbook or ask here.
The figure
The figure is a conceptual one with two conceptual errors. First, the rate in the figure does exceed $V_\mathrm{max}$.

Secondly, the figure shows a linear increase of the rate for substrate concentrations lower than $K_m$. In the graph showing the real shape of the curve, you can see appreciable curvature at lower substrate concentrations.

For the graph, $V_\mathrm{max}$ was set to 1 and $K_m$ to 0.2. The shape of the curve, however, is the same (if you choose the plotting range in an analogous manner) for any combination of the two parameters.
The problems with the figure are common, as the following gallery shows:


Proving a relation for fugacity of component i in term of independent variables (T,V) In Molecular Thermodynamics of fluid phase equilibria by Prausnitz et al. [1] the authors recommend to use the following equation which gives the fugacity of component $i$ in terms of independent variables $V$ (volume) and $T$ (temperature):

$$RT\ln\phi_i = \int_v^\infty\left(\left(\frac{\partial p}{\partial n_i}\right)_{T, V, n_i} - \frac{RT}{V}\right)\mathrm{d}V - RT\ln z$$

I know that I should use chemical potential definition to prove this equation, but there are many different ways to calculate it like using Helmholtz free energy or Gibbs free energy.

I would appreciate someone give me an idea or any references to help me proving this equation.

References

Prausnitz, J. M.; Lichtenthaler, R. N.; Azevedo, E. G. de. Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd edition; Prentice Hall: Upper Saddle River, N.J, 1998. ISBN 978-0-13-977745-5.

I haven't been able to work out all the details of the mathematics yet, but the derivation of this equation must start out from the following equation:

$$nRT\mathrm{d}\ln{\phi} = (V - V^{ig})\mathrm{d}P,$$

where $\phi$ is the fugacity coefficient of the mixture, $n$ is the total moles of the mixture, $V$ is the volume of the mixture $znRT/P$ and $V^{ig}$ is the volume of the mixture under ideal gas conditions: $nRT/P$.  Once the equation is integrated to get $RTn\ln{\phi}$, we get the fugacity of the $i$th component of the mixture by evaluating the following partial derivative:

$$RT\ln{\phi_i} = RT\frac{\partial (n\ln{\phi})}{\partial n_i}$$

Hope that this helps a little.

ADDENEUM

If we write, $$d(nV)=\left(\frac{\partial (nV)}{\partial n_i}\right)_{p,n_k}dn_i+\left(\frac{\partial (nV)}{\partial p}\right)_{ni,n_k}dp$$

then it follows that:  $$\left(\frac{\partial (nV)}{\partial n_i}\right)_{p,n_k}=-\left(\frac{\partial (nV)}{\partial p}\right)_{ni,n_k,V}\left(\frac{\partial (p)}{\partial n_i}\right)_{n_k,V}$$

The rest is easy.  The subscript $n_i$ in the OP equation is a typo.  It should be $n_k$, representing all other species being held constant.


Nucleophilic Substitution when I add $\ce{RCl + H2O -> ROH + HCl}$

but it's also true in reverse $\ce{ROH + HCl -> RCl +H2O}$

however this is not in case of $\ce{FeSO4 +Zn -> Zn + FeSO4}$

$\ce{ZnSO4 + Fe -> no rxn}$

How can a reaction be true both ways ?

It should be better to write equations as

$$\ce{RCl +H2O <=> ROH +HCl}$$

$$\ce{Fe^2+ +Zn <=>> Zn^2+ + Fe }$$

with the latter having the equilibrium shifted to far right.

But is case of high concentration of $\ce{Fe^2+}$ and very low concentration of $\ce{Zn^2+}$, the net reaction would be ongoing leftwards.

If the left side of equation below is smaller than right one, the net reaction in inclined toward iron being replaced by zinc. And vice versa.

$$E^{\circ}_{\ce{Zn/Zn^2+}}+0.0295\log {a_{\ce{Zn^2+}}}=E^{\circ}_{\ce{Fe/Fe^2+}}+0.0295\log {a_{\ce{Fe^2+}}}$$

Note that forward and backward reaction are ongoing simultaneously, having the same rate at the equilibrium.


Do we really have technology with the resolution to distinguish between layers of single atoms of materials? My textbook, Solid-State Physics, Fluidics, and Analytical Techniques in Micro- and Nanotechnology, by Madou, says the following:

  X-ray analysis reveals the symmetries of crystals (lattice type), distances between atomic planes (lattice parameter), the positions of atoms in crystals, the types of atoms from the intensities of diffracted x-rays, and the degree of crystallinity (ordering).

I found this quite fascinating. Do we really have technology with the resolution to distinguish between layers of single atoms of materials? In other words, we can tell when one layer of single atoms ends and the next layer of single atoms begins, and analyse them as such?

X-rays do not allow you to look  at the material just like an microscope. You get the spacing information from the diffraction pattern of only crystalline materials and a lot of mathematics. On the other hand, atomic force microscopy may actually be able to tell you where one atomic layer starts and where it ends on a surface. For example, imagine a gold surface, onto which you deposit a layer of molecules (say on local spot by some means). AFM would allow you to create an image (again by indirect means, as it is not an optical technique) of that molecular layer.
See google images generated by AFM and you will see how fancy they are.

To the OP: A good way to studying is to rely on multiple references of the subject. Do not take everything literally. The authors are human beings after all with limited knowledge and of course they are not the experts of $every$ topic covered in the textbook. When I was a student, I used to spend a lot of time trying to decipher what the author was saying in a particular sentence. With maturity you learn that the better way (read: more scholarly way) is to look at the same topic in another book and see what the other author is saying.


Can acid-base reactions occur in a non-aqueous medium? To better illustrate my question, allow me to refer to the case of ammonium chloride, formed through a classic acid-base reaction.

The reaction between ammonia and hydrogen chloride to give ammonium chloride is a neutralization reaction, since ammonia acts as a base and hydrogen chloride acts as an acid.

The reaction is usually written as:

$$\ce{NH3 + HCl -> NH4Cl }\tag{1}$$

Usually, in the synthesis of ammonium chloride, the reagents in aqueous solution (ammonium hydroxide and hydrochloric acid) are used. In aqueous solution, they behave differently than when they are in anhydrous form.

On the one hand, when ammonia dissolves in water, the ammonium cation, its conjugate acid, is formed:

$$\ce{NH3 + H2O <=> NH4+ +OH-}\tag{2}$$

On the other hand, a similar, but opposite, reaction occurs when hydrogen chloride dissolves in water, forming the chloride anion, its conjugate base:

$$\ce{HCl + H2O <=> H3O+ + Cl-}\tag{3}$$

Then, when mixing both solutions, the following reactions occur:

$$\begin{align}\ce{NH4+ +Cl- &-> NH4Cl}\tag{4.1}\\[0.5em]
\ce{H3O+ +OH- &-> 2H2O}\tag{4.2}\end{align}$$

However, apparently, it is not necessary to mix the solutions to obtain ammonium chloride, just by mixing the vapors of the solutions, it is enough to give the expected product.

But when the ammonium hydroxide evaporates, the ammonia ceases to be in the form of ammonium and returns to its anhydrous form. The same goes for hydrochloric acid. Then, the vapors would be ammonia and hydrogen chloride, but wouldn't this affect the reaction since they are not in the form of conjugate acid or conjugate base now?

I still don't understand acid-base chemistry very well and I would like to know if such reactions, such as that of ammonia with hydrogen chloride, can occur without the reagents being in the form of ions in aqueous solution.

If you dissolve $\ce{NH4Cl}$ in water, your solution contains $\ce{NH4+}$ and $\ce{Cl-}$ ions. If you dissolve $\ce{NaOH}$ in water, your solution contains $\ce{Na+}$ and $\ce{OH-}$ ions.
Now if you mix those two solutions, nearly all $\ce{NH4+}$ and $\ce{OH-}$ will react and produce $\ce{NH3}$ and $\ce{H2O}.$ This shows that $\ce{NH4OH}$ cannot exist. It is decomposed into a solution of $\ce{NH3}$ in water, with only a few percent of $\ce{NH4+}$ and $\ce{OH-}$ ions.

You will certainly object and say that $\ce{NH4OH}$ is a commercial substance, that you can buy anywhere in the market. You are right. But you may not know that the reason of this mistake is commercial and goes back to the 19th century. In the 19th century, the ions were unknown. All substances had a chemical formula, whatever pure or in solution. $\ce{NaOH}$ in solution was made of $\ce{NaOH}$ and that's all. Charged ions like $\ce{H3O+}$ and $\ce{OH-}$ did not exist.

At this time, chemists were often studying the neutralisation reaction

$$\ce{Acid + Base -> Salt + Water}$$

They succeeded in carrying out this reaction with mineral or organic acids, and with many hydroxides (hydroxide was synonymous to base). When a new organic acid was discovered, they tried to make salts with all available hydroxides. Each new salt deserved a publication. And this was fine!

Unfortunately, ammonia was a problem: $\ce{NH3}$ reacts with any acid, but it does not produce any water. $\ce{NH3 + HCl}$ makes $\ce{NH4Cl}$ and that's all: no water! This was difficult to admit. So the chemists decided that, when dissolved in water, $\ce{NH3}$ is transformed into $\ce{NH4OH}.$ This makes sense, because $\ce{NH3}$ was very soluble in water, as if a chemical reaction happens. It was fine. When neutralizing a solution of ammonia by reaction with any acid, the equation of the reaction was:

$$\ce{NH4OH + HCl -> NH4Cl + H2O}$$

$\ce{NH3}$ was only existing in a gaseous state. It was considered as a sort of "dehydrated ammonium hydroxyde". A little bit like other hydroxides, which can easily lose water. At high temperature, $\ce{Cu(OH)2}$ becomes $\ce{CuO}$ and $\ce{Al(OH)3}$ becomes $\ce{Al2O3}.$

So the merchants started to sell ammonia solution with the label $\ce{NH4OH}.$ It was soon discovered that this solution is helpful unblocking blocked up pipes. Everything was OK!

Unfortunately at the end of the century, Arrhenius discovered the ions, and showed that $\ce{NH4+}$ and $\ce{OH-}$ ions cannot exist simultaneously in solution. Disaster! The so called "ammonium hydroxyde" was mainly containing $\ce{NH3}$ and $\ce{H2O},$ with only a few percent of remaining ions $\ce{NH4+}$ and $\ce{OH-}.$ So the merchants changed their labels and started to sell "concentrated ammonia solution". It was OK for the chemists.

But this change was not accepted by the caretakers, managers and contractors. They protested to the manufacturers against this new "ammonia solution" which was useless to unblock blocked pipes. They said they have tried the new and the old solutions. They demanded to get the old ammonium hydroxide, which was "much better" than this stupid ammonia solution.

Well! The merchants are not stupid. They want to sell and get money. So they renounced to modernize their labels, and continued to sell $\ce{NH4OH}$ solutions. This was accepted by chemists and non-chemists.

That is why $\ce{NH4OH}$ is still on sale today. I am an old man, 82. This story was reported to me by my father, who was a chemist who remembered this old time.


Working with typical glass: What temperatures would I be dealing with? I want to be able to melt down typical glass (the kind you usually find at home) and mold it into other shapes. What temperatures would achieve this?

For comparison; if I were to take some sand and convert it to glass (as we know it), what temperatures would I have to generate?

Furthermore, would "heat-treating" glass in some way make it tougher? I mean "Tougher" along the lines of "bullet-proof".

The softening point of glass depends on its composition. For example,

1650°C for fused silica (pure $\ce{SiO2}$)
821°C for Corning Pyrex 7740 borosilicate glass
713°C for Wheaton 900 soda-lime glass

Since you want to use glass found at home, it might be soft lead-glass (e.g. "crystal" tumbler), soda-lime glass (e.g. bottles), or borosilicate glass (e.g. coffee carafe). Melting together different types of glass is usually not successful because unless perfectly mixed, the difference in thermal coefficient of expansion will cause the object to shatter on cooling. One way to test the coefficient of expansion is to combine two different types of glass into a thread to see if it bends on cooling. This can be used to make a glass thermometer, similar to the bimetallic strip in oven thermometers. Calibration is left to the student.

If you start with fairly pure silica sand, then, you need at least 1650°C to get a glass. Although an inexpensive air-propane torch (not oxy-propane) can theoretically reach 1,995 °C, my own experience is that it takes a long time to melt incompletely just a few grams of sand.

Adding other material to silica lowers the softening point. Soda-lime contains alakali metal oxides, $\ce{CaO}$, etc. Rather than using the caustic oxides, cheaper and safer carbonates (or bicarbonates, e.g. $\ce{NaHCO3}$) can be used, though water of hydration and $\ce{CO2}$ must be driven off by heat. You could try a small amount of very fine sand mixed with baking soda, which would eventually melt to make a lower-temperature glass... but be patient, because the glass is so viscous that it takes a long while to dissolve together.

Lead oxide is used to make lead "crystal", with a lower softening point and higher index of refraction than soda-lime glass, but working with lead in a home lab is definitely not advised!


Equilibrium constant in terms of mole fractions $$\ce{CO + H2O <=> CO2 + H2}$$
  A stoichiometric mixture of $\ce{CO(g)}$ and $\ce{H2O(g)}$ is allowed to reach equilibrium. $x$ is the mole fraction of $\ce{H2(g)}$ present.

I am meant to find the result that $$K_p=\frac{4x^2}{(1-2x)^2}$$

However, I only get: $K_p=\frac{x^2}{(1-x)^2}$

Here's what I am thinking:

$$K_p=\frac{x_{\ce{CO_2}}x_{\ce{H_2O}}}{x_{\ce{CO}}x_{\ce{H_2}}}$$

But: $x(\ce{CO})=1-\alpha=x(\ce{H_2O})$ and $x(\ce{CO_2})=\alpha=x(\ce{H_2})$.

Then substituting this in gives me my result. What am I doing wrong?

The initial mole fraction of $\ce{CO}$ present must be $y=0.5$.  Then the initial mole fraction of $\ce{H2O}$ present is also $y=0.5$, and the initial moles of both $\ce{H2}$ and $\ce{CO2}$ are 0.

After equilibration, both $\ce{H2}$ and $\ce{CO2}$ are present in the amount $x$ and so by stoichiometry, $\ce{H2O}$ and $\ce{CO}$ must be present in the amount $y-x=0.5-x=\frac{1}{2}-x$.

Since the reaction doesn't change the number of moles of gas, it doesn't matter much if we use $K_p$ or $K_c$, so $K_p=\frac{x^2}{(\frac{1}{2}-x)^2}=\frac{x^2}{\frac{1}{4}-x+x^2}=~...$

I'm sure you can figure out the rest.


Minimum amount of water in pressure cooker When cooking I would like to know the minimum amount of water I can add to my 6.2 L pressure cooker, worst case, so without any food. At second ring cooking the pressure is 0.8 bar.

I assume I need to solve $pV = nRT$ and know the density of water vapour at 0.8 bar, but that introduces $m = \rho V$, where I suppose $m$ and $V$ are unknowns?

Question

My high school chemistry is really rusty. Can someone help me in the right direction, and perhaps also tell if it at all something that can be solved without knowing the temperature?

In a pressure vessel, the volume $V$ is approximately constant. It is given as $V = 6.2\ \mathrm l$. However, this volume includes

liquid water
steam
air
food

in unknown amounts.

The gauge pressure is given as $p_\mathrm e = 0.8\ \mathrm{bar}$; i.e. the absolute pressure is approximately $p = 1.8\ \mathrm{bar}$.

We may use so-called steam tables to look up the properties of water at the given pressure (in the following, I use parameter values taken from the REFPROP – NIST Standard Reference Database 23, Version 9.0). We may find that the saturation point (equilibrium of liquid water and steam) at a pressure of $p = 1.8\ \mathrm{bar}$ corresponds to a temperature of $T = 117\ \mathrm{^\circ C}$.
At this point, the density of liquid water is $\rho_\mathrm l = 946\ \mathrm{g/l}$ and the density of steam is $\rho_\text{steam} = 1.02\ \mathrm{g/l}$.

If we ignore the volume of the remaining liquid water, air, and food, we may consider the limiting case in which the entire volume $V$ is filled with steam. Thus, the mass $m_\text{steam}$ is given by
$$\begin{align}
m_\text{steam} &= \rho_\text{steam} \cdot V \\[3pt]
&= 1.02\ \mathrm{\frac gl} \times 6.2\ \mathrm l \\[3pt]
&= 6.3\ \mathrm g
\end{align}$$
Therefore, only $6.3\ \mathrm g$ of water (which corresponds to about $6.3\ \mathrm{ml}$ of cold water at normal pressure) are required to fill the entire volume with steam at the given pressure.

However, a significant amount of excess liquid water should remain in the cooker in order for it to function properly. The liquid water is heated in the lower part of the cooker and boils at about $117\ \mathrm{^\circ C}$. The steam is distributed in the entire volume and condenses on colder surfaces (food and in particular the walls of the cooker). The condensed water runs down and it is collected in the lower part of the cooker. This circulation is essential for the uniform distribution of heat within the cooker; hence, the cooker should not dry out. The additional amount of liquid water should be sufficient to wet all surfaces (food and cooker) thoroughly.

Up to now, we have ignored the air that remains in the cooker. At initial (cold) conditions (pressure $p_0 = 1\ \mathrm{bar}$ and temperature $T_0 =20\ \mathrm{^\circ C}$), the density of dry air is approximately $\rho_{\text{air},0} = 1.189\ \mathrm{g/l}$. If we now ignore the volume of the liquid water, steam, and food, we may consider the limiting case in which the entire volume $V$ is filled with air. The mass $m_\text{air}$ is given by
$$\begin{align}
m_\text{air} &= \rho_{\text{air},0} \cdot V \\[3pt]
&= 1.189\ \mathrm{\frac gl} \times 6.2\ \mathrm l \\[3pt]
&= 7.37\ \mathrm g
\end{align}$$
The air cannot escape. At the final pressure of $p = 1.8\ \mathrm{bar}$ and temperature of $T = 117\ \mathrm{^\circ C}$, the trapped air is compressed to a density of approximately $\rho_\text{air} = 1.607\ \mathrm{g/l}$. The corresponding new volume is
$$\begin{align}
V_\text{air} &= \frac{m_\text{air}}{\rho_\text{air}} \\[3pt]
&= \frac{7.37\ \mathrm g}{1.607\ \mathrm{g/l}} \\[3pt]
&= 4.6\ \mathrm l
\end{align}$$
We see that a large part of the volume is taken up by the trapped air. The amount of water that is actually required to create steam at the given pressure is reduced accordingly.


Why is acetyl cyanide not obtained from acetyl chloride? Why doesn't acetyl chloride or any other aliphatic acyl chloride react with cyanide ions and undergo nucleophilic addition elimination to form aliphatic acyl cyanide, even though this reaction occurs for aromatic acyl chloride?

This was the related question in my workbook:

  Which of the following compounds will react with ethanolic $\ce{KCN}$?
  A) ethyl chloride B) acetyl chloride C) chlorobenzene D) benzaldehyde

My answer is A, B, D undergoing SN2, nucleophilic addition-elimination, and benzoin condensation respectively.
Answer is given as only A,D.

Searching the internet led me to this article, but I am not able to understand why.

Acetyl chloride on reacting with $\ce{KCN}$ in ethanol will give ethyl acetate. If acetyl cyanide forms at all, it will immediately react with the ethanol solvent as it is a more reactive species than acetyl chloride.

Acetyl cyanide is known and may be formed by catalytic reaction of hydrogen cyanide and ketene at elevated temperature. Details.

Aromatic acyl cyanides are less reactive than alkyl acyl cyanides much as aromatic acyl chlorides are less reactive. They may be formed by reaction of the neat aromatic acyl chloride with cuprous cyanide (details) and are stable to a wash with cold aq. sodium bicarbonate.


Percent difference: What to place on the denominator? I'm genuinely confused on what to put on the denominator for the formula $$\%~\mathrm{difference} = \mathrm{\left|theoretical~-~experimental\right|\over theoretical}$$ or some other similar formula with different variable names that mean the same thing.

Specifically, I have a reading on the iron content per pill written on the bottle, and I have an amount that I calculated from doing some math with the experimental data. The problem is, I know that the calculated value is experimental, but the iron content per pill isn't really theoretical because it's known. Anyway, which should I put in the denominator?

The experimental value is the one you measured/calculated yourself, from your own experiments. The theoretical value is the one that you obtained from some third party, to which you are comparing your experimental value.

So, the number on the bottle is the theoretical, and the number from your data and calculations is the experimental.


Do atoms form either a positive or a negative charge, but not both? I do not mean at the same time, of course, but I mean it appears from an overview of the common charges formed from ionizing various elements that each element forms one or more of either positive or negative ions, but they never have the opposite charge. For example Fe may have +4 or +3, but never -anything.
However, I am aware that this is from a common ion chart. Perhaps an atom can be a cation and an anion under specific circumstances?

Actually, in theory almost all of the elements can be found with both positive and negative oxidation numbers: it's just a matter of finding a system with the proper reagents and conditions to force it. If you isolate chemical species which have a very strong tendency of displaying some specific behaviour (accepting electrons, donating electrons, coordinating ions, releasing a leaving group, bonding to metals, releasing a proton, adopting a specific molecular geometry, or any other myriad properties), then you can often obtain strange results by clashing them with substances which also have the same tendency, but not quite as strong. This often causes the substance with the weaker behaviour to "operate in reverse".

Let me give a vivid and related example. As we all know, the alkali metals (group 1 elements) are exclusively present as cations with an oxidation number of +1, except in the pure metals, where it's zero, right? Well, here's something which might shatter your world: most of the alkali metals (with the exception of lithium, for now) also form alkalides, that is, stable salts containing discrete, clearly observed $\ce{Na^{-}}$, $\ce{K^{-}}$, $\ce{Rb^{-}}$ or $\ce{Cs^{-}}$ anions, with alkali metals displaying oxidation number -1.

How is this done? All you need is to find a neutral substance with a much stronger tendency to donate an electron than a neutral alkali metal atom (easier said than done). Since neutral alkali metals atoms form quite stable cations upon loss of an electron, this implies you need to search for a neutral substance capable of donating an electron and forming a cation with exceptional thermodynamic and/or kinetic stability. This can be achieved, for example, by the use of cryptands, which are cyclic molecules capable of coordinating very strongly to cations, strongly enough that they even coordinate alkali metal cations very well. The cryptand-coordinated cation is both thermodynamically and kinetically stable enough that alkalide anions, which would be extremely reactive otherwise, are not reactive enough in this case to immediately cause charge transfer and neutralize the negative charge.

Amusingly, it is actually possible to prepare a single compound which contains both alkali metal cations and anions, as exemplified by $\mathrm{[Na(2,2,2- crypt)]^{+} Na^{-}}$, which contains a sodium cation coordinated by a cryptand as a counterion to a natride/sodide ($\ce{Na^{-}}$) anion. One can imagine this compound being made by putting together a neutral sodium atom ($\ce{Na^{0}}$) and the neutral cryptand species $\mathrm{[Na(2,2,2- crypt)]^0}$. As Brian mentions in the comments, this latter species is actually an electride, which can be written as $\mathrm{[Na(2,2,2- crypt)]^{+}e^{-}}$ and thought of as a salt where the anion is a lone electron (!). Both the neutral sodium atom and the electride have a strong tendency to lose an electron in chemical reactions, but this tendency is much stronger for the electride. Thus, the electride ends up having its way, forcing its very loosely bound electron onto the neutral sodium atom, causing the neutral sodium atom to "operate in reverse" and accept an electron rather than donate it, resulting in the $\ce{Na^{-}}$ anion.

This table on Wikipedia is far more complete than most "common oxidation number" tables out there, and it lists many negative oxidation numbers for elements, including iron! For many of the transition metals, negative metal oxidation numbers are achievable by using the carbonyl ($\ce{CO}$) ligand, which removes electron density from the metal atom via backbonding. This stabilizes negative charges on the metal atom, once again allowing the resultant species to survive with the appropriate counterion.


How do you explain partial charges in molecules that have a dipole? For example if i have hydrogen and fluorine covalently bonded with each other, it would result in a polar molecule with the valence electrons being pulled closer to the fluorine than it is to the hydrogen. At some time the electrons will be at the fluorine end more often as it has higher electronegativity and thus making the hydrogen end partly positive. How does this work? If one electron is pulled to the fluorine end (the one that is shared), doesn't this just revert the hydrogen to its neutral state (as an element) and not be a partly positive charge?

Or does the electrons that are shared also including the lone pairs are going to orbit the shell of the other covalently bonded element, so that in some time there won't be any electrons in one end?

At some time the electrons will be at the fluorine end more often as it has higher electronegativity and thus making the hydrogen end partly positive. How does this work?

This is the charge distribution on the "surface" of the HF molecule, estimated by the program molcalc.org:

In a thought experiment, if you placed a negative sample charge (ideally with less charge than the elementary charge of the electron to not mess with the electron distribution) on the surface of the molecule, it would feel a force that is color coded in the picture (blue and cyan - attractive, red and yellow - repulsive). If you go sufficiently far away, the sample charge would not have any interaction because the molecule is net neutral. On the other hand, if you approach either nucleus, the sample charge would experience strong attraction (so the fluorine nucleus is not negative, even though we correctly assign the fluorine atom a negative partial charge).
To get a quantitative measure of the partial charge, you have to divide the space into two parts - that "belonging" to fluorine and that "belonging" to hydrogen. Then, you can sum up the charges due to the electrons and compare it to the nuclear charge.

If one electron is pulled to the fluorine end (the one that is shared), doesn't this just revert the hydrogen to its neutral state (as an element) and not be a partly positive charge?

Two electrons are shared. If you start with atoms, electron density from both atoms moves toward the center of the bond (very little, surprisingly, and drawing a stick for a bond is not very realistic but is great for counting electrons).
Your idea is that one electron is pulled to the fluorine. This is a very simple model, but let's explore it a bit. In this picture, the electron "originating from the hydrogen" would still be involved in binding, so we are not back to atoms.
What really happens is that instead of being spherically distributed around atoms, the electron distribution changes a bit when the bond is formed, with the bonding orbital (in the simplest valence bond or molecular orbital model) having electrons occupy the space near either nucleus, but not in a symmetric way. That leads to the charge separation.

Or does the electrons that are shared also including the lone pairs are going to orbit the shell of the other covalently bonded element, so that in some time there won't be any electrons in one end?

I did not understand this scenario. Maybe edit the question to clarify before it gets closed.


The effect of dilution on the rate of a system in equilibrium I'm attempting to answer a question from my textbook, however the answer that the textbook provides doesn't make sense to me.
The question:

Laura, a VCE chemistry student, loves graphs and wanted to represent an equilibrium reaction by using graphs. The reaction she chose was:
$\ce{Fe^{3+}(aq) + SCN-(aq) <=> FeSCN^{2+}(aq)  \ \ \ \ \ \ \Delta H=-\nu e}$
Draw the rate-time graph for the reaction when the system is diluted at constant temperature.

In my mind, this dilution would instantly decrease the concentration of the reactants and products. But as it decreases the concentration of each species present by the same factor this will not disturb equilibrium, and simply result in a slower rate of reaction.
Given my assumptions I drew a graph where the forwards and reverse reaction both dropped vertically then continued on at the same, lower rate of reaction in equilibrium.
But the answer provided by the textbook is as follows

I just came across another question similar to this, and supposedly the system partially opposes the increase in concentration by favouring the reaction that produces the least amount of particles (in the same way as a change in pressure would cause the system to react). But if this is the case, and the concentrations are still balanced after the dilution, wouldn't it change the equilibrium constant? Which I have been told only temperature can change.
I'm assuming then that my understanding is flawed somewhere in regards to the system remaining balanced after the dilution, but I can't seem to see how diluting the system would unbalance the concentrations as they're all being decreased by the same factor.
Is the answer provided by the textbook wrong? Or is there something that I'm missing?

The rate forwards and back are the same at equilibrium  and differ before this is attained. Let the reaction be $A+B=C$. If each concentration is suddenly halved then the reaction is not necessarily any longer at equilibrium and in this case the rates forwards and back differ until a new equilibrium is reached. The figure shows this. The red curve is the forwards rate $k_1[A][B]$ and the green one $k_{-1}[C]$ the reverse. As you see the new equilibrium takes a short while to re-establish. (You can ignore the scales they depend on the particular rate constants and concentrations used but the behaviour is general.)
The figure you show is not fully supported by the calculation. The red curve (forwards rate)  agrees but the blue one does not.  If you do not halve the product concentration you get the  blue curve in the figure.


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

This question is based on relative rates of reaction of different $\ce{H}$ with halogens. In the free radical mechanism of chlorination of alkane, relative rate of reaction of $\ce{3^\circ : 2^\circ : 1^\circ}$ $\ce{H}$ is in order $5:3.8: 1$.

Using this, we get an idea about product amount. To get a rough estimate of relative amounts of different products, we multiply number of identical hydrogen and its rate of reaction with chlorine.

So there is only $1$ $3^\circ$ $\ce{H}$ and its rel. rate is $5$. So relative amount is $5$. For the $2^\circ$ $\ce{H}$, there are $2$ of them and their relative rate is $3.8$. So its relative amount is $2 \cdot 3.8 = 7.6$. Similar calculation for rest of the hydrogens will show you why given answer is correct.

Refer: Wikipedia


Volume of dibasic acid required to neutralise 0.04 mole of OH- The volume of 0.1 N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04  mole of OH− in aqueous solution is:

  200 mL
  400 mL
  600 mL
  800 mL

Since it is a dibasic acid, it has two ionisable hydrogen atoms and hence its n-factor is 2. I simply used $$V\cdot 0.1\cdot 2=0.04$$ and got volume as 200 mL. But however according to the answer key it is given as 400 mL. Now the answer key is not of the highest quality and often has errors, so it can be challenged (but you need to pay a decent sum of money for the chance that they may look into it). Am I right to challenge this question or am I missing something?

Unfortunately you are missing something. The concentration is already given as an equivalence concentration, thus taking the dibasic nature of the acid into account. Therefore you have been given $c(\ce{H+})=0.1~\mathrm{mol\,L^{-1}}$, not $c(\ce{H2A})=0.1~\mathrm{mol\,L^{-1}}$. In the latter case your calculation would be correct.

The use of normality is ambiguous and it is not recommended to be used any more.


Why do bubbles form when it rains? I noticed sometime ago that bubbles formed in puddles/streets whenever it rained. I am wondering if someone could tell me what could cause such bubbles. I examined the bubbles again today more closely and they didn't look "soapy". I know that is not a really valid answer to rule out that leftover detergent/soap could be the cause of such bubble formation. However, the bubbles are short-lived. Furthermore, there aren't "frothy". In other words, there are no clumps of small bubbles.

Bubbles consist of a gas trapped by a liquid, where the liquid has a surface tension high enough to encapsulate the gas. This gives bubbles three ingredients: the liquid, the gas, and surface tension.

The first ingredient, the liquid, is obviously the rainwater.

The second ingredient, the gas, is (also obviously) air. But what isn't obvious is how the gas becomes entrained in the liquid. Asphalt is porous, meaning the microstructure of asphalt consists of lots and lots of tiny random tubes through the solid material. Thanks to the capillary effect, water is sucked into those tiny tubes, and gas is forced out. This gas being forced out of the porous asphalt material and through the liquid on the surface creates bubbles.

Surface tension is the third ingredient, and controls the bubbles' shape, size, and strength. Surface tension can be enhanced by surfactants (big molecules which, when added to a liquid, agglomerate and increase the surface tension). The surface of a road accumulates lots of surfactants - mostly, oil. Oil consists of many long-chain hydrocarbons, which are perfect surfactants (most detergents are actually long-chain hydrocarbons that are polarized at one end). Surfactants tend to agglomerate (that's why, if you pour a little oil on top of water, it will tend to clump together as it spreads out over the surface of the water). Having oil act as a surfactant increases the surface tension of the rain water on the surface of the road, and makes any bubbles formed in the water stronger. That explains the large bubbles you saw.


(Single) Amino acid rotational degrees of freedom When defining the rotational degrees of freedom for a single molecule, should we just include the psi and phi angles (red lines in the picture) or the rotation around every single bond (yellow lines in the picture) or both?

You should be concerned with bonds where the rotation is non-trivial. For example, rotation about a C-H or O-H bond doesn't do anything. But rotation around the serine side-chain C-O bound or the side-chain central carbon bond or the C-O carboxylate bond all matter.


Ideal gas law? implied volume? Working on a practice exam. The question is

  A volume of gas contains
  0.25 moles of $\ce{CO2}$
  0.30 moles of $\ce{N2}$
  0.15 moles of $\ce{O2}$

  What is the partial pressure of $\ce{O2}$ if the total pressure was $690\ \mathrm{mmHg}$ and the temperature was $25.0\ \mathrm{^\circ C}$

I know the formula is $p_a=\dfrac{n_aRT}{V}$

So far I know that

$n_a = 0.25\ \mathrm{mol}$

$R  = 0.0821\ \mathrm{l\ atm\ K^{−1}\ mol^{−1}}$

$T = 298\ \mathrm K$

But I can't seem to figure out volume. I was reading somewhere that volume is directly proportional to The amount of moles, but maybe it has something to do with the total pressure. I'm a little lost.

The only thing I can think of would be getting volume via the Ideal Gas Law as well. Would that work?

You don't need the Ideal Gas Law. You can use it (and I'll show you how). The question asks for partial pressure. You want Dalton's Law of Partial Pressures.

  Dalton's Law

Dalton's Law states that the total pressure in a system is equal to the sum of the partial pressures of the components:

$$P_T=P_A + P_B + P_C + ...$$

At constant volume and temperature (which you have, even if you don't know the volume), the ideal gas law can be rearranged so that:

$$\frac{P}{n}=\frac{RT}{V}$$
$$\frac{P_T}{n_T}=\frac{P_A}{n_A}$$
$$P_A=\frac{n_A}{n_T}P_T$$

  Ideal Gas Law

You can find the volume of the system because you have the total moles $n_T$ and the total pressure $P_T$:

$$V=\frac{n_T RT}{P_T}$$

You can then replace the number of moles with $n_A$ and solve for $P_A$;

$$P_A=\frac{n_A RT}{V}$$

This approach is basically equivalent to the previous except:

Using the ideal gas law you have to do two calculation (volume and partial pressure), while you only need to do one with Dalton's Law
Fewer math steps means less rounding, which means less error and higher precision.
If you use Dalton's Law, you don't have to worry about the units of temperature or pressure. Temperature is not in the equation, and the pressures are a ratio. You can use whatever pressure unit you want without converting.
You don't need to worry about $R$ and its units in the Dalton's Law approach.

  Solve problems as ratios whenever you have the chance.


Does molarity change if you decrease volume? For example if the solute of the solution contains $\pu{0.5 mol}$ and the total volume is $\pu{0.5 L}$, the molarity would be $\pu{1M}$. But if I take a portion of that solution, let's say $\pu{0.1 L}$, would that molarity be still $\pu{1M}$? If it were that way, the solute would be $\pu{0.1 mol}$ too.

Or, do the moles of the solute stay the same $\pu{0.5 moles}$ and the molarity would change to $\pu{5M}$?

So my question is, does the solute decrease when the volume decreases too?

No, solute will not decrease if volume will decrease.But molarity will increase for sure, cause molarity is the amount of solute in litre of solution.

Consider container A in which you took 1 litre of water and added two spatulas of NaCl and in another container B you took 0.5 litre of water and added two spatulas of NaCl. Now which container has greater amount of solute , the second one that is container B ,hence molarity of container B solution will be greater than container A.

Hope it will work to clear concept.


What is the difference between reacting and dissolving? When one chemical is added to another how would we come to know whether the compound dissolves in it or reacts with it? For example $\ce{AgNO3}$ dissolves in $\ce{CCl4}$ whereas it could also have been a double displacement reaction. And what is the major difference between the two  processes at atomic level?

CCl4 is a covalent compound (having no free ions). Thus it cannot enter into an ionic reaction/double displacement reaction.

Basic Requisites for checking reactivity

Reactivity Series (Elements higher up in the series replace the lower elements)
Checking whether bonds are covalent or ionic (You might just need to remember which compounds are which type. Some compounds can have both type of bonds)

Bonds

Ionic = "Ionic bonding is a type of chemical bond that involves the electrostatic attraction between oppositely charged ions. These ions represent atoms that have lost one or more electrons (known as cations) and atoms that have gained one or more electrons (known as anions)." Wikipedia
Thus the elements can take or give electrons. In a solution they exist as ions.
Covalent = "A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs and the stable balance of attractive and repulsive forces between atoms when they share electrons is known as covalent bonding." Wikipedia
Thus it has no free electrons to give or take.


Is it 'Phenyl' or 'Phenol' that is used as a cleaning agent? I often see Phenyl written on cleaning agents in stores. Here are some examples:

One day, a guy standing beside me at a store picked up a bottle of phenyl/phenol and he said, "The big corporation got it wrong. It's phenol, not phenyl."
I looked up the definition of Phenol, and it reads:

A toxic white soluble crystalline acidic derivative of benzene; used in manufacturing and as a disinfectant and antiseptic; poisonous if taken internally

This is what it says for phenyl:

A hydrocarbon radical (C6H5) regarded as the essential residue of benzene, and the basis of an immense number of aromatic derivatives

I am not a student of chemistry, so I don't know what's right or wrong here. But I want to know whether the big corporation got it right or wrong.
It would be nice if anyone here can explain this in the layman's terms.

This is what the pack says:
Ingredients: Coaltar acids and oils, phenolic compounds.
Pesticide/bactericide/fungicide/germicide.
It is a highly concentrated Phenyle, dilute 1 part phenyle with minimum 100 parts of water, resulting in a pinkish white solution.
Black Disinfectant Fluid - Phenolic type
Class Black Type Normal Grade 3

It's definitely not phenyl in there. This isn't stable and would immediately react to something else. Phenyle seems to be a name for a certain disinfecting agent out of pine oil.

It says "phenolic type" on the bottom, so my best guess, without seeing the actual ingredient list, is that it is a derivative of phenol. Derivative means that phenol is the base structure and it's somehow modified. Phenol itself is toxic but in the past it was used as a disinfectant, especially by doctors. It actually turns your skin white and that color goes away after some hours. Pine oil seems to have phenols in it, but I cannot find information on the structure of those.


Why does basicity of group 15 hydrides decrease down the group? In my textbook it is written that the order of basic strength of pnictogen hydrides is

$$\ce{NH3 > PH3 > AsH3 > SbH3 > BiH3}$$

I tried but could not find any explanation as to why this happens. What is the explanation?

Each of these molecules has a pair of electrons in an orbital - this is termed a "lone pair" of electrons.  It is the lone pair of electrons that makes these molecules nucleophilic or basic.  As you move down the column from nitrogen to bismuth, you are placing your outermost shell of electrons, including the lone pair, in a larger and more diffuse orbital (the nitrogen lone pair is contained in the n=2 shell, while the bismuth lone pair is in the n=6 shell).  As the electron density of the lone pair is spread over a greater volume and is consequently more diffuse, the lone pair of electrons becomes less nucleophilic, less basic.

Commment: But greater the EN of the central atom , lesser would be its tendency to donate electrons right?

Response to Comment: The ENs of the central atoms are N (3.04), P (2.19), As (2.18), Sb (2.05), Bi (2.02). All of these atoms, except for nitrogen have similar ENs and I think the electron density argument is valid for them. However, in the case of nitrogen and phosphorous there is a significant EN difference that would tend to argue in the direction opposite to my answer. However there is another important difference between N and P. The lone pair in ammonia is in an sp3 orbital. In all of the other cases the central atom is essentially unhybridized (~90° H-X-H angles) and the lone pair exists in an s orbital. Therefore the lone pair electron density in ammonia (being in an sp3 orbital) is effectively increased compared to phosphine where the lone pair is in an s orbital. So even though the EN difference between N and P is significant, when hybridization differences of the central atom are taken into account the electron density argument still explains the trend in basicity.


Unique carbons on long carbon chains If we have a long carbon chain like octane for example, do we determine if a carbon is unique by just looking at the immediate surrounding atoms of each  carbon (in this case 3 unique carbons), or do we need to take into account the symmetry of the whole molecule (many more unique carbons)? Edit: This is for NMR purposes

In general, the problem is a mathematical one. What you are asking for is the 'equivalence classes' of the atoms under the symmetry group. As pointed out in the comments, this may or may not be chemically important.

One simple way to divide the atoms into equivalence classes is to use Morgan numbers. The algorithm is particularly simple:

Label the atoms with their degree (so terminal carbons are "1")
Update each atom's label by adding the labels of the neighbouring atoms
Repeat until the number of different labels is constant

So, for a linear hydrocarbon, the labels start out as [1, 2, 2, ..., 2, 1] and update to [3, 5, 6, 6, ..., 6, 5, 3]. After a couple more rounds, you get a pattern of labels like [a, b, c, d, d, c, b, a].

This works reasonably well, although there are more complex algorithms for determining non-equivalent atoms. As you have worked out, the atoms beyond the immediate neighbourhood are necessary.


Liquefying butane in a freezer If place a can of pressurized liquid butane (such as a lighter refill canister) into a freezer to get it below its boiling point of $-1\ \mathrm{^\circ C}$,
and then release it into a container kept at a freezing temperature, will it remain in the liquid state without boiling off (until it reaches its boiling point), or will it boil off immediately?

I had a cold metal container kept at a stable temperature of $-7\ \mathrm{^\circ C}$, but when I released room temperature butane into it, it boiled off instantly.

Why did that happen, and what will happen when it is released from the can at a temperature at or below $-1\ \mathrm{^\circ C}$?

Since the room temperature butane didn't instantly cool when it hit the container, it was still above the boiling point.

If you had a large freezer that was roughly at equilibrium at a temperature below $-1^\circ C$, and you put both the pressurized container and the empty container in there and let them reach equilibrium, then when you opened the pressurized container it would not boil and you could conceivably pour it into the other container.

However, butane has a large vapor pressure (the enthalpy of vaporization is relatively small). This means that even at low temperatures, it will evaporate very quickly. This is especially true because the atmosphere does not normally have any butane in it - so even a small partial pressure combined with convective transport in the air will cause it to evaporate even faster. If you imagine pouring some rubbing alcohol out on a table and then putting a fan next to it, it is the same idea. Butane molecules randomly have enough energy to escape the liquid, and air currents carry them away. Since there is no butane vapor nearby to establish an equilibrium with the liquid, more butane evaporates to take its place.

The net result of this is that even if you did the experiment in a freezer and made sure everything was below $-1^\circ C$ (which includes you wearing well-insulated gloves and not breathing on it), it might still evaporate so fast that it would look as if it were boiling.

From wikipedia, here is a plot of vapor pressure vs T for butane:

The boiling point is the point where the vapor pressure line exceeds atmospheric pressure. Atmospheric pressure is usually around 760 mmHg. This is a log plot for the P axis, which means one large grid step is ten times larger than the previous one. From this graph, you can see that at $-30^\circ C$, the vapor pressure is still ~200 mmHg. This means that roughly 26% of the air at equilibrium would be butane. Unless you have a very tiny freezer, that is a lot. This means that you might have to wear a gas mask and get a 55 gallon drum of butane before you could really do any "pouring" in the traditional sense.

At $-100^\circ C$ you are down to 1 mmHg vapor pressure. At this point I would say that even a small amount (like what is in a lighter) would stay liquid long enough that you could pour it.


Why replace c with v in Broglie hypothesis I am in XI and I am currently studying atomic orbitals, I stumbled up and am stuck at the following step of de Broglie's hypothesis.

I did a little search on-line and found almost same derivation everywhere, the only place I found a more complex and supposedly accurate formula was the Wikipedia itself, though it did not explain the derivation.

Few such sites:

http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/De_Broglie_Wavelength
http://entrancechemistry.blogspot.in/2012/09/de-broglie-equation.html

In the derivation of the formula $$\frac{1}{\lambda}=\frac{h}{m\color{green}c}=\frac{h}{m\color{red}v}$$

we replace $c$ with $v$. I want to ask why?

As of Wikipedia it gives a better formula with the Lorentz factor $\gamma$ :-

$$\frac{1}{\lambda}=\frac{h}{\color{red}\gamma m_oc}=\frac{h}{m_ov}  \sqrt{\color{red}1\color{red}-\frac{\color{red}v^\color{red}2}{\color{red}c^\color{red}2}}$$

Can anyone give the derivation or a hint or a logic that helps understand why the derivation from $E=mc^2$ and $E=h\nu$ gives the De Broglie equation.

C is for radioactive rays while v is for the matter waves..here c is for a particle travelling with light speed, i.e., radioactive rays, so we replace it with v velocity of any particle for its matter wave

$E=mc^2$ and $E=hv$.
Therefore,  $$mc^2=hv$$
Or $$mc^2=\frac{hc}{k}$$, where k=wavelength
$$mc=\frac{h}{k}$$
Or, $$k=\frac{h}{mc}$$
   Replacing c by v we get
$$k=\frac{h}{mv}$$
Note the difference between matter waves and electromagnetic waves. E.W. are associated with electrical and magnetic field while matter waves are not so E.w. travel at 3×10^2 m/s. While matter waves travel with different velocity. The equation of de broglie with c is valid, it can be treated for finding the wavelength of electromagnetic waves, or electrons. But the velocity of matter waves are quite smaller as compared to light speed. So he used a general v in place of c V can be equal to c when the particle is travelling at light speed


Why is change in entropy negative and change in enthalpy negative for the reaction of Magnesium and Hydrochloric acid? So

$$\ce{Mg(s) + 2 HCl(aq) -> H2(g) + MgCl2(aq) + Heat}$$

From a conceptual standpoint, shouldn't the change in entropy be
positive for this reaction because there are more moles of gas in
the products vs the reactants? Why is this not the case? Can this
be justified through a conceptual view of the problem rather than
a pure mathematical approach(simply calculating delta s as
negative)?
From a conceptual standpoint, why is the change in entropy negative for this reaction?

Screenshot of reference data (source):

For starters, you should be suspicious when the reference page is formatted as it is.

The real issue I see is that the page lists $\Delta S_{f}$. This is where you should immediately call BS.

We don't use this quantity because entropy is not an energy, therefore, you cannot set the standard state entropy to 0. That means that entropy of formation is pretty useless because you need to know the actual entropy of the standard state of components to know what the entropy of the species is. For elements, the standard molar entropy is positive. However, for solutions (and this is what I got wrong in my now-deleted comment), the standard molar entropy can be negative thanks to solvation (specifically, solvation shells of water can be highly "ordered").

In order to compute the entropy of this reaction, you need to know the standard molar entropy of each species. The easiest thing to do is to look it up in a trustworthy reference source, for example a NIST database or CRC Handbook of Chemistry and Physics.

$$
\begin{array}{|c||c|}
\hline\hline
\text{substance} & S^{\circ} (\mathrm{J}\ \mathrm{mol}^{-1}\ \mathrm{K}^{-1}) \\
\hline
\ce{H2(g)} & 130.68 \\
\ce{Mg(s)} & 32.67 \\
\ce{HCl(aq, 1 M)} & 56.6 \\
\ce{MgCl2(aq, 1 M)} & -25.1 \\
\hline
\end{array}
$$

Sources: hydrogen, magnesium, hydrochloric acid CRC handbook under "Thermodynamic Properties of Aqueous Ions", magensium chloride CRC handbook under "Thermodynamic Properties of Aqueous Ions"

These are the same values in your reference, but I had trouble believing those data. So, basically, the answer is correct. This reaction has negative entropy change.

Why? Because solutions are kind of evil. Solubility and other solution phenomena can be really hard to predict because they depend so much on how the solvent interacts with the solute. In this case, it's probably driven by some idiosyncratic size of the chloride anion, which somehow allows water molecules to form very nice solvation spheres around the anion. I'm rationalizing this because the standard molar entropy for solutions of magnesium iodide (84.5) and magnesium bromide (26.8) are all positive. These ions are probably of a size that prevents the water molecular from fitting together nicely into a single shell. The imperfections in the solvation shell increase entropy.

That means that the large entropy of hydrogen gas is balanced by some stupid solvent effect with negative entropy, so the overall change in entropy is negative.


Carbon monoxide formation in broken boilers It's common knowledge, that if you're boiler is broken, it might produce carbon monoxide ($\ce{CO}$) as a byproduct, which is very toxic, even in small quantities. My question is, why do broken boilers do this?

The combustion of hydrocarbons works as follows:

$\ce{hydrocarbon + O2-> CO + H2O}$

The reaction can then continue to the complete combustion:

$\ce{CO + 1/2 O2-> CO2}$

An example of balanced equation is the combustion of methane:

$\ce{CH4 +  3/2O2 -> CO + 2H2O}$
$\ce{CO +  1/2O2 -> CO2}$

So, carbon monoxide is a product of the incomplete combustion of a hydrocarbon.

It is produced when there is too much fuel (eg: methane), but not enough oxidant (eg: oxygen from air).

I guess that a boiler which cannot mix properly air and fuel produces carbon monoxide.


Can someone induce cancer in another person Is it possible to poison someone's food on a daily basis in order to give them cancer?

The thing here is that i'm worried someone at my workplace might be poisoning other people's water (each employee has his own packs of water bottles at their desks, also, this person seems to have some mental issues).
It is possible to put chemicals in a bottle with no water leaking via a syringe. So my fear is that this bad person (this person IS bad., believe me) may have poisoned, in a daily basis for 3 or 4 weeks, our water.

So the question is, could this person poison the bottles to make others get cancer? Or another highly dangerous disease?

[…] So the question is, could this person poison the bottles to make others get cancer? Or another highly dangerous disease?

Yes, in principle.

Information on known cancerogenic compounds can be found in CMR (Canceogenic, Mutagenic and toxic to Reproduction) lists published by national organizations responsible for workplace safety.

In Germany, this would be the IFA (Institut für Arbeitsschutz der Deutschen Gesetzlichen Unfallversicherung).

List with old nomenclature

List with new nomenclature

I have added these lists just for the sake of completeness!

If you suspect any of the activities that you described, contact the local authorities.


Why does polonium form acidic oxides? All metals form basic oxides. Metallic oxides are basic in character.

Polonium is a metal, so it should form basic oxides but it forms acidic oxides. Why are its oxides acidic in character?

All metals form basic oxides.

That's true. I cannot remember metal, that does not form basic oxide

  Metallic oxides are basic in character.

Not necessary. Polonium is not the only one metal, forming acidic oxides. Mn, Cr and some others do it as well.

Oxide is considered acidic, when it forms hydroxide, that is acidic (well, it is simplification, but let's stick with it for a while). Hydroxide is acidic, when it can lose proton, giving stable anion. Acidity of hydroxide and metallicity of element formed it does not connect any way. Acidity depends of ability of remaining part of molecule to stabilize negative charge, which usually correlate with $\ce{O/OH}$ ratio and electronegativity of central atom. So, $\ce{H6TeO6}$ is quite weak acid and $\ce{H2CrO4}$ or $\ce{HMnO4}$ are quite quite strong acids.


Exothermic or Endothermic reactions Can we determine whether a reaction is endothermic or exothermic? For example if we are given the following reactions:

$$\ce{2SO2 + O2 -> 2SO3}$$

$$\ce{H2 + I2 ->2HI}$$

$$\ce{N2 + 3H2 -> 2NH3}$$

Can we determine weather these reactions are endothermic or exothermic or do we have to just memorize this? Are there any general trends that could make memorizing this information more manageable?

In general just by looking at these equations there is no surefire way to know which is endo or exothermic. However you can look at the bond strength or the ∆H of formation for each molecule. In short you have to know something.

The energy stored in the molecules(specifically in the bonds) is lower before the reaction in an endothermic reaction and higher in an exothermic reaction. This means if you know relatively how strong the bonds or how much energy it takes to form each of the molecules then you can know whether the reaction is endo or exothermic.

If you have a few ∆H of formation values you can use Hess's Law


Why doesn't hydrogen displace metals from their oxides? As I understand it, a displacement reaction is where a more reactive element kicks out a less reactive element from a compound.

For example, chlorine is more reactive than iodine, so chlorine will displace iodine from potassium iodide, yielding potassium chloride and iodine.

If magnesium reacts with water you get magnesium oxide and hydrogen. But hydrogen is generally a lot more reactive than magnesium, so isn't this backwards? The above paragraph predicts that hydrogen would displace the magnesium from magnesium oxide.

So why does it actually happen the other way round?

If magnesium reacts with water you get magnesium oxide and hydrogen.

True. So you can't put out a magnesium fire with water. It just burns hotter.

$$ \ce{Mg(s) + H2O ->T[aqueous] MgO(s) + H2(g) ^}$$

    But hydrogen is generally a lot more reactive than magnesium, so isn't this backwards?

No, in the above reaction you're reacting water with magnesium. The hydrogen is given off as a gas, though a tiny amount will stay dissolved in the water.

    So why does it actually happen the other way round?

It will under suitable conditions. So with heat and under a pressurized flow  of hydrogen gas the reverse reaction will occur.

$$ \ce{ MgO(s) + H2(g) ->T[heat, pressure] Mg(s) + H2O(g) ^} $$


Temperature and reaction quotient Consider an exothermic reaction. If we double the temperature at which the reaction is run, will the reaction quotient increase?

I know that an increase in Q relative to K causes the reaction to favor the reactants as the system tries to return to equilibrium.

If a reaction is exothermic, and we take heat to be a product, wouldn't an excess of heat cause the reaction to favor reactants?

The question I'm having trouble with is this one; apparently the answer is choice 2. I wonder if this is a typo or am I fundamentally misunderstanding something.

Just to clear things up first (I might be saying something you already know, but I just want to be sure): the intent of the question is not to ask you what happens to the reaction quotient after equilibrium is re-established.

If a system is originally in equilibrium, with $Q = K_1$, and then you disturb it such that the equilibrium constant changes to $K_2$, then after the system re-reaches equilibrium you will have $Q = K_2$. If you interpret the question as asking whether $Q$ has changed after equilibrium is re-established, then the question basically amounts to asking whether the equilibrium constants $K_1$ and $K_2$ are different.

What the question is asking is whether $Q$ is affected instantaneously by the disturbance. Re-reading your question I think you got it, so...

Anyway, heat is not a product.

The reaction quotient for that reaction is defined as

$$Q = \frac{a_{\ce{CO2}}}{(a_{\ce{CO}})^2} \approx \frac{p_{\ce{CO2}}}{(p_{\ce{CO}})^2}$$

and $K$ is defined as the value of $Q$ at equilibrium.

Nowhere does heat appear in the equation. The idea of using heat as a "product" is nothing more than a mnemonic to help you remember the direction in which the equilibrium shifts. As long as the pressures are fixed, the temperature does not affect the instantaneous value of the reaction quotient.

When you increase the temperature, what happens is that the equilibrium constant $K$ drops. Let's say that before changing the temperature $Q = K = 0.01$ (just an example). If you increase the temperature, $K$ might fall to $0.001$ but $Q$ at that instant will still be $0.01$. As the system re-equilibrates, $Q$ will drop to $0.001$.

So, it is not that $K$ remains constant and $Q$ increases. It is more like: $K$ drops and $Q$ remains constant (at least instantaneously).


Electrolysis on aqeuous solution I know when electrolysis on a concentration aqeuous solution which contains halide ions($Cl^-$ , $Br^-$ , $I^-$). The halide ions is discharged although $OH^-$ is more easily to discharged.

For example, electrolysis of concentration aqueous sodium chloride. The cation $Cl^-$ and $OH^-$ will attract to the anode. The $Cl^-$ ions will be discharged although according to the electrochemical series that the $OH^-$ will be more easily discharged.

Why do that happen?

Let's explain this using an example. Consider a cell in which an aqueous solution of sodium chloride is electrolyzed. The $\ce{Na^+}$ ions migrate toward the negative electrode and the $\ce{Cl^-}$ ions migrate toward the positive electrode. But indeed at the anode, there are  two substances that can be oxidized at the anode: $\ce{Cl^-}$ ions and water molecules.
$$\ce{2Cl^-   \rightarrow Cl_2 + 2e^-}$$ $$\ce{2H_2O   \rightarrow O_2 + 4H^+ + 4e^-}$$
The standard-state potentials for these half-reactions are $E^0_{ox}= -1.36 V$ for the first couple and  $E^0_{ox}= -1.23 V$ for the second couple.

At first glance, it would seem easier to oxidize water ($E^0_{ox}= -1.23 V$) than $\ce{Cl^-}$ ions ($E^0_{ox}= -1.36 V$).

It is worth noting, however, that the cell is never allowed to reach standard-state conditions. The solution is typically 25% $\ce{NaCl}$ by mass, which significantly decreases the potential required to oxidize the $\ce{Cl^-}$ ion. The pH of the cell is also kept very high, which decreases the oxidation potential for water. The deciding factor is a phenomenon known as overvoltage, which is the extra voltage that must be applied to a reaction to get it to occur at the rate at which it would occur in an ideal system.

Under ideal conditions, a potential of 1.23 volts is large enough to oxidize water to $\ce{O_2}$ gas. Under real conditions, however, it can take a much larger voltage to initiate this reaction. (The overvoltage for the oxidation of water can be as large as 1 volt.) By carefully choosing the electrode to maximize the overvoltage for the oxidation of water and then carefully controlling the potential at which the cell operates, we can ensure that only chlorine is produced in this reaction.


Setup for Sensing change in concentration of NaCl I am making a sensor to sense the change in concentration of NaCl. My background is in electronics and chemistry is not my strong area. I got some screen printed electrodes, with Silver (Working Electrode) and Ag/AgCl (Reference Electrode). My initial test is to put the electrodes in separate solutions, with Reference Electrode in a reference solution of known concentration of NaCl and the Working Electrode in the test solution of varying concentration of NaCl. Will this give me a potential difference and will I be able to determine the change in Concentration of NaCl in the working electrode? If not can you make any suggestion on the setup? Thanks.

This will work with the correct set up. You're constructing a galvanic cell, which will result in a cell voltage. The cell voltage will change when you alter the concentration of NaCl on either side. Important though that your two half cells (each side) are connected both electronically and ionically. Do you have a salt bridge or a membrane between them? https://en.wikipedia.org/wiki/Salt_bridge


How to esterify phenol with oxalic acid? Is it possible to fully esterify phenol with oxalic acid in order to get diphenyl oxalate? What catalyst should I use and is the synthesis possible to perform at home with hobby equipment?

I don't know for sure, but it could work provided some conditions be fulfilled

use of oxalyl chloride
use low temperature, in order to try to suppress fragmentation of the intermediate (semi-ester) into the simple monoacyl chloride (by CO loss): in such cases $\ce{(COCl)2}$ behaves much like phosgene ($\ce{COCl2}$.
use the phenol as a dissociated salt (sodium, potassium) in polar solvents such as sulfolane or acetonitrile, this is optional and possibly risky, due to enhanced competitive "C-acylation" if ortho/para positions are free.
use Piridine-based solvents as reactivity enhancers (they convert acyl chloride to acyl immonium salts). Optionally, even stronger catalysts such as DMAP (p-dimethylamino pyridine) could work

Anyway, to sum up, the major risks stem from:

"C-acylation" (which is generally kinetically slower but irreversible and thermodynamically favored over "O-acylation"), if some suitable position is avail.
formation of diaryl carbonates (on extrusion of CO).
I guess this last drawback coul be at least reduced by working at low temperature.


Isopropyl alcohol to clean germs on countertops My wife recently found some website recommending mixing 91% isopropyl alcohol, white vinegar, water, and corn starch and using it to wipe down countertops. This would be where we put our raw chicken and pork after washing them.

I'm concerned that this won't kill the germs as well as harsh chemicals would, although I am as eager as she is to replace those hard chemicals. I need to know, from a chemist, whether isopropyl alcohol really will kill all those E. coli and Salmonella and whatever else the raw poultry is leaving behind.

As a bonus, can anyone recommend any better solutions that we can make on our own than the one mentioned above?

A 70% “rubbing” alcohol solution will kill those pesky bacterium usually in ten seconds or less. It will also kill a host of viruses, but not all of them.

For your common household disinfecting, a 70% solution will do the trick. Especially for e. Coli, salmonella, staph, and a few others that would be risk bacterium when preparing poultry, etc.

A final bit of information: 70% will be a more effective carrier than the 90-99% equivalent. Why? Because the water mix (30%) will act as a carrier or transport mechanism to get the alcohol into the bacterium cells in order to terminate them! The 90% mix simply is not very effective at penetrating the cell membranes.

See this for additional information:  https://www.cdc.gov/infectioncontrol/guidelines/disinfection/disinfection-methods/chemical.html


Binding Energy and Enzyme Rate I was preparing for the Canadian Chemistry Olympiad and I ran into the following question on one of the past exams:

  If enzyme A binds to the substrate 25 times stronger than enzyme B, what is the ratio of the catalytic rate between enzyme A and enzyme B if the energy of the two transition states is identical? What is the difference in activation energy between the two reactions?

I feel as if I can make a qualitative prediction, but cannot derive a quantitative prediction. I suspect, since enzyme A bind more strongly to the substrate than enzyme B, it would have a greater activation energy, and thus result in a slower reaction.

I suspect that it is necessary to use the equation
$$ \begin{align}
k = \frac{kT}{h} e^{-\Delta G^\ddagger/RT}
\end{align} $$

But the relationship between the two activation energies does not seem simple

$$\begin{cases}
\Delta G^\ddagger_A = \Delta G_{E + S \rightarrow T} - 25\Delta G_\text{bind} \\
\Delta G^\ddagger_B = \Delta G_{E + S \rightarrow T} - \Delta G_\text{bind}
\end{cases}$$

Plugging this into the the original equation, and solving for $k_B/k_A$, I get the following:

$$ \frac{k_B}{k_A} = \frac{e^{\Delta G_\text{bind}/RT}}{e^{25\Delta G_\text{bind}/RT}} = e^{-24\Delta G_\text{bind}/RT}$$

Since $\Delta G_\text{bind} < 0$, $k_B/k_A > 1$, supporting my qualitative prediction. Yet this does not give me a useful ratio, since it is dependent on the binding energies of enzymes A and B. This makes me suspect I am missing/overlooking something, any insight would be appreciated.

Your intuition that the enzyme binding more strongly to substrate will be slower is correct, this is called ground-state stabilization.  I think where you might be making an error is how you handle the factor of 25.  The factor of 25 probably refers to the ratio of equilibrium constants between E and ES for the two enzymes, not the ratio of binding energies.  So you need to start with delta G = - RT ln Keq, and compare for 2 different Keq's.  Then use your equation to estimate the difference in rate constants.


Equilibrium constant when adding more of a reactant Suppose I have the reversible reaction:

$$\ce{A +B⇌ C} $$

The reaction is at equilibrium with equilibrium constant $K$

I am told that if I increase the concentration of $\ce{B}$, the rate for the forwards reaction will exceed the backwards one. Fair enough.

I am also told that $K$ will necessarily increase. Why though? Its true that Forward reaction > back ward reaction until we reach a new equilibrium such that more of $\ce{C}$ is produced but I don't see why this implies in any way that the final quotient $\frac{[\ce{C}]}{\ce{[A][B]}}$ will necessarily be any greater.

Certainly if we have a simple reaction:

$$\ce{A⇌ C} $$

and we add more of $\ce{A}$, then the equilibrium constant for the new final final state will remain as it was, ceteris paribus.

What am I getting wrong, because my textbook suggests that $K$ will always increase, no matter what type of reaction I am dealing with (of course, as long as all reactants are in a suspended form, e.g. dissolved or gaseous and the reaction is subject to Le Châtelier's principle).

Please, if possible, keep the answer as a level intelligible to a high-school student. I don't know much about advanced university chemistry.

You're correct.  The equilibrium constant will not change.  The equilibrium constant is only defined at equilibrium.  Your analysis of the situation was flawless.


What do curly brackets {} mean? I learned that square brackets ex: $[\ce{H2O}]$ can be used to denote the concentration of water molecules in a fluid. The notation  used much in equilibrium problems and such.

However, sometimes I have seen curly brackets {} being used in similar scenarios, but I never really understood the difference. When are curly brackets used and what do they mean?

I found this example from an exam where the problem is to see if a solution of $\ce{Ca(OH)_2}$ in water is concentrated enough to result in precipitation.

Following is what how they solve it in the example:

Solubility constant for $\ce{Ca(OH)_2}$ is given as $K_\mathrm{s} = \{\ce{Ca^{2+}}\}\{\ce{OH^-}\}^2$ (100% saturation point I think)
$Q = \{\ce{Ca^{2+}}\}\{\ce{OH^-}\}^2 \approx [\ce{Ca^{2+}}][\ce{OH^-}]^2$ (actual saturation I think)

Where $Q$ is the actual saturation of the solution. And the answer is that no precipitation occurs since $Q < K_\mathrm{s}$ in this specific example.

So they seem to suggest that {} is something different than the concentration [] and that it can be approximated to the concentration.

The curly brackets denote "activity of" the species therein.

See the Wikipedia section: Basic definitions and properties of Equilibrium constant


Where does the name ‘aliphatic’ come from? Generally, carbons and the hydrogens bonded to them are classified as one of the following:

aromatic
olefinic
aliphatic
(something for triple bonds I don’t know)

The meaning of aromatic is rather clear to anybody: They belong to aromatic systems (even though the Hückel rule has nothing to do with a certain type of smell). Olefinic is also clear enough: Olefins were originally those gases which could generate an oily substance by reaction with e.g. chlorine ($\ce{C2H4 + Cl2 -> C2Cl2H4}$; the latter being 1,2-dichloroethane); the name olefin derives from French gaz oléfiant — oil-creating gas.

But I have no clue where the adjective aliphatic or the noun aliphat derive from, which are applied to saturated carbons. What was its original meaning and why do we use it?

Olefins are hydrocarbons that contain at least a double bond. The term is used —sometimes misleadingly— interchangeably with alkenes.

  Although the nomenclature is not followed widely, according to IUPAC, alkenes are acyclic hydrocarbons with one double bond between carbon centers. Olefins comprise a larger collection of cyclic and acyclic alkenes as well as di- and polyenes. — Alkene nomenclature – Wikipedia

The Goldbook page Wikipedia cites says the final words:

  Olefins

  Acyclic and cyclic hydrocarbons having one or more carbon–carbon double bonds, apart from the formal ones in aromatic compounds. The class olefins subsumes alkenes and cycloalkenes and the corresponding polyenes.

While Oxford dictionary seems to agree with the origin thingy:

  Origin

  Mid 19th century: from French oléfiant 'oil-forming' (with reference to oily ethylene dichloride).

Funny thing is, "aliphatic" also means "producing fats". Goldbook says:

  aliphatic compounds

  Acyclic or cyclic, saturated or unsaturated carbon compounds, excluding aromatic compounds.

Some definitions exclude alicyclic compounds from the list of aliphatic compounds, but IUPAC doesn't agree, so. Note that aliphatic compounds can be saturated or unsaturated.

So, back the main answer part,

  Aliphatic; Etymology

  From Ancient Greek ἄλειφαρ (áleiphar, “oil, ointment”). — Wiktionary

There are literally no articles on the Net speaking of the first discovered or documented aliphatic compounds, but I read in some textbook somewhere which mentioned that since these compounds were first referred to as what animal fats used in primitive or medieval lamps were consisted of, they're referred to as "aliphatic".

The only thing about a "first aliphatic" I could find is mentioning "Muconic acid is the first aliphatic compound formed as a result of aromatic ring splitting, which is further oxidized to fumaric acid and then participates in the cell metabolism as a standard endogenous substrate." by this paper.


Stoichiometric defects in silver bromide Stoichiometric defects (Schottky and Frenkel defects) in ionic crystals are formed due to irregularities in the ionic arrangement which maintain the stoichiometry and electroneutrality of the solid. It is known that $\ce{AgBr}$ shows both Frenkel and Schottky defects while most other compounds  show only one of these defects; for example, $\ce{AgCl}$ shows only Frenkel defects.

The ratio of the sizes of the cation and anion determines which type of defect is shown by the crystal. If the ratio is closer to unity, Schottky defects are more common, and if it deviates significantly from one, Frenkel defects are more common. Intuitively, this is correct since Frenkel defects involve the movement of the cation (or the smaller ion) into interstitial sites, which is indeed easier if the ion is very small as compared to the other ion, making it more mobile. Generally, the activation energy for cationic movement is lower. And if the ionic sizes are comparable, filling the interstitial sites is not feasible, and hence Schottky defects are more likely to be exhibited by the crystal.

The $\ce{Cl-}$ ion is smaller than $\ce{Br-}$ (i.e. the ratio $r_+/r_-$ for $\ce{AgCl}$ is closer to 1 than it is for $\ce{AgBr}$). Yet, $\ce{AgBr}$ displays Schottky defects while $\ce {AgCl}$ does not. Why is that so? (The size and hence the mobility of $\ce{Ag+}$ should be the same in both cases.)

Fundamentally, although ionic radii may offer some guidance as to defect formation, the bottom line is that the silver halides have complex bonding energetics that have to be taken into consideration.  I might suggest 'Defect structures in the silver halides', D.J. Wilson et al., Phys. Rev. B77, 064115 (2008) (presuming you have APS journal access). There the authors present DFT calculations on the formation energies of both Frenkel and Schottky defects in AgCl and AgBr.


How do I obtain Chrysoidine from Benzene? So, basically I can convert Benzene to Phenol using Dow's synthesis process and then use $\ce{NH3}$, $\ce{ZnCl2}$, $\ce{HNO2}$, $\ce{NaNO2}$ under 0-5 °C controlled temperature and convert the phenol to Diazonium Chloride Salt. Then with Coupling Reaction, I can convert the Diazonium Chloride to Aniline Yellow. But, Chrysoidine has another $\ce{-NH2}$ group attached to the Ortho position. So, an explanation would be greatly appreciated as to how to obtain Chrysoidine from there on.

I think aniline yellow is likely to nitrate ortho to the aniline giving you the wrong structure. Also the Sn/HCl reduction may well reduce the azo double bond.

Here is a route from benzene.
From benzene a simple nitration gives nitrobenzene. Divide your nitrobenzene, reduce one portion to aniline by hydrogenation over Pd or Pt catalyst. Nitrate the second portion to 1,3-dinitrobenzene procedure here. Reduce the 1,3-dinitrobenzene by hydrogenation over Pd or Pt catalyst to give m-phenylenediamine. Diazotise your aniline to the diazonium and react this with the phenylenediamine. This route was used industrially to prepare Chrysoidine in the 19th century reference here.


How to convert an amide to an imine? We haven't learned this in class yet, but after looking online, and through my textbook, I can't seem to find a definite answer. I am hopelessly and utterly lost.

The problem can be solved using retrosythesis.

You look at the product and imagine how that was made.

It's like looking at Lego construction, trying to identify the pieces - or looking at a pair of jeans, trying to identify how it was stitched together.

Then you go back and do the same for the previous building blocks.

The final product apparently is an imine. I'm sure we have seen that before among one of the exercises you worked on - in that case it was an intramolecular reaction.

Imines are usually made from carbonyl compounds and primary amines under acidic conditions. This one is made from propanal and ethylamine.

Looks good here, I see $\ce{H2NCH2CH3,\ H3O+(cat)}$ among your reagents.

What about the aldehyde then? The number of carbon atoms in the aldehyde and the amide (propionamide is the starting material) are the same, just the (formal) oxidation of the carbonyl/carboxyl group are different.

Let's think about a direct reduction of the amide with $\ce{LiAlH4}$:

OK, that isn't it. Just two steps would not have been enough anyway ;-)

But we do remember that

esters can be reduced to alkanols with $\ce{LiAlH4}$
alkanols can be oxidized to aldehydes with PCC (pyridiniumchlorochromate)

All this is among the list of reagents. Nice, we know to make the propanal from a propionic ester!

How to turn propionamide into a propionic ester then?

Direct reaction with methanol under acidic conditions does not work, but after alkylation of the amide with an alkyl halide, we should be ok.

Lets's summarize our five steps then:


Using PV = nRT to solve an electrolysis problem Question: Two electrolytic cells are connected in series (the same current passes through each cell). One cell for the electrolysis of water produces $\pu{100 cm^3}$
of oxygen, measured at $\pu{273 K}$ and
$\pu{1.01
×
10^5
 Pa}$
. The second cell contains molten lead(II) bromide,
$\ce{PbBr2}$.
Determine the mass, in $\pu{g}$, of lead produced.
My working: $$PV=nRT$$
$$n={PV\over RT}$$
$$n\pu{={(1.01 \times 10^5)(100 \times 10^{-6})\over (8.31 \times 273)}} $$
$$\pu{= 4.45 \times 10^{-3} moles}$$
$$\pu{mass = 207.2 \times ANS = 0.922 g}$$
The given solution is:

Why do I have to multiply by 2 - what is wrong with my method?

Both methods yield $\pu{4.45E-3}$ mol $\ce{O2}$. But to get $\pu{1 mol}$ $\ce{O2}$, you need $\pu{4 mol}$ electrons, as the half-reaction at the anode is : $$\ce{2 H2O -> 4H+ + O2 + 4 e-}$$So to get $\pu{4.45E-3 mol}$ $\ce{O2}$, the needed amount of electrons is $4$ times the amount of $\ce{O2}$, or $4$· $\pu{4.45E-3}$ = $\pu{1.78E-2}$ electrons. Now $2$ electrons are needed to produce $\ce{1 Pb}$. So $\pu{1.78E-2}$ mol electrons will produce half of it, or $\pu{8.9E-3}$ mole $\ce{Pb}$. And this amount weighs $\pu{8.9E-3} ·\pu{207.2 g}$ = $\pu{1.84 g}$