text:
CO poisoning - What I've been taught is a hoax?
In elementary school, I happened to ask my teacher:

What is $\ce{CO}$ they talk about these days?

I was then taught that Carbon monoxide "competes" with the oxygen in our blood (we were  taught that oxygen gets to our cells via blood at that time) and replaces it and thus, doesn't let enough O get to cells and this will be fatal in extreme cases.
I didn't do any research about it after that because I later made the implication that $\ce{CO}$ is able to replace O and react with Hb instead; but
I recently became doubtful about my previous lessons and decided to do a research to see what really was going on:

Carbon monoxide mainly causes adverse effects in humans by combining with hemoglobin to form carboxyhemoglobin (HbCO) in the blood. This prevents hemoglobin from releasing oxygen in tissues, effectively reducing the oxygen-carrying capacity of the blood, leading to hypoxia. Additionally, myoglobin and mitochondrial cytochrome oxidase are thought to be adversely affected. Carboxyhemoglobin can revert to hemoglobin, but the recovery takes time because the HbCO complex is fairly stable. Wikipedia, CO poisoning

Wikipedia states that the problems are:

$\ce{CO}$ is able to combine with Hb to form HbCO.
This compound is fairly stable, so we'll have a hard time removing $\ce{CO}$ from hemoglobin.

Other sources, such as How stuff works, Answers.com and CDC.gov have given either only more explicit explanations of wikipedia, or the explanations weren't "chemical-biology" enough.
I haven't seen a single reference in which the first assumption was declined, or agreed with. Is what I'd learnt non-scientific?

Check this about CO affinity to HB.

Each hemoglobin molecule has 4 subunits (4 hemes each having one iron atom). Each time a molecule (such as oxygen or CO) binds to one subunit, it changes the binding properties of the other subunits. This is referred to as positive "cooperative binding" in the case of hemoglobin, because binding one molecule makes the other sites bind more easily.
Initially, considering hemoglobin with no CO or oxygen bound, CO competes with oxygen for the same binding sites (ligating one of the iron atoms).  CO has a binding constant 300 times lower than oxygen, meaning a factor of 300 lower CO concentration will result in an equal percent of bound protein.
Considering CO has reached a level such that 2 out of the 4 sites can have CO bound, the remaining 2 sites would still provide enough oxygen carrying capacity for a person to live if the sites were function as normal; however, the 2 remaining sites then bind oxygen too tightly and don't release it.
Other iron-containing proteins such as myoglobin can also bind CO and are also inhibited.
See Pharmacokinetics and Mechanisms of Action of Carbon Monoxide for more infomation.

+1 Thanks Dave. Can you give me a reference to study more about the issue?

@MARamezani here is a reference, it was more difficult to find a reference than I thought.  http://nershco.com/case_studies/co%202.pdf You should read about cooperative binding of hemoglobin subunits generally also.


text:
What makes E1 more favourable than SN1?
Each goes through a carbocation intermediate which is enormously reactive so what would promote attack towards a proton (E1) rather than towards the carbocation for substitution. Besides an increase in temperature I can't think of much.

The most important factor is the nucleophilicity versus basicity of the attacking species.
Taking the trimethylcarbenium cation as a classic example:

Sterically hindered bases such as lithium diisopropyl amide (LDA) will favour elimination because it is physically much easier for them to remove a small proton from the outside of the cation than to reach the carbon atom at the centre surrounded by three methyl groups and form a new bond there. On the other hand, small species such as cyanide and water will favour substitution because because it is physically easier for them to reach the central carbon.
Additionally the strength of the base influences which reaction takes place. LDA is a strong base and the methyl hydrogens are quite acidic means that the elimination reaction is much more likely. However, weak bases such as iodide and water will favour substitution because they are less basic and so they are less likely to abstract an acidic proton from one of the methyl groups.
It should also be noted that the nature of the solvent has a strong effect both on the stability of the cation but also on the nucleophilicity of the attacking species. This page has some discussion on the effects of solvents on nucleophilicity and the website also has good discussions of elimination and substitution mechanisms.
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/alhalrx1.htm#hal1

source: Stack Exchange

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

sort: votes


text:
Constructing an equation for the precipitation of a salt
I was asked to write a balanced equation, as well as the net ionic equation, for the formation of $\ce{CaCO3}$ by precipitation. The equations should also include state symbols.
Unfortunately, I am not familiar with the concept of precipitation. Could anybody please show me a method for constructing these equations?

Precipitation is usually a single or double displacement reaction, and refers to the formation of an insoluble, solid salt from soluble ions. For instance, silver bromide, $\ce{AgBr}$, is insoluble in water. However, it is made up of $\ce{Ag+}$ and $\ce{Br-}$ ions, which we can introduce in the form of soluble salts:
$$\ce{LiBr(aq) + AgNO3(aq) -> LiNO3(aq) + AgBr(s)}$$
In an aqueous solution, lithium bromide and silver nitrate underwent a reaction, forming lithium nitrate and silver bromide. In the products, silver bromide is denoted with the state symbol (s), which stands for "solid". This reflects its insolubility in water and hence its tendency to "precipitate".
The net ionic equation is when we only write the ions that participate in the reaction. The ions that do not participate are called spectator ions. For example, in the reaction above, only $\ce{Ag+}$ and $\ce{Br-}$ participated in precipitation. Thus, the net ionic equation for the reaction above is:
$$\ce{Ag+(aq) + Br-(aq) -> AgBr(s)}$$
One way of deriving this net ionic equation is to realise that salts such as $\ce{LiBr}$ and $\ce{AgNO3}$ do not exist as molecules in water, but rather dissociate to give their constituent ions:
$$\begin{align}
\ce{LiBr(s) &->[dissolution] Li+(aq) + Br-(aq)} \\
\ce{AgNO3(s) &->[dissolution] Ag+(aq) + NO3-(aq)}
\end{align}$$
However, $\ce{AgBr}$ is insoluble and does not dissociate into ions. If we go back to the first equation and replace all the soluble salts with their constituent ions, we get
$$\ce{\underbrace{Li+(aq) + Br-(aq)}_{from LiBr} + \underbrace{Ag+(aq) + NO3-(aq)}_{from AgNO3} -> \underbrace{Li+(aq) + NO3-(aq)}_{from LiNO3} + \underbrace{AgBr(s)}_{does not dissociate}}$$
The spectator ions which do not take part in the reaction are $\ce{Li+}$ and $\ce{NO3-}$, as they are unchanged in this reaction. Subtracting them from both the left- and right-hand sides, we obtain the net ionic equation:
$$\ce{Ag+(aq) + Br-(aq) -> AgBr(s)}$$
If we take this idea and apply it back to $\ce{CaCO3}$, we first need to come up with soluble sources of the ions $\ce{Ca^2+}$ and $\ce{CO3^2-}$. A possible choice is $\ce{CaCl2}$ and $\ce{Na2CO3}$, although that is hardly the only option.
The overall equation may be written as
$$\ce{CaCl2(aq) + Na2CO3(aq) -> 2NaCl(aq) + CaCO3(s)}$$
and the net ionic equation as
$$\ce{Ca^2+(aq) + CO3^2-(aq) -> CaCO3 (s)}$$
where $\ce{Na+}$ and $\ce{Cl-}$ are the spectator ions.
Finally, here are some web pages you might find helpful:
http://www.occc.edu/KMBailey/Chem1115Tutorials/Net_Ionic_Eqns.htm
http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Precipitation_Reactions

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


text:
Why must the leaving group in E1cb be poor?
Typically the leaving group for E1cb is poor (like -OH or -OR) but why must this be the case? The substrate appears in the rate equation so surely a good leaving group would be beneficial?

There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another.  For example, in some elimination reactions the E1 and E2 pathways can operate in competition with one another.  An activation energy is associated with each of these 3 reaction pathways.  Whichever pathway has the lowest activation energy will be the major pathway followed.  By changing solvent, reaction temperature, relative strength of the nucleophile, relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other.

Typically the leaving group for E1cb is poor (like -OH or -OR) but why
  must this be the case?

The E1 mechanism involves ejecting a leaving group in its first step, while the E1cb mechanism involves removing a proton in its first step. Let's consider how changing the leaving group can shift an elimination reaction towards one pathway or the other.  To a first approximation, changing a leaving group will not affect how hard or easy it is to remove the proton. So it is reasonable to assume that the activation energy for the E1cb process doesn't change as we vary our leaving group.  If we use a better leaving group (make our leaving group more stable) that means that we have made ejecting the leaving group a lower energy pathway and the E1 process will become more favorable relative to the other elimination mechanisms.  If we change our leaving group to one that is an extremely poor leaving group (make our leaving group less stable), then ejecting it becomes a higher energy process and the E1 reaction becomes less competitive with the other reaction pathways. Saying this last sentence differently, when we use a poor leaving group we raise the activation energy for the E1 mechanism.  Since the rate of the E1cb process is not affected by the leaving group, its rate remains unchanged.  Consequently, a poor leaving group will disfavor the E1 process making the E1cb process more competitive. If the leaving group is bad enough, we can disfavor the E1 process so much that we wind up pushing our reaction all the way over to the E1cb side.

Your answer cleared my confusion. Thanks a lot!!

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


text:
Finding specific activity of an enzyme from Km values for a reversible enzyme reaction
I have an enzymatic equation in the form:
$$E + S\xleftrightarrow[k_{-1} = 2.6\text{ x }10^{-2} \text{s}^{-1}]{k_1 = 3 \text{ x }10^8 \text{M}^{-1}\text{s}^{-1}} EX \xleftrightarrow[k_{-2} = 3.8\text{ x }10^{-8} \text{M}^{-1}\text{s}^{-1}]{k_2 = 1.45 \text{ x }10^3 \text{s}^{-1}} E + P$$
For the forward reaction, I have calcluated the $K_m$ value to be:
$${K_m}_f = \frac{k_{-1}+k_2}{k_1} = 4.83\text{ x }10^{-6}\text{M}$$
And for the reverse, I calculated $K_m$ to be:
$${K_m}_r = \frac{k_{-1}+k_2}{k_{-2}} = 3.82\text{ x }10^{-6}\text{M}$$
From this info, I need to somehow calculate the specific activity of the enzyme in both the forward and reverse reactions.
The weight of one enzyme unit is 50,000 kDa.
Specific activity is defined as the number of 'Units' per mg of protein.
One enzyme unit is defined as the amount of enzyme that catalyzes the formation of 1 μmol of product per min under optimal assay conditions, which are assumed for this question (saturation of substrate in forward reaction / saturation of product in reverse reaction).
Can anyone point me in the right direction? I have a feeling that this is related to the $V_{max}$ term in the Michaelis-Menten equation:
$$v = \frac{V_{max}\text{[S]}}{K_m + \text{[S]}}$$
In that $K_m = \text{[S] at}\frac{1}{2}V_{max}$.
So, $$[S]_{f} \text{ at } V_{max} = 2(4.83\text{ x }10^{-6}\text{M}) = 9.66\text{ x }10^{-6}\text{M}$$
So, $$[S]_{r} \text{ at } V_{max} = 2(3.82\text{ x }10^{-6}\text{M}) = 7.64\text{ x }10^{-6}\text{M}$$
but I dont know how to relate it to specific activity.

v only equals Vmax in the limit of infinite substrate concentration.

@DavePHD I suppose that makes my substrate concentration calculations kind of pointless, but do you have any other ideas for what I can do to get these values?

The Michaelis-Menten equation that you have is derived assuming there is no reverse reaction (from product to the enzyme-substate complex).  See http://sys-bio.org/wp-content/uploads/downloads/2012/03/CB_Chapter3.pdf for the corresponding equation considering reversiblity.  Also, I never heard of an assay with saturated product.  The way you say "saturation of substrate in forward reaction / saturation of product in reverse reaction" sounds like an equilibrium situation, where there is no net formation of product.

@DavePHD This is part of an assignment question. The exact question is: "If the subunit molecular weight of the enzyme is 50,000 kDa, calculate the specific activity if: a) there are conditions of saturating substrate, and b) if there are conditions of saturating product.

If the enzyme is staturated with substrate, the reaction rate will be $k_2$.
1450 moles per second, per mole of enzyme.
87,000 moles per minute, per mole of enzyme.
87,000,000,000 micromoles per minute, per mole of enzyme.
Based upon "50,000 kDa" (which looks like a typo, 50 kDa sounds more reasonable), one mole of enzyme is 50,000,000 grams or 50,000,000,000 milligrams.
So 87/50 = 1.74 units per milligram.
Then for when product is staturated, use $k_{-1}$ instead of $k_2$.

Thank you so much. If you wouldn't mind, could you explain some of your thought process behind your solution? This one really stumped me.

well, if you like math the reasoning would be consider equation 3.2 http://sys-bio.org/wp-content/uploads/downloads/2012/03/CB_Chapter3.pdf in the limit of infinite substrate, and it becomes k2 times the enzyme concentation (Et).  If you don't like math the reasoning would be if the enzyme is staturated with substrate, only what happens to the enzyme-substrate complex affects the rate.


text:
Radon gas in earthquake prediction - Why Rn?
I recently studied an article about predicting earthquakes and how correctly realizing that an increase in $\ce{Rn}$ gas is a sign for the earthquake saved a whole city. The following is derived from the wikipedia page for earthquake prediction:

There are reports of spikes in the concentrations of such gases prior to a major earthquake; this has been attributed to release due to pre-seismic stress or fracturing of the rock. One of these gases is radon, produced by radioactive decay of the trace amounts of uranium present in most rock.
Radon is useful as a potential earthquake predictor because being radioactive it is easily detected, and its short half-life (3.8 days) makes it sensitive to short-term fluctuations. A 2009 review found 125 reports of changes in radon emissions prior to 86 earthquakes since 1966.

However, wikipedia then attempted to prove that most of the assretions related to radon are false statements,
but all this leaves me with one question: Why radon? (Why isn't another chemical species used for identifying the danger of feasible earthquake?) Maybe things I studied were too technical for me to understand, so please be as simple as you can be.
(Feel free to edit tags)

I wrote an answer with references for a similar question on Earth Sciences. (Had to delete a similar answer here).

Uranium is found in low-levels in all rocks and soil.  Radon is a gaseous radioactive decay product of uranium.  As the uranium undergoes radioactive decay, radioactive radon is generated and trapped in the rocks that contain the uranium.
The earthquake theory involving radon suggests that prior to the actual quake, there is some subterranean movement where rocks are crushed, soil is uncompacted and the trapped radon is released producing a pre-quake spike in radon concentration.
So radon presents the following attributes:

it is present in all rocks and soil
if a rock is broken or if soil is disturbed, radon will be released
it is gaseous and air currents / thermal gradients will carry it up to the earth's surface producing a detectable plume
it is radioactive, this makes detecting small amounts or small changes in radon concentration relatively routine due to well-developed and very sensitive methods for detecting and accurately measuring radioactivity.
radon has a very short radioactive half-life, a bit under 4 days.  Being a gas and having a short half-life is very useful in terms of measuring radon emissions.  If a radon emission spike occurs, the gas will dissipate quickly and after about 10 half-lives (40 days) normal background levels of radioactivity will return.

Another important characteristic is that radon is a noble gas, so it won't react chemically with anything on the way to the surface.

@thomij  Excellent point!


text:
Is a hydrogen bond considered to be a van der Waals force?
Is a hydrogen bond considered to be a Van der Waals force?

According to the IUPAC gold book a van der Waals force is:

The attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The term includes: dipole–dipole, dipole-induced dipole and London (instantaneous induced dipole-induced dipole) forces.

Hydrogen bonding is a type of dipole-dipole interaction, so it would fit the definition of a van der Waals force.
The way I think of it is: van der Waals forces are anything that make a gas non-ideal, since that's how they were originally discovered and defined.

But doesn't hydrogen bonding contain a rather sizable covalent component which would mean it doesn't fall under the IUPAC definition?

@Phillip - Sure, you could argue that. I guess it would depend on if you consider the covalent nature or the ionic nature more significant. Traditionally, it is classified as a particularly strong dipole-dipole interaction, which makes sense because we normally talk about van der Waals forces in the context of gases.

The provisional IUPAC recommendation for the definition of hydrogen bond requires "evidence of bond formation" and refers to "partial covalent bond formation".  This would seem to exclude the "other than those due to bond formation" requirement of the van der Waals definition. http://media.iupac.org/reports/provisional/abstract11/arunan_prs.pdf

I agree with @DavePhD and the source he cited (maybe make it an answer for a more permanent record?). The goldbook however recommends, to regard an Hydrogen bond as an "electrostatic interaction", which would make it fall under the mentioned vdW description. This is one of those topics, where a pure classification is not possible - in some cases the covalent character would outweigh the ionic and in some cases it would be the other way around.

@Martin, that's why I look at it in the context of gases, which is really the only context in which it makes sense to talk about van der Waals forces as a group. In any other context, the different interactions are treated separately.

I would not consider bifluoride, ammonium, and hydronium ions to be formed through dipole interactions.

There are two definitions of a hydrogen bond in the Gold Book: hydrogen bond (1994) and hydrogen bond (1999). The later stresses both the covalent bond nature and the electrostatic interaction component, which both can make it fall outside of the van der Waals forces as of the 1994 definition.
EDIT: Answer edited following the remarks made by @tomij below. See also the interesting discussion below his answer as to the context (historical and experimental) leading to the introduction of the notion of van der Waals forces/interactions.

The later definition is not saying hydrogen bonds are covalent, it is saying that a covalent bond between H and an electronegative atom results in electrostatic interactions between the H and a different electronegative atom. Also, the van der Waals definition is excluding charge interactions between ions, but H and the other bonding atom are not ions. It also says "The term is sometimes used loosely for the totality of nonspecific attractive or repulsive intermolecular forces," which would include hydrogen bonding. It also mentions dipole-dipole, which includes hydrogen bonding.

@thomij In the latter definition, it says "forms an additional weaker bond" and then "[the] energy of hydrogen bonds [...] results from the electrostatic interaction and also from the orbital interaction [between] the hydrogen donor [and] the hydrogen acceptor molecule." Isn't orbital interaction a feature of covalent bonds? Also, hydrogen bonds then form dipolar bonds and result from coordination which also involves covalent bonding... In acides/bases isn't the link at least partially covalent?

It is true that hydrogen bonds are partially covalent, which is why they are distinguished from other dipole-dipole interactions. However, they are also not fully covalent - which is why they are distinguished from fully covalent bonds as well. Historically van der Waals forces were any force that accounted for non-ideal gas behavior of real gases. I think it's fair to say that hydrogen bonding is one of those forces. I also think it's fair to say that because the other forces have little to no covalent character, hydrogen bonding should be thought of as different from them.

@thomij Indeed, the historical perspective is quite enlightening. I've made a limited bibliographic survey of recent publications and although some authors argue that HBs and vdW interactions are distinct, either their argument is quantitative, thus making the distinction arbitrary, or they assign a priori attributes to vdW interactions chosen so as to set them apart from HBs, but I doubt that those attributes were stated by van der Waals himself (I need to read his original work still). But so now I tend to think that HBs should actually be considered to belong to the set of vdW interactions.

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


text:
Relationship between surface area of electrode and reaction rate of hydrogen in fuel cells
I am looking at the effects of increasing the surface area of which the flowing hydrogen fuel is in contact with the anode in a fuel cell.
At the anode the hydrogen is catalytically split into protons and electrons. The catalyst used is most commonly platinum. So, for a fixed flow rate of hydrogen, increasing the contact surface area with the anode will increase the number of hydrogen atoms in contact with the catalyst, and the duration that they are in contact with it. Therefore the number of oxidization half cell reactions will increase. But what is the mathematical relationship that governs the number of reactions that will occur? Is it just a linear relationship, i.e doubling the surface area doubles the number of reactions, or is it more complex?
If it is impossible to get an exact mathematical relationship, what is a good approximation on how these variables will scale with each other? Also, how do chemist usually predict the number of reactions/reaction rate, and could this be applied to my example?

Any galvanic cell produces its equilibrium voltage at open-circuit.  Let's say your cell is at +3 V before you allow any current to flow through it.  As soon as you throw the switch, you allow that 3 V to push a current through the cell and some external load that you connect to it. There are several resistances within the cell that act to reduce the cell voltage from the initial 3 V.  The electrolyte resistance, concentration dependent resistances at the electrodes and electron transfer resistances all contribute to the total voltage drop in the cell, and hence, effectively limit the amount of current it can generate.  Typically, one of these resistances is much larger than any of the others depending on the exact configuration and operating conditions of the cell.  It is the electron transfer resistance that you are concerned with in your question.  The traditional way of expressing the voltage drop due to electron transfer resistance is the Tafel equation:
$$\eta _{a}=a+b\, log\, i$$
where $a$ and $b$ are empirical constants, $i$ is the current density and $\eta _{a}$ is what is called the activation overpotential.  We see from this expression that the amount of voltage drop in the cell due to electron transfer increases linearly with the logarithm of the current density.  It is desirable then to minimize $\eta _{a}$ for a battery or flow cell application so that we have the maximum amount of voltage available for pushing current through the load.  A very good way to do this, as you pointed out, is to simply increase the electrode area.  By doing so, we decrease the current density which decreases $\eta _{a}$.
So, the exact mathematical relationship you seek would be the empirical Tafel expression given above, or even better, the Butler-Volmer equation which is really just a more fundamental way of expressing the Tafel relationship.  All you should need to do is locate some references on platinum polarization in aqueous solutions.  There just happen to be tons of papers on this subject.  Here is one to get you started:

Sheng W. et al, Journal of The Electrochemical Society, 157 11
  B1529-B1536 2010

This is a fantastic answer, Its the exact information I need! Thank you very much! I'm pretty certain I understand, but I might have some follow up questions in the morning once I fully wake up :)

I have one question, I should probably know this, but it confuses me slightly: If the activation overpotential is proportional to the log of the current density, that means for a cell delivering a constant current the activation overpotential is proportional to log(1/A). Since power is current * voltage, that means for a cell delivering constant current the power the cell is capable of producing is also proportional to log(1/A). Is this correct?

I guess what I'm asking is: Is the power of the cell proportional to the activation overpotential?

The power output depends on both the internal resistance of the cell and the resistance of the load it is connected to.  Play around with this widget to get a feel for how it works: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/powtran.html  . If you know the activation overpotential as a function of current density, you could calculate an estimate of the cell resistance over a range of current.  Plug it into the widget and see what you come up with!

Thankyou. One last question. This document, page 5, says the theoretical max voltage of a cell is $E_0 = 1.229V$. Does this mean the actual volatge of the cell is $E = E_0 - \eta_a - \eta_e - \eta_c$ where $\eta_e$ and $\eta_c$ are the  electrolyte resistance and concentration dependent resistance?

Those are overpotentials rather than resistances, but yes, you are correct in their designations. The electrolyte and concentration overpotentials are typically much more difficult to estimate than the activation overpotential, as they are highly dependent on the cell geometry and mass transport conditions. Without some kind of computational modeling, you'll just have to neglect these two overpotentials.


text:
Synthesis of 2-Phenyl-2-Butanol from Grignard Reagent
The question is this:
2-Phenyl-2-Butanol can be synthesized by three different combinations of a grignard reagent and a ketone. Show each combination.
I can only think of one.

Butanone with Br-Mg-Ph

I can't think of anything else.
I'm not too sure if this will work:
phenylbutanone withe Br-Mg-Me
now i'm stuck. i can't think of the last one.

Is a variation on the Grignard reagent acceptable, such as Ph-Mg-I? BTW, putting grignard "2-Phenyl-2-Butanol" synthesis into Google found many good references.

Yeah i figured. but i'm curious if all any variation would work. Like changing Br to and I, will it affect the reaction in anyway? that's what i'm unclear about as well. that's why i tried to stick with Br.

Hint: The carbon atom bearing the OH group in the product has three substituents--phenyl, methyl, and ethyl. Any one of these groups could be introduced as a Grignard reagent, with the appropriate ketone. You have already correctly deduced one of the three possibilities.


text:
Is it possible to separate aluminum hydroxide?
I want to know a way to separate aluminum hydroxide into aluminum and hydroxide ions, with a chemical reaction, or another process if necessary.
I have $\ce{Al(OH)3}$ and I want to separate it into $\ce{Al}$ and $\ce{{}^{-}OH}$.

What do you mean by separate? Do you want to isolate the ions on their own or simply break apart the lattice in the solid?

There are countless possibilities in what you could have meant: 1: You're seeking homework help and asking for the net ionic reaction. 2: You're looking for an easy "home" experiment, with consideration to (1.either safety 2. or possibility 3. or economical applicability). 3: You're curious about how they perform such an "action" in industrial chemistry. 4: You want info in the field of geochemistry, looking for a possible reaction that happens in a possible mineral. etc. etc. etc. And this is only when I understand that you mean break apart the lattice. Unclear what you're asking, really.

I have Al(OH)₃. I want to separate the following into Al and OH-

High school student here, I immediately thought of more active metals in the activity series table. Is it possible to have the aluminum hydroxide react with a more active metal to replace the aluminum and yield elemental aluminum and the substitute metal hydroxide?

It isn't really possible to have isolated hydroxide (-OH), because it is charged.  Only neutral compounds can be isolated, not ions.


text:
How to find the alcohol content of a probiotic liquid?
Probiotics are microorganisms that are believed to provide health benefits when consumed. Water Kefir is one such probiotic drink. This has blend of bacteria, yeast (BY), sugar water and the result of action of BY on sugar water. There is alcohol. One of my problem is to find out the alcohol content in the solution. Less expensive, simpler methods/protocols are much appreciated.

Complex instrumental analytics (HPLC, photometry using a $\ce{NAD+/\,NADH}$ testing kit, etc.) is out of question.
In principle, the most simple approach would be the use of a hydrometer to measure the density before and after fermentation.


text:
How are nuclei stable?
We all know that the density of the nucleus is very high.
Nuclei are made up of protons and neutrons, and while protons have the same charge, they are closely packed in a nucleus. How does the repulsion between protons not break apart the nucleus?

Since the gravitational force between two protons is negligible there must be another force holding the nucleus together. This is the strong nuclear force, which as the name suggests is extremely strong but it is also extremely short range and so it's effects are only felt on the scale of nuclei and baryons. As you can see in the graph, if two protons approach each other they repel each other due to electrostatic repulsion but once they get within about $\mathrm{3~fm}$ of each other the strong nuclear force begins to become significant and at about $\mathrm{2~fm}$ it starts to outweigh the electrostatic repulsion and bind the two protons together.
It is also interesting to note that at even shorter ranges the strong nuclear force is repulsive and so there is an equilibrium position, indicated on the graph, where the strong nuclear attraction and the electrostatic repulsion are equal in magnitude and so the net force on the protons is zero.
Note that the graph shown is for two protons and so in nuclei containing more than two protons the numbers will be different but the principle is the same.

The thing is that once the 2 protons come within 2 fm it starts to outweigh the electrostatic repulsion.. BUT what is responsible for the 2 protons to come near to each other by over coming electrostatic repulsions???

@vamsi two protons only overcome the electrostatic repulsion barrier in a high energy nuclear fusion process, for example in the sun or another star.  Even in the core of a star, this can only happen due to quantum tunneling through the barrier. http://hyperphysics.phy-astr.gsu.edu/hbase/astro/procyc.html

Then what about bon's answer??

what are you confused about?

@bon two protons overcome electrostatic repulsions only when they are at a distance less than or equal to 2fm.what is the source that is bringing 2 protons closer to each other so that there will be nuclear force between them?How this is happening in all nuclie?

@vamsi nuclei are formed in the cores of stars where the temperature is high enough that the protons can be forced close enough together to be bound by the strong nuclear force as DavePhD said. once a nucleus has formed it is stable because a large amount of energy is required to overcome the strong nuclear force and separate the two protons again. (see the graph)

@bon then all the nuclei which are existing in the atoms of all living and non living things in this world are formed as a result of Nuclear reactions occurring in sun and stars?

@vamsi yes all nuclei except hydrogen and some helium formed in the cores of stars. hydrogen and most helium formed shortly after the big bang. also a few nuclei have been transformed from one type to another through nuclear decays and cosmic ray interactions

Protons and neutrons in a nucleus are constantly emitting and absorbing
little particles.   When one nucleon emits a little particle called a "meson" that another nucleon absorbs, a strong force between the two nucleons results. This is called (strangely enough) the strong nuclear force1, and it is strong enough to counteract the powerful electrostatic repulsion of the protons in a nucleus.
Note: The modern explanation is richer and more complex than this. Since the 70's these "mesons" have been understood to be made up of smaller particles called quarks and gluons, which are also the building blocks of nucleons. In fact the strong force doesn't just bind the nucleus together; it holds the quarks in protons and neutrons together, too.  The contemporary explanation of the strong interaction is in the realm of quantum chromodynamics, and really beyond the scope of chemistry. You can read more about it here.
Production and destruction of the messenger mesons violates the law of conservation
of mass and energy! However, if the messenger particle has a very short
lifetime, and so exists only for a very short time within a very small space,
the particle can exist within the limitations set by the uncertainty principle.
Particles like this are called virtual particles.
Nuclei are very small (on the order of femtometers in radius), so the range of
the strong nuclear force must be very small. You can make a back-of-the-envelope estimate of the range as follows.
The uncertainty principle says that you can’t exactly determine the position
and momentum of a very small particle simultaneously:
$$\sigma_x \, \sigma_p\ge \frac{\hbar}{2}$$
where $\sigma_x$ and $\sigma_p$ are the uncertainties in the position and momentum of a particle, and $\hbar$ is the reduced Planck constant.  Given that the mass
of the meson that mediates the strong force is $m \approx 2.4\times 10^{−28}$ kg, and the uncertainty in the velocity can’t be any larger than the speed of light $c$, you can compute $m c$ as a bound on $\sigma_p$.
You can then estimate $\sigma_x$, which should be related to the maximum distance that a meson can exist from the nucleon that generated it without violating
the uncertainty principle, or the universal speed limit:
$$\sigma_x \approx \frac{\hbar}{2 m c} = 0.7\ \text{fm}$$
That’s the right order of magnitude for the range of the strong nuclear force; it's also about the distance between nucleons in the nucleus, and about 6/10ths of the classical radius of a proton.

1. There is a different nuclear force called the weak nuclear force which plays a role in fission and radioactive decay, so calling it just "nuclear force" might be a little ambiguous.

Coulomb's electrostatic repulsion between two protons is extremely large as compared to the gravitational force of attraction between them. Therefore, as you understand correctly, if Coulomb's repulsive and gravitational attractive forces are the only forces operating inside the nucleus, it cannot be stable. The stability of nucleus has been attributed to the existence of a third type of force inside the nucleus called Nuclear force.

Reference: Modern's abc of Physics by Satish K.Gupta, Part 2, Class 12.


text:
Which Basis Set is suitable For Mercury-complex in DFT calculations?
Which Basis Set is suitable For Mercury-complex in DFT calculations? Please provide the answers with any journal reference.

What research have you done on things far?

you should provide 1)package you are using 2)properties of interest 3)other atoms 4) size of the complex (ideally the structure too) 5) machine you are running it on.

Gaussian 09, optimization of mercury complex contains nearly 40 atoms..........cluster computer

Welcome to chemistry.se! Learn more about this site, taking the [tour] and visiting the [help]. Could you [edit] in an example of what you want to do. Is it a cluster of 40 Hg atoms or is it a cluster of 40 atoms containing one Hg? What are the other atoms? Does it have a charge? What DFT method would you like to do? What accuracy is necessary?


text:
Saytzev and Hofmann elimination in E1
From what I have read, E1 largely produces the more substituted alkene because of the fact that more substituted alkenes are more stable as a result of hyperconjugation. However, in E2 the situation is different for a number or reasons. One of these reasons is steric effects of potentially large substituents on the $\beta$- carbon which would promote formation of the less substituted alkene. This effect does not seem to be mentioned for E1 but surely it must still be valid, right? Do large subtituents or bulky bases promote the formation of less substituted alkenes in E1? If not why not?

In the E2 reaction, the size of the base can affect the selectivity of the deprotonation, because the steric interaction of a large base with a hindered proton will go over a higher transition state than the interaction of a large base with a less-hindered proton. This is kinetic control.
I'm not aware of any practical case where the selectivity of an E1 reaction can be controlled by the size of the base. The reason is that the deprotonation of the carbocation intermediate will have a transition state that is very close in energy to the carbocation itself. I agree with you that the size of the base should have some influence on the transition state energy, but in practice, it seems that it has less influence than the stability of the resulting alkene.
Also note that often E1 is run under equilibrating conditions, where even if a less-substituted (less-stable) alkene is formed, it can be protonated to reform the carbocation, and the material will funnel to the most stable product. This is thermodynamic control.

Addtionally, the for the E1 mechanism, the first step (formation of the carbocation) is the slower step. More importantly, the carbocation is planar at the cationic carbon and thus has a very different geometry leading to different steric considerations.


text:
Why does a matchstick burning in a container cause suction?
As you see in the picture, a burning match in a container causes suction and this is a traditional therapeutic method to relieve bone pain.

What is the reason? I think it should be related to gas changes ($\ce{O2}$ being used for combustion and etc.), but I am not sure about it.

When the container is open to the atmosphere and has the match burning in it, the temperature of the air inside is higher than the temperature of the surrounding air.  In order for pressure to be equal in the container and outside it, the density of the air in the container is less.  When the match is removed and the container placed on the person, two things happen.  One, no air can enter or leave the container, and two, the air in the container begins to cool.  When the air in the container cools its pressure drops.  The atmosphere pushes on the air in the container indirectly through the person, creating what is called suction.
This is similar to drinking through a straw.  You are not sucking the liquid through the straw as much as you are removing the opposing force that prevents the atmosphere from forcing the liquid through the straw.  The opposing force being the air pressure in the straw.

I'm familiar with this user from biology.SE, and for sure he's gonna be asking you for reference! Maybe leaving a link or two wouldn't be such a bad idea! :D

@brinnb MARamezani is right and I won't accept it without a reference.

Beat me to it, +1.


text:
What does the prefix "dihydro" in the systematic name of luminol refer to?
I'm doing a research project on the fascinating chemiluminescent molecule luminol, and figured it would be a good idea to start breaking down the systematic name of the chemical and linking the terms to the structure. The systematic name of luminol is 5-amino-2,3-dihydrophthalazine-1,4-dione, and I get it all apart from the "dihydro" - any help? Thank you very much in advance! :)

On the left is 5-amino-2,3-dihydrophthalazine-1,4-dione and on the right is the molecule which the name is based on, phthalazine. The 2,3-dihydro refers to the fact that at the 2 and 3 positions (the two nitrogens) hydrogens have been added (hydro is the prefix for hydrogen).

Fantastic, thank you! I suspected this but didn't want to go ahead without checking. Also, for brownie points: is there a techincal name for the molecule that a molecule is based on - i.e. an OH group is called a functional group, so in the case of the luminol the phthalazine would be called its...? This word may not exist, just curious!

@AndrewCatherall I think they are just called base names

@AndrewCatherall Late to the party on this one, but in my experience luminol would be said to be derived from a phthalazine core.


text:
What is the white substance left behind after boiling down water
I was boiling down water in a stainless steel pot as an experiment the other day. When the water was completely vaporised, a white substance was left behind on the bottom of the pot. I tried rubbing it off the bottom, but it would not come off. I'm unsure what this substance is. What is this substance?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

I had the same problem. Just added enough white vinegar to cover the bottom of pot and let it sit for about 5 min. Then I added some Ajax type kitchen pot cleaner and it all came out. The cleaner alone did not work. The vinegar did the trick.

These are minerals that are naturally present in "tap" water. Chemically, they are most likely a mixture of calcium carbonate and magnesium carbonate, both of which are only very lightly soluble in water.
As you boil away the water, these dissolved calcium/magnesium carbonates remain behind and their concentrations eventually become greater than their respective solubility limits. At this point they begin to precipitate out of the solution, forming the 'scale' you see on the pot.  Interestingly, while most salts become more soluble at higher temperatures, calcium carbonate is anomalous in that its solubility decreases as temperature increases (see the accepted answer here), which accelerates the precipitation process.
The easiest way to get rid of them is to add a cup of white vinegar and let it sit in the pot. You might need to agitate the scale from time to time. You can boil the vinegar in the pot if you want, but it will stink! Over time, the vinegar will dissolve the precipitate, and you will see carbon dioxide bubbles forming. You can then pour out the vinegar and flush the pot with lots of water.

Or you can use $\ce{HCl}$. Oh no no no, wait. Perchloric it is! :D


text:
Is Acrylamide carcinogenic? Why?
Recently concerned with health effects by common chemicals existent in food I've been rather busy reading article after article; and just an interesting one came around:

Acrylamide (or acrylic amide) is a chemical compound with the chemical formula $\ce{C3H5NO}$. Wikipedia
Some evidence suggests exposure to large doses can cause damage to the male reproductive glands. Acrylamide is considered a potential occupational carcinogen by U.S. government agencies. Still wikipedia
A series of case-control studies have investigated the relationship between dietary intake of acrylamide and the risk of developing cancers of the oral cavity, pharynx, esophagus, larynx, large bowel, kidney, breast, and ovary. These studies generally found no excess of tumors associated with acrylamide intake. Cancer.gov

Well, though the references are generally controversial, you can conclude that though nothing is proven by the studies, take care:
FDA: States that acrylamide is a carcinogen of animals, is a human concern, but "should I worry when eating high-temperature fried or roasted food?" "NO."
Food standards agency: Lab proves acrylamide a carcinogen, but real life didn't give up many victims, though this isn't a new issue.
Along with researches done by FAO and WHO, still the carcinogenic concerns aren't very serious.
As a chemist, what reasons would you have provided that acrylamide is carcinogenic or not? Should I really worry about it when I eat french fries?

Acrylamide is a Michael acceptor. All Michael acceptors are potentially carcinogenic because DNA can act as a Michael donor. This Michael reaction can damage the DNA, which can ultimately develop into cancer. Of course how well a certain Michael acceptor reacts with DNA depends on its chemical structure (i.e. the carbon chain and other function groups as well as its conformation), and hence not every Michael acceptor is equally carcinogenic.
If you worry about eating too much French fries I would first be concerned with my caloric  and sodium intake. For that there is solid evidence that it greatly reduces human health if taken in excess.


text:
Find amount of acid and base required to make buffer
I am wondering if it is possible to know exact amount of acid and base that we can put in together to make a buffer.
I want to make a 0.05 M extraction buffer using disodium hydrogen phosphate (DHP) and sodium dihydrogen phosphate (SDP). I want to make 100 mL of buffer so I worked out that I would require 0.78 g SDP and 0.88 g DHP.
The issue is that it sometimes takes a lot of time by mixing one solution into the other and keep it checking with the pH meter, there has to be some mathematical way to get the exact amounts which would save time and my chemicals.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

Usually when you are preparing a buffer, you can about the pH and the strength (total concentration). Let's say you want to combine acid $\ce{A}$ and its conjugate base $\ce{B}$ to make a buffer with total concentration $C=[\ce{A}]+[\ce{B}]$ and a set $\text{pH}$.
The Henderson-Hasselbalch equation provides a relationship between the ratio of the two concentrations to the pH:
$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{[\ce{B}]}{[\ce{A}]}\right)$$
Because $[\ce{B}]=C-[\ce{A}]$, we can write, and solve for $[\ce{A}]$ (and then for $[\ce{B}]$).
$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)$$
$$\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)= \ce{pH}-\ce{p}K_a$$
$$\dfrac{C-[\ce{A}]}{[\ce{A}]}=10^{\ce{pH}-\ce{p}K_a}$$
$$C-[\ce{A}]=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]$$
$$C=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]+[\ce{A}]=[\ce{A}]\left(1+10^{\ce{pH}-\ce{p}K_a}\right)$$
$$[\ce{A}]=\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}$$
$$[\ce{B}]=C-\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}=\dfrac{C\left(10^{\ce{pH}-\ce{p}K_a}\right)}{1+10^{\ce{pH}-\ce{p}K_a}}$$
So... if you have a total concentration $C$ and a $\ce{pH}$ in mind for an acid-base pair with a certain $\ce{p}K_a$, just plug them in.

The above derivation of course assumes that $[\ce{A}]\ne0$, which would not make sense anyway.

The simplest and still an accurate way is to start with the acid form of the buffer component weighed out to the desired molarity and adjust the pH with 1N NaoH. If your desired pH is greater than the pKa, then you can expect that it would take quite a bit of NaOH. In this case you should start with 0.05M NaH2PO4. But what is your desired pH? When you talk about making a buffer, you should have information on molarity, volume, pH and pKa ready.

You can't weigh something to the desired molarity. You can measure a mass, weigh a force, and calculate a concentration (molarity). Also if you start with the desired concentration, and then add acid/base to adjust the pH, then you are diluting the buffer and end up at a much lower concentration than intended.


text:
offset in measuring decomposition potentials
as far as I've understood to measure the decomposition potential of an electrolyte using a galvanic cell we just put two electrodes, connect them to a battery, and vary the external potential till when we observe a current. The potential at which the current starts to grow linearly is the decomposition potential of the electrolyte. There is not appreciable current before the decompositon potential because the electrodes are supposed to be inert and do not provide the ionic charges needed to "close the circuit". Is it right?
An other question. How the role of the electrodes is taken into consideration in the experimental setup? Is it considered an offset due to an eventual nonzero cell potential? I made this question because I'd like to be sure that I'm understanding the experimental setup.
Tirrel

Ok I will answer your  last question first. If you have two identical electrodes(i.e. same electrode material), the cell potential will be zero and there won't be any issues there.
You have to use inert electrodes so that you are not measuring the dissolution of the electrode itself.
The one part I am not sure you understand properly is the current growing. You want to vary the potential until the current raises exponentially when continuing to sweep. This will be a very sharp rise that will be hard to miss.


text:
How are these compounds strong bases, when they are sparingly soluble in water?
How is it possible that some insoluble compounds in water, e.g. $\ce{Mg(OH)2}$, $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$ are considered strong bases?
Are they even strong bases? How can I tell if a compound is a strong base or not?
Is it possible to explain with equations and/or formulas what happens when these compounds are dissolved into water?

$\ce{Ca(OH)2}$ is not insoluble in water --> 0.173 g/100 mL (20 °C). I also do not quite understand your question.

@Jori s\he asks if there exists a general way of predicting if a compound will be a strong base in water. At least that's what I got from it.

The typical definition of strong base that you will see at an introductory level is that there is complete dissociation to hydroxide ions, for example that NaOH completely dissociates into Na+ and OH-.
This is a terrible definition because reactions never go to completion.  The definition contradicts the concept of chemical equilibrium.
In reality:
$\frac{[Na+(aq)][OH-(aq)]}{[NaOH(aq)]} = K_B = ~4M$ (from table discussed below)
So even dissolved aqueous NaOH is not all dissociated.  It can exist as ion pairs or molecules in solution without being dissociated.
$K_B$ is quantification of how strong or weak the base is.  Strong base could be defined as $K_B$ exceeding a certain value. Let us say $K_B>1$ and therefore $pK_B<0$
So the strength of the base only refers to the equilibrium between the undissoicated dissolved species and the dissociated ions.  It is independent of how much or how little solid dissolves. Some hydroxides, like NaOH and KOH have high solubilities and some have low solubilities, but $K_B$ is a different equilibrium.
For quantitative data on the $K_B$ values of hydroxides, see:
DISSOCIATION CONSTANTS OF INORGANIC ACIDS AND BASES IN AQUEOUS SOLUTION
Look up the corresponding metal ion alphabetically, and a $pK_B$ value is given.
However, keep in mind that for each of $\ce{Mg(OH)2}$, $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$ there will be two $K_B$ values, since there are two hydroxides that can dissociate.
The $pK_B$ values for the second hydroxide dissociating are Mg 2.6, Ca 1.4, Sr 0.8, Ba 0.6 becoming stronger bases as you go down the column in the periodic table, increasing radius stablizing the charge of the ion.  The first $pK_B$ would be even lower, but I don't see any quantitative values.  It would be so low that it would be difficult to measure.  So, yes these are strong bases.
As far as how to tell, other than experimentally, how strong a base of a hydroxide is, there is a good lecture The Acid-Base Character of Oxides and Hydroxides in Aqueous Solution by Pilkington.  As you go from left to right across the periodic table (excluding the noble gases), the hydroxides progress from being strong bases to being acids.

Can you please explain how, on moving from left to right across the periodic table, "the hydroxides progress from being strong bases to strong acids"?

@KaumudiHarikumar NaOH (strong base), Mg(OH)2 (base), B(OH)3 (weak acid), CO(OH)2 (weak acid), N(O)2(OH) (Ka =21), S(O)2(OH)2 (strong acid), Cl(O)3(OH) (even stronger acid)

OK, but you haven't explained the reason behind this trend...

@KaumudiHarikumar The bonds between oxygen and the X element become increasingly covalent as you go to the right on the periodic table.  In water, Na+ and OH- can easily dissolve as separate ions,  but a C-OH bond won't break just by adding a compound to water.

Ohh, but it was my understanding that ionic bonds are generally stronger than covalent bonds :/ Secondly, why do the bonds become increasingly covalent? Fajaan's rules?

@KaumudiHarikumar Oxygen and Chlorine have similar electronegativity, so going left on the period table from O or Cl the electronegativity difference becomes greater and greater.

Hm, OK. Are covalent bonds really stronger than ionic bonds?

I click the link and get a "Not found" error.  Please update, thanks.

@OscarLanzi try this: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.577.7419&rep=rep1&type=pdf


text:
Hydrogen bond comparison
How do you compare the strength of H bonds when it's the same atom on both sides of the H? Like $\ce{N-H\bond{...}N}$ vs $\ce{F-H\bond{...}F}$? Is one still stronger than the other, or are they both the same since in both cases the pull on both sides is equal?

In some cases, such as [F-H-F]- (biflouride), there is a symmetric hydrogen bond, with both H-F bonds equally strong.
However, just because the two atoms bonding to the H are the same doesn't mean it is a symmetric hydrogen bond.  For example water dimer does not have a symmetrical hydrogen bond, the H is closer to one O atom than the other.
There can be cases where the hydrogen being exactly in the middle is a potential energy minimum, and other cases where the middle is not a local minimum but there are two minima offset from the middle.
See Symmetry of hydrogen bonds in solution for more information.


text:
Is Saliva an acid or base?
I have a debate with some friends who believe that saliva is just basic, but with the enzymes which make them reactive with objects in our mouth, shouldn't it be an acid in this case?

I can count four reasons that say saliva pH isn't the same among humans:

As people age, the composition of saliva changes in their body; this is easily explainable with the fact that the body isn't performing the actions the same way as you grow up or become old.
People have different eating habits; this will put the body in impression of producing especial kinds of enzymes and reduce\increase the production of certain factors in the saliva.
People aren't the same when it gets to taking care of the insides of their mouth. You can be brushing your teeth twice as much as a close friend of yours does.
Human digestive system isn't always performing its duties consistently and without "ups" and "downs" during the day, which has an important effect on enzymes produced.

And a sea of other reasons....
But, if we are to answer this question:

Saliva is a watery substance located in the mouths of animals, secreted by the salivary glands. Human saliva is 99.5% water, while the other 0.5% consists of electrolytes, mucus, glycoproteins, enzymes, and antibacterial compounds such as secretory IgA and lysozyme. Saliva, wikipedia

I'll also say that according to here, a healthy range of pH of saliva varies between 6.5 and 7.5 during the day, but is optimally between 6.5 and 6.8.

Source
Thus, saliva is a very weak acid, or sometimes a very weak base. Considering something like that an acid or a base is like saying that the water we drink is a base and the water of the raindrops that fall from the sky of a peaceful country is acidic.
Saliva is really important for it's being a buffer.

The pH value of resting whole saliva is slightly acidic, which varies between pH 5.75 and 7.05, and it increases with increasing flow rate up to pH 8. Besides flow rate, the pH also depends on the concentration of salivary proteins, bicarbonate and phosphate ions that have considerable buffering capacity. Bicarbonate concentration is ∼5–10 mM/L in resting conditions, and it may increase up to 40–60 mM/L with stimulation, whereas phosphate concentration is ∼4-5 mM/L in saliva rather independently from the flow rate. Besides bi-carbonate and phosphate, a significant amount of other ions are present to maintain the slightly hypotonic osmolarity of saliva. The most important ions are sodium (1–5 mM/L resting; 100mM/L stimulated), chloride (5 mM/L resting; 70 mM/L stimulated), potassium (15mM/L resting; 30–40 mM/L stimulated), and calcium (1.0 mM/L resting; 3–4 mM/L stimulated).Many other ions such as ammonium, bromide, copper,fluoride,iodide, lithium, magnesium, nitrate, perchlorate, thiocyanate, and so on can be found in the whole saliva in lower concentrations.
And appropriate ph is set up for optimum activity of enzyme especially during digestion process.


text:
Can we do combustion caused by limited supply of air in a lab? How?
For example Ethyne in combustion by limited supply of air

$$     \ce{ 2C2H2 + 3O2 ->[\Delta] 2CO2 + 2H2O + 2C } $$ vs.
$$     \ce{ 2C2H2 + 5O2 -> 4CO2 + 2H2O + Energy } $$

What are the limits of air availability to make these combustions?
How to do these experiments in lab?

Hmm, I see a couple of problems in here:
Energy in the reaction is misunderstood:
In general, breaking bonds is endothermic, and making them is exothermic. In any combustion reaction we have bonds broken ($\ce{O2}$'s for example) and bonds created (forming $\ce{H2O}$, in this instance). The overall change in the energy defines a reaction to be endo- or exothermic; and in the case of combustion (any type) it's exothermic. If you want to write "heat" and "energy" in both sides of the reactions you face, it's going to look a little awkward. Note that in the biochemical reactions in which you see terms like "energy" and such in only the side of reactants or the products (e.g.: ATP energy releases) it's because they want to put emphasis on that particular intake or release of energy.
The equation you've provided for incomplete combustion is incomplete:
I don't see any balanced equations in there, do you? And, I can't get the reason for why you've written two $\ce{CO2}$s.
However, your second equation is about the complete combustion of ethyne, and is balanced and correct.
As for your answer, take a look at what wikipedia says:

The incomplete (partial) combustion of a hydrocarbon with oxygen produces a gas mixture containing mainly $\ce{CO2, CO, H2O}$, and $\ce{H2}$.

Depending on how much your burning sample will be out of reach of air, the mixture will vary in weight percent of the products of the reaction.
In a lab, you normally will get incomplete combustion reactions, for reasons:

Your sample is hardly pure hydrocarbon.
The air in the lab isn't in wild currents and overall, is "stale" in simple words.
The hydrocarbon doesn't get enough "requirements" for combustion and ignites incompletely.
etc.

Source

Source
A good sign (or alarm) of the incomplete combustion is the orange hue of fire, especially if it's accompanied by a visible "smoke". If the experiment requires complete combustion, we'll make use of burners such as Bunsen burner.
So, let's summarize: The incomplete combustion is the normal thing happening in the burning of ethyne. You usually do some modifications if you're to have a complete combustion reaction. When the examiner claims "with contact with adequate supplies of air", s\he only means that a complete combustion takes place, though this is rare in real life. Plus, ethyne is not a very good hydrocarbon to start with, because as CDC mentions, it has a potentially negative effect on the nervous system, causing headaches, asphyxia etc., as well as painful frostbites.

Also, avoidance of ethyne is recommended. It is a nasty gas.

Great point, @permeakra!


text:
Hydrazine as a reduction agent
When using hydrazine as a reduction agent to reduce graphene oxide, which is the product hydrazine oxidises to?

I'm sorry for the wrong answer initially given; you might consider to unaccept this one.
I'm not sure whether this has ever been examined in detail. It is likely that nobody bothered to collect the (volatile) reaction products of hydrazine.
According to a procedure provided by the group of Rodney Ruoff (DOI), the reduction was performed in an open flask equipped with a reflux condenser. An aqueous suspension GO was first sonicated (to support exfoliation) until it became clear, then hydrazine hydrate was added and the mixture refluxed at 100 °C for 24 hours. During this time, the reduction product precipitated as a black solid.
No remarks on what actually happened to the hydrazine were given in this article.

If hydrazine is acting as a reducing agent, then it should be oxidized during the reaction.

@jerepierre You're absolutely right. I was thinking in oxidative degradation of carboxylates, which has nothing to do with the question asked.

The decomposition reaction of hydrazine ($\ce{N2H4}$) is as below:
$$\ce{N2H4 \rightarrow N2H2 + H2}$$
The forming $\ce{H2}$ will reduce the graphene oxide and hydrazine itself oxidises to $\ce{N2H2}$.

So, why not use H2 in the first place?

To avoid contaminating the graphene with metals I suspect.


text:
Why neutrons are neutral?
If electrons have a negative charge and protons have a positive charge
then how come neutrons have zero charge without consisting of protons and electrons?

Nice question. I'm not sure if it's physics or chemistry, though. Wikipedia says that the fact that neutron has no charge is proven by experiments and is empirical, no proofs or explanations.

Neutrons, are neutral but they are composed of quarks that have partial charges that balance

this question is off-topic, it belongs to physics

I'm voting to close this question as off-topic because it's not really about chemistry and would be better suited for Physics.SE.

Protons and neutrons are both baryons, subatomic particles composed of three quarks. Quarks are smaller fermionic particles that make up protons, neutrons, and other hadrons.
Electrons are leptons, fermions that are not composed of quarks.
Quarks have mass, spin, charge, and (confusingly) color charge. Color charge is a quantum property that quarks have, and all hadrons need to have a net zero color charge.
The proton is composed of three quarks: two up quarks and one down quark. Each up quark has a spin of $+\frac{1}{2}$, a mass of $2.3~\text{MeV/c}^2$, and a charge of $+\frac{2}{3}$. Each down quark has a spin of $-\frac{1}{2}$, a mass of $4.8~\text{MeV/c}^2$, and a charge of $-\frac{1}{3}$. The total charge of the proton is $2(+\frac{2}{3})+(-\frac{1}{3})=+1$. The total spin of the proton is $2(+\frac{1}{2})+(-\frac{1}{2})=+\frac{1}{2}$. The masses of the quarks do not add up because most of the proton's mass is in its confinement energy.
We can do the same analysis for a neutron, which is composed of one up quark and two down quarks. The charge of the neutron is $(+\frac{2}{3})+2(-\frac{1}{3})=0$. The total spin of a neutron is $(\frac{1}{2})+2(-\frac{1}{2})=-\frac{1}{2}$. Again the masses do not add up.
Protons and neutrons are different because the are composed of different quarks. However... enough physics. Now for the nuclear chemistry.

Interconversion of protons, neutrons, and electrons

The proton is a stable particle with a lifetime at least $2.1\times 10^{29}$ years.
However, protons can be converted into neutrons (and electron neutrinos) by electron capture:
$$\ce{p+ + e- -> n +\nu}_e$$
This requires an investment of energy because the rest mass of the neutron is higher than the rest mass of a proton by $1.29~\text{MeV/c}^2$.
Free neutrons are not stable, undergoing beta decay to a proton, electron, and electron antineutrino with a halflife of about 10 minutes.
$$\ce{n -> p+ + e- +\bar{\nu}}_e$$
Most importantly, and fortunately, most neutrons bound in nuclei are stable. The protons and neutrons in a nucleus can be treated quantum mechanically analogously to electrons from several atoms bound in a molecule. There are not a collection of independent particles that do not interact. The nucleons occupy quantum mechanical energy levels. Neutrons can only decay if the proton it would become can occupy a lower energy level, which is usually not possible except for those radioisotopes which tend to undergo beta decay, thanks to the Pauli Exclusion Principle! Thus, the bound neutron is stabilized.

What I understand from you explanation that protons are not positive charge  but they're composed of quarks and those quarks give the proton its positive charge and neutron a neutral charge. Thanks sir for your time, I promise you that I'll check the references you give me so I can completely understand your answer

I don't think it is accurate to say up quarks have spin 1/2, down quarks -1/2 , and that this determines the spin of the proton and neutron.  I think both up and down have spin 1/2 and, just as two or more spin 1/2 electrons could be parallel or antiparallel, so it with quarks.  The old theory was that such alignment of the quarks resulted in the spins of the proton and neutron, but even this is not the current theory: http://www.scientificamerican.com/article/proton-spin-mystery-gains-a-new-clue1/

@DavePhD - I will admit to not being an expert on particle physics. I went with what was in the Wikipedia articles and mapped them to my much better understanding of electrons

@BenNorris protons, neutrons, electrons, and all types of quarks (up, down, strange, etc.) all have spin 1/2 even according to Wikipedia.  https://en.wikipedia.org/wiki/Spin_(physics) and https://en.wikipedia.org/wiki/Spin-%C2%BD  Their anti-particles are also spin 1/2.


text:
The sponge isn't foaming up, why?
When I add dish soap to a sponge it is able to foam up. But if I leave the sponge alone for an hour, it doesn't foam up, not even if it is wetted.
Why? The sponge should still contain dish soap an hour later.

This might help.

Tap water contains calcium and magnesium ions; more so if the water is hard. These ions can bind to soap over time and make an insoluble soap curd.
Is the sponge clean? If not, another reason is that the surfactants in the soap that make the foam might be forming micelles around whatever grime is in your sponge, so that there aren't many left to make the foam.


text:
How to calculate the pressure of an ideal gas that undergoes isothermal expansion?

A 2.35 mole sample of an ideal gas, for which $C_{\mathrm{m},v}=3R/2$ initially at $\pu{27 ^\circ C}$ and $\pu{1750 kPa}$, undergoes a two stage transformation. For each of the stages described in the following list, calculate the final pressure as well as $q$, $w$, $\Delta U$ and $\Delta H$. Also calculate $q$, $w$, $\Delta U$ and $\Delta H$ for the entire process.
(a) The gas is expanded isothermally and reversibly until the volume triples.
(b) Beginning at the end of the first stage, the temperature is raised to $\pu{105 ^\circ C}$ at constant volume.

Since the process is isothermal, $\Delta H = 0$. In order to calculate the final pressure. I use the formula,
$P_\mathrm{i}V_\mathrm{i}^{x} = P_\mathrm{f}V_\mathrm{f}^{x}$,
where $x = C_{\mathrm{m},p}/C_{\mathrm{m},v}$.
Hence $x = 5/3$. I then substitute this into the formula to get
$$P_\mathrm{i}V_\mathrm{i}^{5/3} = P_\mathrm{f}V_\mathrm{f}^{5/3}.$$
Since the volume triples, $V_\mathrm{f}/V_\mathrm{i}=3$, hence manipulating the formula I got
$$P_\mathrm{i}/P_\mathrm{f}=3^{5/3}$$
and since $P_\mathrm{i}=\pu{1750 kPa}$, I substitute this into the above equation to get $P_\mathrm{f}$, which answers the first part of the question.
However I don't seem to get the right answer, hence I could not continue doing the question. Could anyone explain?
Also, am I right to say that we use
$P_\mathrm{i}V_\mathrm{i}^{x} = P_\mathrm{f}V_\mathrm{f}^{x}$
for an reversible adiabatic process only.
$w = -nRT\ln\left(T_\mathrm{f}/T_\mathrm{i}\right)$ for an isothermal process and
$\Delta U = nC_{\mathrm{m},v}(T_\mathrm{f} - T_\mathrm{i}) = -P_{\text{external}}(V_\mathrm{f} - V_\mathrm{i})$
for a non-reversible adiabatic process.
Im rather confused as to decide under which conditions should I use each of these three formulas.

Hint: In a) the gas is expanding isothermally ($\Delta T=0$). You're computing it as though it's expanding adiabatically ($q=0$). The two processes are quite different! You can use Boyle's law to compute the final pressure for a), and then Amonton's law $P_1/T_1 = P_2/T_2$ to compute the final pressure after process b).

So can i use that work formula, $w=-nRTln(V_{f}/V_{i})$

Yes, that applies to isothermal ideal gas processes (process a) only).

@weijit Your computation for adiabatic expansion is correct btw. Note however, as Fred (is it okay to say Fred?) posted in his answer that for adiabatic expansion $q=0$, not $\Delta H = 0$. +1

Am i right to say that we use $P_{i}V_{i}^{x}=P_{f}V_{f}^{x}$ for an reversible adiabatic process only. $w=-nRTln(T_{f}/T_{i})$ for an isothermal process and $ΔU=nC_{v,m}(T_{f}-T_{i})=-P_{external}(V_{f}-V_{i})$. for a non reversible adiabatic process. Im rather confused as to decide under which conditions should i use each of these three formulas.

Yes, also all of those relationships apply to ideal gases only. To clear up what the conditions are, derive those equations from scratch---they're not long derivations, and if you can derive them, you'll see exactly what assumptions are being made to get them.

I am going to use the physics conventions of work according to which, work done by the gas is positive. So if you want the answers as per chemistry conventions than just flip the sign of the work done, keeping the magnitude same.
Part (a): Reversible Isothermal Expansion
\begin{align}
\Delta U_a &=nC_v\Delta T\\
\Delta U_a &=\pu{0 J}\\
\end{align}
\begin{align}
\Delta H_a &=nC_p\Delta T\\
\Delta H_a &=\pu{0 J}\\
\end{align}
\begin{align}
W_a&= \int_{V_i}^{V_f} P.dV\\
W_a&= \int_{V_i}^{V_f}  \frac{nRT}{V}.dV\\
W_a&= nRT\ln\frac{V_f}{V_i}\\
W_a&= \pu{2.35 \times 8.314 \times 300 \times \ln3 J}\\
W_a&= \pu{6439.37 J}\\
\end{align}
\begin{align}
q_a&= W_a+ \Delta U_a\\
q_a&= \pu{(6439.37 + 0)J}\\
q_a&= \pu{6439.37 J}\\
\end{align}
By Boyle's Law,
\begin{align}
P_iV_i&=P_fV_f\\
P_f&=P_i\times \frac{V_i}{V_f}\\
P_f&= \pu{\frac{1750}{3} kPa}\\
P_f&= \pu{583.33 kPa}\\
\end{align}
Part (b): Isochoric Heating
\begin{align}
\Delta U_b &=nC_v\Delta T\\
\Delta U_b &=\pu{2.35 \times 1.5 \times 8.314 \times 78 J}\\
\Delta U_b &=\pu{2285.93 J}\\
\end{align}
\begin{align}
\Delta H_b &=nC_p\Delta T\\
\Delta H_b &=\pu{2.35 \times 2.5 \times 8.314 \times 78 J}\\
\Delta H_b &=\pu{3809.89 J}\\
\end{align}
\begin{align}
W_b&= \int_{V_i}^{V_f} P.dV\\
W_b&= \pu{0 J}\\
\end{align}
\begin{align}
q_b&= W_b+ \Delta U_b\\
q_b &= \pu{(0 + 2285.93)J}\\
q_b&= \pu{2285.93 J}\\
\end{align}
By Gay-Lussac's Law,
\begin{align}
\frac{P_i}{T_i}&=\frac{P_f}{T_f}\\
P_f&=P_i\times \frac{T_f}{T_i}\\
P_f&= \pu{\frac{1750}{3} \times \frac{378}{300} kPa}\\
P_f&= \pu{735 kPa}\\
\end{align}
Overall Process:
\begin{align}
\Delta U&= \Delta U_a+ \Delta U_b\\
\Delta U &= \pu{(0 + 2285.93)J}\\
\Delta U&= \pu{2285.93 J}\\
\end{align}
\begin{align}
\Delta H&= \Delta H_a+ \Delta H_b\\
\Delta H&= \pu{(0 + 3809.89)J}\\
\Delta H&= \pu{3809.89 J}\\
\end{align}
\begin{align}
W&= W_a+ W_b\\
W&= \pu{(6439.37 + 0)J}\\
W&= \pu{6439.37 J}\\
\end{align}
\begin{align}
q&= q_a+ q_b\\
q&= \pu{(6439.37 + 2285.93)J}\\
q&= \pu{8725.30 J}\\
\end{align}


text:
Why is predicting products of chemical reactions difficult?
I was going to ask whether there are software that could be used to predict the products of any given chemical reaction. However, I then noticed these two earlier questions

Chemical software for solving reactions
Software for predicting chemical reactions

where it is said such predictions are too difficult to make. What makes the prediction difficult?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

I can give you a short answer: Mainly because there isn't only one type of chemical reaction and there aren't only a few factors that have to be taken into account when predicting. But I think "good answers are going to be too long for this format".

I don't get why this was put on hold. The answer does not necessarily need to be overly long. DavePhD already provided some good basic information.

Dave provided a simple (no, not really!) example of how predicting the result would be hard. He didn't give all the reasons, but one. And I'm not the only guy with this idea, I suppose.

Because you need would need to solve the Dirac equation (relativistic Schrodinger equation) for all the particles in the system to get a completely accurate result.
Even solving the time independent non-relativistic Schrodinger equation for $\ce{H2+}$ in isolation, where there is only one electron and two protons involves approximating the protons as fixed relative to each other.
Now imagine trying solving a system of $10^{23}$ molecules each having multiple nuclei, and dozens or even hundreds of electrons, and including time dependence!

While technically it is true, there are easier ways to have good guesses than this. Just as in Physics macroscopic Newtonian equations are good approximation for the behaviour of a pendulum, and you don't need to solve the Dirac equation for the atoms in the pendulum neither.

Yes, but we could argue the other way too, that the Dirac equation doesn't even completely explain the hydrogen atom (Lamb shift), or para versus orthohydrogen, or the London dispersion interaction (the switch from 1/r^6 to 1/r^7 dependence at long distance that Cashmir explained).


text:
Temperature of an atom
I read somewhere that the temperature of an atom is not defined. The definition of temperature is only for larger systems. Why is this so?

Thermodynamic functions are strictly defined only for macroscopic systems (systems that have an essentially infinite number of atoms). You can't apply them to individual atoms because that would be confusing large-scale averages with individual microscopic values.
Here's an analogy: the average speed of cars on a stretch of highway might be 55 mph, but it's possible that no individual car is traveling at that exact speed.
Another analogy: the average global temperature is rising over long periods of time, but if I look at the thermometer outside my window, the local temperature is falling over a short period of time... I can't say that my local temperature drop disproves global warming, because weather is a different thing from climate.
Suppose that you knew the speed of a single atom. Now, for a macroscopic ideal gas, the average speed of molecules $\overline{v}$ is $$\overline{v} = \sqrt{\frac{8 R T}{\pi M}}$$ It would be wrong for you to plug in the speed of your single atom and solve for its temperature, because that equation was derived by assuming that you had a macroscopic system (with a certain speed distribution over many molecules); it only applies to average velocities, not individual velocities.
This applies to other thermodynamic functions, too. Let's look at a simple chemical example. For a macroscopic crystal, the Gibbs free energy might be written as $$G = Ng(P,T)$$ where $N$ is the number of atoms and $g$ is the Gibbs free energy per atom. Little $g$ is a function of pressure and temperature.
So can we say that little $g$ is the Gibbs free energy of a single atom? No. It's the average Gibbs free energy per atom for a huge number of atoms. To see the difference,  consider a small cluster of atoms from that same crystal, the previous equation will need some correction terms:
$$G = Ng(P,T) + a(P,T)N^{2/3} + b(T)\ln{N} + c(P,T)$$
where the "a" term is a surface free energy and the last couple of terms might come from free energy contributed by things like rotation of the cluster (Source: T. L. Hill, Thermodynamics of Small Systems, Dover, 1962.) You can't say that $g(P,T)$ is the Gibbs free energy of a single atom in the cluster because of those correction terms.
TL;DR: Thermodynamic functions are averages over very large numbers of particles; random fluctuations are tiny compared to the mean values. But with small systems, those fluctuations become quite important, and there are additional effects that cause macroscopic averages to poorly describe microscopic systems.

Very nice answer. But wouldn't the ergodic hypothesis (which in short states that the ensemble-average is equal to the time-average) imply that thermodynamic functions would also be valid for a microscopic system (e.g. a single particle) if I consider the behavior of the system averaged over a very(!) long period of time?

@Phillip - I agree with you - I think maybe it would be more correct to say that the instantaneous temperature of a single atom is undefined, even though the time-averaged temperature may be defined. This is also assuming that there is a suitable reference frame for determining the velocity of a single atom in the first place. Also, "very long" for atoms might not be that long for humans. As a rough guess, I think that at room temp a couple of nanoseconds would probably give a good average.

@Philipp and thomij, yes, you'll be comparing two averages, not an average and a single value.

Aah i see. I remember reading the derivation of the average kinetic energy of the atoms of an ideal gas, and it involved quite a few assumptions. Thanks for the great answer sir!

While generally temperature in textbooks is connected with the speed of atoms/particles, strictly speaking not the velocity, but the distribution of velocity what counts. If you microscopically observe a bunch of particles, their velocity can be whatever as your frame of observation can be any inertia system. When you want to calculate temperature, you choose  reference system which has the same velocity as the center of mass of those particles, and then you calculate the average velocities form that reference. The average will be small, if the distribution of velocities is sharp, and will be large if the distribution is broad. The calculated temperature can be low even for fast particles e.g. in a gas flow, if all the particles moving in the same directions, same velocity!
Let's apply this to one atom: you don't have a distribution of different velocities! Once you adjusted your reference point to the center of mass, ie the particle, it has zero velocity and meaningless to look for a distribution of different velocities.

There are electrons in the atom...

@jinawee Electronic temperature can only be relevant is there are electronically excited states below kT, and a trivial coupling to the kinetics of the atom (generally no). Even so, their behavior is not classical, per def, so we don't discuss them in classical theories.

I'll just rephrase @Greg's answer:
Thermodynamics distinguishes heat and mechanical work. From a microscopic point of view, heat is the unordered movement of atoms/molecules, while mechanical work is the ordered movement (all the atoms of the piston move at the same velocity = same speed & direction; free energy). As Greg says, the actual movement is decomposed into those two components: a macroscopic translation and an unordered part where the sum of all velocities is zero. This second part constitues heat (in terms of total energy of all those microscopic movements), and temperature (distribution of speed = |velocity|).
A single atom has only one velocity, so we cannot decide which part of it is ordered and which is unordered. We can thus speak of the kinetic energy and velocity of the atom, but not of distinguishable free energy and heat, nor of temperature.


text:
Cathodic protection of tubes in Shell & Tube Heat Exchanger
Our $\ce{Cl2}$ liquifier is of the following type:

It is entirely made of mild steel (shell as well as tubes plus the connecting pipelines). At shell side there is Freon (R-12) with its refrigeration cycle (not shown here); at the tube side there is dried & compressed $\ce{Cl2}$ gas from compressor section (~2bar max). One must realise that drying of $\ce{Cl2}$ is done physically via passing it through packed glass wool columns & then washing it in $\ce{H2SO4}$ scrubbers & then again through glass wool columns...Compression is done via $\ce{H2SO4}$ liquid ring compressors.
Problem we face is excess corrosion of tubes which thus lose their thickness & may in time compromise...For instance every 2-3 weeks we open up this liquifier & wash the tubes (via high pressure demi/RO water) & we found quiet an amount of dark-green/brown sludge indicating ineffectiveness of our drying section. Lab test are yet to be done so that I can clearly say what this sludge is made of. From the WIKI what I understand the compounds of such colour can either be oxides or chlorides of Iron because sulphates of iron are yellow in colour...
In a nutshell what I can do is increase the number of drying towers or increase the number of glass wool columns, but the most active measure I am thinking of cathodic protection via impressed currents. Have any user here applied impressed currents to a S&T heat exchanger & how can it be done? I would love any literature on it.
As a side note our liquifier can be electrically isolated using gaskets & insulated teflon bolts on any flange fitting.

Cathodic protection requires an electrolyte to work. You could just pass current through the tubes but how is it going to return to the power supply? If there is no electrolyte to conduct ions it cannot provide any protection to your system. Have you considered replacing your heat exchanger with glass or glass-coated metal? If the capital costs of this are prohibitive, I would think your best bet would be to try drying the gas some more, as you mentioned. Do you happen to know the water content of the gas going into the exchanger now?

Pure dry chlorine should not attack the tubes. After your washing of the system, do you thoroughly dry it to remove any residual water, and heat it to beyond 100C? Also, glass-wool (Brink) demisters will remove droplets, but you might need to chill them to lower humidity of the Cl2 gas, as well.

yes via instrument air the liquifier is thoroughly dried

@Qubit1028 ok sir I got your point what about sacrificial protection. http://marineengineeringonline.com/shell-and-tube-heat-exchangers/ read the Water Boxes category


text:
Are there any good examples of commonly ingested molecules that contain particular toxic individual elements?
In the house on my own and an article popped up on my feed with the following statement:

While the Government has insisted the chemical is safe, cadmium is
  recognised as a cause of lung cancer and during the Second World War
  was considered by the Allies as a chemical weapon.

The proceeding paragraphs are about zinc cadmium sulphide.
full article
I think this is a flawed argument, essentially inferring that any given molecule can be said to be toxic just because one of it's component parts is toxic but I can't think of a good example to back this up.
What is a good example of a common compound that is ingested often (daily?) but contains a particularly toxic element if that element were ingested in some way on it's own?

Apple seeds - they contain cyanide, but not enough to actually harm you unless you harvest ONLY the seeds for ingestion.. and while most people don't eat the seeds, do you think that the companies making apple juice/cider bother with removing them?

@user2813274: I don't think the apple juice is a good example: the cyanide is released when the amygdalin is hydrolysed. But apple juice is sour, so HCN gas would leave the juice. When in contact with air, HCN is oxidized to HOCN (cyanate; far less toxic than cyanide). Humans do something similar: we detoxify cyanide by making thiocyanate (SCN⁻) which goes out via urine. The capacity of this pathway is limited, it covers apple seeds in "typical" amounts, very few bitter almonds (which are bitter!), and is close to its limit for people eating lots of manioc (chronic cyanide poisoning).

Many antidepressants also are fluorine-substituted, and elemental fluorine kills any living thing.

Even almonds contain some quantity of cyanide. Toothpaste contains fluorides. Fluorine is faaaar more powerful than an acetylene torch, and can destroy your body within seconds. It's going to rip off the hydrogen from whatever it can find, and form hydrofluoric acid.

Table salt! What would be worse than putting sodium (it can spontaneously combust if you get it wet) and chlorine (used as a war gas in WWI) all over your food?
Then there's water which always contains some $\ce{OH-}$ (an active ingredient in Drano) and $\ce{H+}$ (that stuff eats through metal, man). Although those aren't elements.
And hiding inside every $\ce{O2}$ molecule are two deadly oxygen atoms; once freed, they'll wreak havoc on just about anything they come into contact with.

Haha, this is funny and true... (and this comment is absolutely unnecessary.)

As extreme example take Barium. Toxic under all circumstances but one, $\ce{BaSO_4}$, because this one is insoluble and is used in medical imaging.

@ssavec: which is pretty close to the difference in free Cd²⁺ vs. CdS

@cbeleites: exactly, the solubility (or better bioavailability) is the key. I wanted to point out the funny fact, that the doctor could ask you to drink a glass of "milk", containing extremely poisonous metal.

Also insert the usual joke about the chemist and his friend at the restaurant.

A few things:
1) There is a saying that "the dose makes the poison." This reflects the fact that most substances (even water!) become toxic at a high enough dosage.
2) As to your question about individual elements, surprisingly many of them are highly problematic in elemental form. But very reactive elements (Group 1, 2, 16 (some), and 17) are often not present in their elemental form in nature, and once reacted, they are quite benign (as Fred Senese noted about table salt). These are the best examples for your specific question.
3) The other problematic elements tend to be so-called "heavy metals", and they can be problematic because they can substitute for lighter metals in the same group and interfere with biological processes (generally slows them down). For example, both zinc and iron are necessary for health, but cadmium and mercury (same group and zinc) are very toxic, and the same goes for ruthenium and osmium (similar to iron). This goes to your example: these heavier metals can often be problematic regardless of the form.

I want to clarify that, although as the other answers demonstrate many elements are harmless in some but not all possible compounds, cadmium is not like that.  Cadmium is a heavy metal; heavy metals should be assumed toxic in all their forms.  This is because they easily substitute for lighter elements in the same column of the periodic table (for cadmium, that would be zinc) and many of those lighter elements are critically important to normal biochemistry.  (Zinc, for instance, is used in key enzymes for aerobic respiration, protein synthesis, and DNA replication.  Substituting cadmium will make many if not all of those malfunction.)  For the same reason, heavy metals are easily absorbed by the body (the digestive system is actively trying to take up as much zinc as possible) and not easily gotten rid of.
Some heavy metal compounds are so unreactive and/or insoluble that they're harmless, but that's the exception rather than the rule.
That said, the microorganisms mentioned in that article were probably orders of magnitude more dangerous than the cadmium compound.

As already mentioned in Fred's answer, Sodium (and other group 1 and 2 metals) are dangerous in elemental form, but are flammable and corrosive rather than toxic. But the halogens in elemental form are, as he says, much more toxic than the corresponding anions.
Another element which is very important for life in the form of phosphates but rather toxic in elemental form is phosphorus. My high school chemistry teacher claimed that phosphorus poisoning could be diagnosed by glow in the dark vomit.
Carbon, which is another extremely common element essential for life, will do you no good at all if inhaled in the form of coal dust.

Finally, it's worth noting that toxic heavy metals like lead and mercury are often far worse in their compounds than as the free elements.
There are various reasons for this, one of which is bioavailability. Lead pipes were once  commonly used and only really a hazard when they corroded and thus dissolved in the water they were carrying. (BTW corrosion products of copper are also pretty bad for you.) Even if you eat lead shot, most of it will pass through your system unchanged.
On the other hand, the organolead compounds (tetraethyl lead etc) that were once used in gasoline as an antiknock were extremely toxic required special handling procedures. The combustion products were much less toxic (though still a health hazard.)

But bioavailability works the other way round as well: if the bioavailability is extremely low (lead in lead pipes covered under $\ce{CaCO_3}$ or the $\ce{BaSO_4}$ in ssavec's comment to FredSenese's answer), the actual danger can be very low.

Also remember that while white phosphorous ($\ce{P_4}$) is toxic (and self-iginiting), but red phosphorous is not nearly as dangerous.

@cbeleites that's a good point, I hadn't thought of that one. But it's not straightforward bioavailability of "elemental barium", because Ba reacts with water to form Ba(OH)2 which is a soluble (and therefore bioavailable) compound. Ba(OH)2 is classified as "harmful" with an LD50 of 308mg/kg in rats according to the first source I found. That's toxic but not massively so. I tend to think of group 1 and 2 metals as "not heavy metals" because the really nasty ones tend to be in the transition and posttransition blocks, though it is debatable.

I'd even have said that that say, eating, elemental Ba is massively dangerous because of the basic reaction and the formation of hydrogen, and my guess would be that you may have so much trouble with the alkali burn that toxicity doesn't matter... Maybe I was thinking too large scale: toxicity will play a role if the amount is small enough that you have enough pH buffering capacity. Then the question is probably whether enough sulfate is availble. So not straightforward at all. Whereas BaSO4 is straightforward: not easy at all to release the Ba²⁺ once the sulfate is formed.

@cbeleites I wouldn't say it was just dangerous to eat elemental Ba, I'd say it was impossible. You'd burn your mouth and stop before you'd ingested sufficient to kill yourself by toxicity. The question asks for "toxic" elements, and you'd really need barium in a more, ahem, "palatable" form to be able to ingest enough to cause any internal organ damage.

The obvious example is roxarsone, an arsenic based growth hormone commonly found on the shelves in chicken. We are assured that it is non toxic. I guess the body handles toxic elements differently if they are bound to other elements?
http://www.iatp.org/documents/arsenic-widespread-in-chicken-testing-finds

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

Salvarsan (or arsphenamine), a medicine for syphilis, is also an arsenic based compound.

Soap is another good example, if you include corrosive materials. And while you're not ingesting it, you're likely absorbing some amount via your skin on a regular basis.
One of the necessary components of soapmaking is lye, or sodium hydroxide, which is so alkaline that it will eat through flesh. It also has an exothermic reaction with water, to the point that it can be explosive if you pour the water onto lye instead of the lye into the water (the former mixes it too fast, while the latter allows for a slower mix and more time for things to settle a little bit).
However, due to the chemical reaction that lye undergoes with fatty acids, you're left with an inert and safe compound.
Additionally, sodium hydroxide is often used on pretzels before baking, but the (much more subtle) reaction with the dough renders it nonhazardous, as the end result is a change in the pH of the surface of the dough, which gives you that nice brown, chewy exterior. (Sodium bicarbonate or sodium carbonate can be used for this purpose, as well.)
Borax is another one, though human exposure is generally skin or indirect contact. Boric Acid, one of the chemical components of Borax, is toxic in both acute exposure (LF50 = 1-20g/kg in humans, depending on age), and chronic exposure (as little as 32mg/kg for chronic lethal dose). However, Borax, itself, is a salt of Boric Acid, and so, doesn't carry the same toxicity unless broken down with something like hydrochloric acid. In fact, Borax is a great detergent for all sorts of household cleaning purposes (laundry, dishes, all-purpose cleaning).
Baking soda (sodium bicarbonate) -- most commonly used as a leavening agent in baked goods -- is the product of a reaction of sodium chloride, ammonia, and carbon dioxide in water. Its cousin, washing soda (sodium carbonate) can be produced using sodium hydroxide. Which one you get from that initial reaction depends on how much carbon dioxide you add to the mix.

I don't think soap is a very good example, since the finished product (at least nowadays) doesn't really contain any sodium hydroxide in any meaningful sense: it may well contain some sodium ions, but there's no hydroxide left to speak of, as what isn't consumed in the saponification process is washed away afterwards. That said, back in the old days before modern industrial chemistry, some cheap crude soaps did often contain considerable amounts of left-over NaOH, making them somewhat caustic and unpleasant to use.

@IlmariKaronen - the same can be said of any reaction involving compounds as opposed to elements. Compound reactions will usually result in one or more other compounds, plus one or more elemental thing (often a gas), and sometimes heat. In the case of soap, the hydrogen is released. However, that doesn't change the fact that it's required in the soap making process and is caustic on its own. Also, it's not the industrial process of making soap that gives less leftover lye, but the use of crystalline lye instead of potash, resulting in more accurate measuring of the lye content.

This may be one of the simplest molecules to answer the question.
Barium sulphate is widely used, in the form of an emulsion in water, for the "barium meal" X-ray procedure for diagnosing problems in the digestive tract. Barium is highly toxic, as are most of its compounds, but the sulphate is insoluble in water and stomach acid, so does not pass into the bloodstream and kill the patient.
Barytes, the mineral, is slightly impure barium carbonate (oftem pink colour) and is found in most parts of the world where mineralisation is present, e.g. in and adjacent to veins of metals such as lead and copper ores. It is safe to handle, and obviously has not been dissolved by groundwater over millions of years, demonstrating that it is insoluble.
But ingest almost any other barium compound, and you die. Handle with extreme care!
Oh, and as a quick edit in response to another post above, barium is indeed a heavy metal. I agree that almost all heavy metals (exceptions such as gold and platinum) are toxic, however there will probably be a few of their compounds which, like barium sulphate, can't pass into the body fluids due to lack of solubility, and as such will be non-toxic.

A better example would be Vitamin B12, which contains a Cobalt atom at the center of the molecule. The cobalt ion $\ce{Co^{2+}}$ is cancer-producing when alone. But when included in a complex like Vitamine B12, it is one of the most important molecules for human beings.