text:
Why does paper turn yellow over time?
Yesterday I saw a bill of some book which I bought 6 months back. Then this question struck my mind: "Why does paper turn yellow over time"?
I remember my teacher saying some organic compound was used to make paper white. And this effect is reversed over long periods of time. Can anybody explain?

from "Stuff Matters" by Mark Miodownik:

Paper yellows with age for two reasons. If it is made from cheap low-grade mechanical pulp, it will still contain some lignin. Lignin reacts with oxygen in the presence of light to create chromophores ( meaning, literally, 'color carriers'), which turn the paper yellow as they increase in concentration. This type of paper is used for cheap and disposable paper products, and is why newspapers yellow quickly in light.
It used to be fairly common to increase the textural quality of paper by coating it with aluminum sulfate, a chemical compound that is used primarily to purify water, but what wasn't appreciated at the time was that this treatment creates acidic conditions. This causes the cellulose fibers to react with hydrogen ions, which results in another form of yellowing. It also decreases  the strength of the paper. Large number of books from the 19th and 20th centuries were printed on this so-called acid paper and can now be easily identified in book shops and libraries by their bright yellow appearance. Even non-acid paper is susceptible to this aging, just at a slower rate.


text:
Why do fluorescence intensities decline with replicate number?
I am performing a determination of riboflavin with fluorescence spectrophotometry. I've dissolved my riboflavin in a dilute GAA solution, created a calibration curve and measured my unknown's fluorescence.
For these types of analyses I usually perform triplicate measurements, however I've noticed that spectrofluorometric measurements always decrease with replicate number. I assume this is due to 'photobleaching' of my samples. Since the phenomenon occurs with replicate rather than over time, and since my sample is in a dark environment, I would imagine this can only be the result of the excitation light emitted by the instrument itself. I find it surprising that this causes such a dramatic change in fluorescence from measurement to measurement. I also notice that with enough replicate measurements, eventually the intensity decline seems to bottom out. Can anyone explain the specifics of this phenomenon to me?

Am I right in assuming that this is photobleaching (deactivation of the chromophore) rather than photodegredation (breaking the molecule), or are these the same?
How exactly is the fluorophore deactivated?
Can the degree of fluorescence intensity decline be predicted quantitatively?
Would my calibration be more accurate if I used single measurements instead of taking the mean of three replicates?

While I can't answer all of these questions, I doubt you're looking at photobleaching. Photobleaching happens when a significant portion of the electrons have left the ground state, and unless riboflavin has an exceptionally long fluorescence lifetime, you really shouldn't see this effect. Photobleaching is what you see when you wail on a few thousand molecules with hundreds of watts of power for extended periods of time. My best guess is you're looking at some sort of photocatalyzed reaction. Does the fluorescence return if you leave the sample in the dark for ~5min?

I may also not have understood your details about "replicates rather than time" comments correctly--please let me know if that seems to be the case.

I think the sample's intensity is going down as a function of the number of times I hit it with the excitation wavelength (to take a measurement), rather than just over time.

My gut says "photodegredation," (which is not the same thing as photobleaching) but I have no idea what the nature is. It could be that the excited riboflavin forms a complex with the GAA, which could either be permanent or slowly degrade over time. If it's the latter, you could potentially restore the quantum yield by keeping the sample in the dark for a while (or heating it up to break up the complex).

Another thing you could try, of course, is to just split your sample up into 3 separate vials, and test each of them individually, instead of doing 3 sequential tests on the same sample. This would allow you to keep the advantage of multiple reference tests without having to worry about this degradation effect.

these are good ideas @chipbuster.
I initially thought photodegredation, but I was skeptical because I thought it would only happen to small, easily ionizable molecules, so I figured with a large organic like riboflavin it must be some sort of small-scale degredation of the chromophore that leads to deactivation of the fluorescence.  Separate samples would certainly be best.
I just figured since spectrofluorophotometry is such a common technique, more people would know this problem.

I figured as much too. Hopefully someone who actually knows what's going on will be along soon!

what is GAA solution , glacial acetic acid ? Sounds very much like photodegradation, unless the same happens in the dark. Does the absorption or fluorescence spectrum change? What happens in other solvents, maybe changing solvents is all that is needed, say ethanol etc.

That looks like a bleaching to me. The notion that bleaching only occurs with excesive light intesity on small populations of fluorophores is news to me. Might also that in my field we just use the bleaching term loosely. Best way to test is to take a long exposure of your sample and monitor the fluorescence intensity over time.
With our chemistry we usually encounter intensity loss that can be fitted with exponential decay function which would explain the bottoming out that you are describing. We often take multiple such "bleaching curves" and fit them to correct for intesity loss in our experiments that can take multiple hours.
In case you are only exposing your sample for short periods i would just make sure you measure each sample the same number of times that should rule out any error between samples. Thus reducing to only one measurment will not make it more correct.
Bleaching is poorly understood thus far. There is not a way to predict what excatly happens with a specific molecule.

Also how big is the effect for example (Start value)/(3rd measurment value) ?

The difference between the 1st and 3rd measurement at 0.2-1.0 ppm ranges from 0.04 to 0.2 emission untis. It's a pretty small difference, but I did try some with many repeated measurements and it declines for quite a while but eventually bottomed out. I didn't save those data though. I agree with you that it's bleaching. Making a bleaching curve is a good idea.

the (start value)/(3rd value) ranges from 1.008 to 1.012 for the standards of 0.2-1.0 ppm. For only three measurements, it's not a big change.

Could you also address the other subquestions so that this can be a complete answer?

so you are looking at and 0.8 to 1.2 % change in intensity which is imho an okay value depending on your setup. To rule out any other processes one could also alter excitation intensity to check if the relative intesity loss over multiple measurments dependes on the excitation intensity. I updated the answer to adress every aspect you brought up. But I am at a loss with regards to the mechanism of bleaching effects.

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sort: votes


text:
How will metallic lithium react with atomic oxygen?
Very high in Earth's atmosphere, there's a lot more atomic oxygen than anything else. Although I'm sure the energies involved in keeping oxygen in this state are an important variable, I'm interested (for now) in what compounds would form.
Mistake here: should be targeting $\ce{Li2O}$, not $\ce{LiO2}$. Would you get $\ce{LiO}$, then $\ce{LiO2}$? If so, will $\ce{LiO}$ be stable enough to bond with another $\ce{O}$ atom later? I can easily believe that lithium metal will simply erode away faster than a film of oxide could develop, given temperatures in the thermosphere and above. I just haven't seen much chemistry on this, however.

It would form $\ce{Li2O}$?

In those conditions, nearly everything would prefer to stay atomic.

You know, if you say metallic lithium, that implies a concentration of lithium atoms in a small area (or else it would not be able to form metallic bonding).

Adding to the comment from @DHMO - look at a Li-O phase diagram. For an example, K. Chang and B. Hallstedt, CALPHAD 35 160-164 (2011). $Li_{2}O$ is much more stable than $Li_{2}O_{2}$ (as it is named there).

$\ce{LiO}$ does not exist, $\ce{Li2O}$ does.

@NilayGhosh in that area everything exists.

As explained in Long-term ozone decline and its effect on night airglow
intensity of Li 6708 ˚A at Varanasi (25◦N, 83◦E)
and Halley Bay (76◦S, 27◦W) Journal of Earth System Science 120, No. 2, April 2011, pp. 291–300:
$\ce{Li + O -> LiO}$
$\ce{Li + O2 -> LiO2}$
$\ce{Li + O3 -> LiO + O2}$
$\ce{LiO2 + O -> LiO + O2}$
The article gives rate coefficients for each of these reactions.
The article also gives $\ce{Li}$, $\ce{O}$ and $\ce{O3}$ concentrations as a function of altitude for the range 80-99km (85km being considered the lower boundary of the thermosphere)

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text:
What is a "tube lens" in a mass-spectrometer ion source and what does it do?
I came across the Russian phrase "Transporting capillary voltage: 85 V" in a text I was translating. The paragraph was describing the operating parameters of an electrospray ionization-based ion source in an MS system.
The author told me that the proper English translation is "Tube lens voltage: 85 V". I googled for electrospray "tube lens" and found this picture:

I haven't found a down-to-earth explanation of what it is yet. Maybe someone here knows and can explain this in simple words.
Why is it a "lens"? It looks like a simple channel throught which the liquid flows to the needle.

By the time it hits the lens, the liquid had better be a stream of ionized molecules and atoms. Then it looks just like the first lens in almost any ion source including the skimmer as the ground of the electrostatic lens. If a molecule is ionized, consider the field lines between the tube at potential and the skimmer at ground.


text:
Fuel mass required to heat a room

A room in a house measures 3.7 m × 4.7 m × 4.0 m. Assuming no heat or material losses, how many grams of natural gas (methane, CH4) must be burned to heat the air in this room from 15.0 degrees C to 25.0 degrees C. Assume that air is 78% N2 and 22% O2.

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center. This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

Related: How much does a light heat the air around it?

In the first step, we calculate the energy that is needed
to heat the room by ten degrees.
To do that, you need the heat capacity of air:
$C_\text{air} = 1210~ \text{J} \cdot \text{m}^{-3}\cdot \text{K}^{-1} $
This is the energy that is needed per volume unit to increase the temperature by one Kelvin (i.e. one Degree).
To increase the temperature by $\Delta T = 10 ~\text{K}$ for a volume of $V_\text{room}=\text{3.7 m}\cdot\text{4.7 m}\cdot\text{4.0 m}=69.56 ~\text{m}^3$  you would thus need an energy of
$$ E = C_\text{air}\cdot  V_\text{room} \cdot \Delta T =
1210~ \text{J} \cdot \text{m}^{-3}\cdot \text{K}^{-1} \cdot 69.56~\text{m}^3 \cdot 10~\text{K} \approx 842~\text{kJ}$$
We can now calculate the amount of methane that needs to be burned to get that energy. For this, we take the enthalpy of combustion $\Delta H_\text{comb}$:
$$\Delta H_\text{comb} = −882.0 ~\text{kJ/mol} \equiv 55.1~\text{kJ/g}$$
Thus, we calculate the mass of methane to be
$$ m_\text{methane} = 842/55.1 ~\text{g} = 15.3~\text{g}.$$
So about 15 g of methane would be needed to increase the temperature of that room by 10 degrees. By the way, 15 g methane correspond to a volume of about 20 L.

Is that the upper heating value of methane or the lower heating value?

The upper heating value.


text:
clock reaction dilutions for KI
Stock solution 0.1M KI solution 100mL
Volume of KI
30
10.5
5.25
Volume of water
0
19.5
14.75
concentration of KI
0.1
0.05
0.025
How did they calculate these numbers?
 I tried c1 v1 = c2 v2
For example for 0.05M:
(0.1)(v1) = (0.05)(30/1000)
v1= 0.015L = 15mL, what does this 15mL represent. How did they calculate the volume of KI and water.
I want to dilute KI and I want the final solution to be 30mL. And the concentrations that I need are shown above (0.1, 0.05, 0.025, 0.0125).
Please help

The formula
$$c_1\cdot V_1=c_2\cdot V_2$$
is correct.
Since your stock solution has a concentration of $c=0.1\ \mathrm{mol\ l^{-1}}$, the value for the initial concentration $c_1$ is always
$$c_1=0.1\ \mathrm{mol\ l^{-1}}$$
Since you want the final volume of each solution to be $V=30\ \mathrm{ml}$, the value for the final volume $V_2$ is always
$$V_2=30\ \mathrm{ml}$$
You want four individual solutions (labelled $\mathrm a$, $\mathrm b$, $\mathrm c$, and $\mathrm d$) with four different final concentrations ($0.1\ \mathrm{mol\ l^{-1}}$, $0.05\ \mathrm{mol\ l^{-1}}$, $0.025\ \mathrm{mol\ l^{-1}}$, $0.0125\ \mathrm{mol\ l^{-1}}$. These are the individual values for the final concentration $c_2$:
$$\begin{align}
c_{2,\mathrm{a}}&=0.1\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{b}}&=0.05\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{c}}&=0.025\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{d}}&=0.0125\ \mathrm{mol\ l^{-1}}
\end{align}$$
Now you can rearrange the above-mentioned equation to solve for the initial volume of the stock solution $V_1$:
$$\begin{alignat}{2}
V_1&=\frac{c_2\cdot V_2}{c_1}\\[6pt]
V_{1,\mathrm a}&=\frac{0.1\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=30\ \mathrm{ml}\\[6pt]
V_{1,\mathrm b}&=\frac{0.05\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=15\ \mathrm{ml}\\[6pt]
V_{1,\mathrm c}&=\frac{0.025\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=7.5\ \mathrm{ml}\\[6pt]
V_{1,\mathrm d}&=\frac{0.0125\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=3.75\ \mathrm{ml}
\end{alignat}$$
You measure these volumes of the stock solution and transfer them into four separate vessels.
Finally, you have to fill up each solution with pure water to the final volume of $V_2=30\ \mathrm{ml}$. The required additional volume of water $V_{\ce{H2O}}$ is approximately
$$\begin{alignat}{2}
V_{\ce{H2O}}&\approx V_2-V_1\\[6pt]
V_{\ce{H2O},\mathrm a}&\approx30\ \mathrm{ml}-30\ \mathrm{ml}&&=0\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm b}&\approx30\ \mathrm{ml}-15\ \mathrm{ml}&&=15\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm c}&\approx30\ \mathrm{ml}-7.5\ \mathrm{ml}&&=22.5\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm d}&\approx30\ \mathrm{ml}-3.75\ \mathrm{ml}&&=26.25\ \mathrm{ml}
\end{alignat}$$

Are you sure? Because the person doing the experiment got different values.

Are you sure, because they got different results. For c2 =0.05M, they used 10.5mL Potassium iodide, and 19.5mL water. For c=3 = 0.025M, they used 5.25mL KI and 14.75mL water. please help.  how did they get these exact values since its very different than the values you calculated

@Shahad You can check all values yourself using the given equation $c_1\cdot V_1=c_2\cdot V_2$. You can see that your first examples do dont give the concentrations that you want: $0.1\ \mathrm{mol\ l^{-1}}\times10.5\ \mathrm{ml}\neq0.05\ \mathrm{mol\ l^{-1}}\times\left(10.5\ \mathrm{ml}+19.5\ \mathrm{ml}\right)$ and $0.1\ \mathrm{mol\ l^{-1}}\times5.25\ \mathrm{ml}\neq0.025\ \mathrm{mol\ l^{-1}}\times\left(5.25\ \mathrm{ml}+14.75\ \mathrm{ml}\right)$. Also note that $5.25\ \mathrm{ml}+14.75\ \mathrm{ml}$ gives $20\ \mathrm{ml}$ of solution and you want $30\ \mathrm{ml}$.


text:
Are van der Waals coefficients independent of temperature?
In the van der Waals equation for real gases, the coefficients $a$ and $b$ are used as respective correction terms for pressure and volume in the ideal gas equation.
I want to know whether the van der Waals coefficients are independent of absolute temperature or not.

The "formal" constants are just that - constants. However one could certainly have a table of $a$ and $b$ values for different ranges of temperature and pressure. See wikipedia article real gas for other equations.   https://en.wikipedia.org/wiki/Real_gas

The constants a and b in the Van der Waals equation are supposed to be independent of temperature.  But it is important to remember that, even though the Van der Waals equation does a better job of approximating the behavior of real gases than the ideal gas law (over a larger range of parameter values), it too is just approximation that applies only over a limited range of values for the parameters.

The van der waals constant can be directly obtained from the critical temperature and pressure. We have,

${a = \frac{27 R^2 T_c^2}  {64 P_c}}$
${b =  \frac{R T_c }{ 8 P_c}}$

Since, $T_c$ and $P_c$ are constants, we can conclude that the van der waals constants are really "constants."

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text:
Reaction of (RhCp*Cl2)2 with butadiene
This is the reaction that took place:

This is the information from the $\ce{^1H}$ NMR spectrum:
$$\begin{array}{|c|c|c|}
\hline
\delta \textbf{ / ppm} & \textbf{Relative intensity} & \textbf{Multiplicity} \\ \hline
1.65 & 3 & \text{d, } J = 6.2\ \mathrm{Hz} \\ \hline
1.73 & 15 & \text{s} \\ \hline
2.88 & 1 & \text{dd, } J = 11.0,1.5\ \mathrm{Hz} \\ \hline
3.28 & 1 & \text{dd, } J = 6.6,1.5\ \mathrm{Hz} \\ \hline
3.58 & 1 & \text{dq, } J = 7.0,6.2\ \mathrm{Hz} \\ \hline
3.96 & 1 & \text{ddd, } J = 11.0,7.0,6.6\ \mathrm{Hz} \\ \hline
\end{array}$$
What is the product formed?
I know the product only contains carbon, hydrogen, rhodium, and chlorine, and I think its molecular formula is $\ce{C14H21ClRh}$ (perhaps $\ce{H22}$ instead of $\ce{H21}$). My initial thought was that the diene could displace one chloride ligand to form $\ce{[RhCp^*Cl(C4H6)]}$.

There are two approaches to solving this - one coming from NMR analysis, and the other coming from understanding the chemistry. Ideally they lead to the same solution! The NMR data can only give one possible solution here. Do you know what [RhCp*Cl2]2 is used for? This will help you along the right path.

The precatalyst system [RhCp*Cl2]2 was developed by Peter Maitlis and his group at the University of Sheffield in the 1970's and 80's. It has been used as a hydrogenation catalyst under a number of conditions, and is readily commercially available. Under the right conditions, ethanol can act as a suitable transfer hydrogenation source (not quite as good as 2-propanol, but better than methanol), but in this case, in the presence of a suitable base (sodium carbonate), the rhodium dimer will react to give a stable product.
$$\begin{array}{|c|c|c|}
\hline
\textbf{label} & \delta \textbf{ / ppm} & \textbf{Relative intensity} & \textbf{Multiplicity} \\ \hline
\textbf{A} & 1.65 & 3 & \text{d, } J = 6.2\ \mathrm{Hz} \\ \hline
\textbf{B} & 1.73 & 15 & \text{s} \\ \hline
\textbf{C} & 2.88 & 1 & \text{dd, } J = 11.0,1.5\ \mathrm{Hz} \\ \hline
\textbf{D} & 3.28 & 1 & \text{dd, } J = 6.6,1.5\ \mathrm{Hz} \\ \hline
\textbf{E} & 3.58 & 1 & \text{dq, } J = 7.0,6.2\ \mathrm{Hz} \\ \hline
\textbf{F} & 3.96 & 1 & \text{ddd, } J = 11.0,7.0,6.6\ \mathrm{Hz} \\ \hline
\end{array}$$
From the NMR, it is possible to deduce what structure is likely, if we construct a valid spin system from the available data.

signal B: δ1.73 integral of 15 : we still have a Cp*
signal A: δ1.65 integral of 3 : looks a lot like we have a methyl group. It has a doublet coupling of 6.2Hz, which corresponds to signal E at δ3.58.
signal E: δ3.58 is single 1H which appears as a dq. We can account for the q coupling; it is next to a methyl group, which leaves the doublet coupling of 7.0Hz, which corresponds to signal F at δ3.96
signal F: δ3.96 is a single 1H with additional coupling to 2 other single 1H nuclei; one of 11.0Hz (signal C), which is pseudo-trans, and the other at 6.6Hz (pseudo-cis, signal D). This two signals have a mutual coupling of 1.5Hz, typical of a pseudo-gem couping on a double bond.

So that gives a spin system skeleton that looks like:

which, is able to form a stable π complex.
Homework spoiler alert:

and a useful reference from many years ago....

 K. Moseley,   J. W. Kang and    P. M. Maitlis J. Chem. Soc. A, 1970, 2875-2883. DOI: 10.1039/J19700002875


text:
How would you go about solving this pH problem?

What is the $\mathrm{pH}$ of a $0.24\ \mathrm M$ solution of benzylammonium chloride ($\ce{C6H5CH2NH3Cl}$)?
  $K_\mathrm b$ for benzylamine is $2.2\times10^{-5}$
  Your answer must be within $\pm 0.4\ \%$

I was able to determine that $[\ce{H3O^+}]=-1.8\times10^{-5}\ \mathrm M$ using $K_\mathrm b$. Then I used the formula $\mathrm{pH}=-\log[\ce{H3O^+}]$ to determine that the $\mathrm{pH}$ is $0.0000414465$. However, this answer keeps being marked as incorrect.
What am I doing wrong?

Welcome on chemistry.se! How did you come to the negative proton concentration?

The problem is idiotic. It calls for an answer good to +/- 0.4%, but the $K_b$ is only good to +/- 2.3% and the molarity is only good to +/- 2.1%. Absurd. The problem should have specified 0.240 M and $K_b = 2.20 \times 10^{-5}$

The $K_\mathrm{b}$ value of benzylamine, the conjugated base of benzylmmoniumchloride, is given. The acid constant of benzylammoniumchloride can be inferred via the identity
$$ \mathrm{p}K_a + \mathrm{p}K_b = 14. $$
Thus, we have
$$ \mathrm{p}K_\text{b,benzylamin} =  4.65$$
$$ \mathrm{p}K_\text{a,benzylammoniumchloride} = 9.35 $$
$$ \Rightarrow K_\text{a,benzylammoniumchloride} = 10^{-9.35}$$
As furthermore
$$ K_\text{a,benzylammoniumchloride} = \frac{[\mathrm{H}^\oplus] \cdot [\mathrm{PhCH_2NH_2}]}{[\mathrm{PhCH_2NH_3^\oplus]}} = \frac{[\mathrm{H}^\oplus]^2}{[\mathrm{PhCH_2NH_3^\oplus]}}$$
and
we see that
$$ K_\text{a,benzylammoniumchloride} = \frac{[\mathrm{H}^\oplus]^2}{[\mathrm{PhCH_2NH_3^\oplus]}}$$
This leaves us with an equation with only one unknown variable and is easy to solve as $[\mathrm{PhCH_2NH_3^\oplus}] = 0.24~\mathrm{\frac{mol}{L}}$ is given:
$$
[\mathrm{H^\oplus}] = \sqrt{[\mathrm{PhCH_2NH_3^\oplus]}\cdot K_\text{a,benzylammoniumchloride}} \approx 1.0354 \cdot 10^{-5} ~\mathrm{mol/L}
$$
And as the pH is the negative of the logarithm with base ten we have:
$p\mathrm{H} \approx 4.985$

By the way, a pH of about 0 should strike you as odd for such a weak acid. A pH of about 0 you would only get with strong acids like $\mathrm{H_2SO_4}$ or HCl.


text:
Why is it advised to put a Styrofoam plate under an Erlenmeyer flask?
At school today, we were doing experiments with ammonium nitrate and sodium hydroxide, and mixing them with water into an Erlenmeyer flask (separate flasks, of course). My instructor told us that we had to keep a Styrofoam plate under the Erlenmeyer flask at all times.
After the experiment, she told us to find out why we put a Styrofoam plate under the flask by tomorrow or we would get a 0 on that assignment.
Does anyone know why? I'm thinking it's to protect the table from the heat or coldness of the flask, or to protect the table in case of spills. We have to get the question exactly as she wants it, so I am looking for help.

What observation did you make during/after the experiment? The zero sounds harsh but the question is a good one.

The first flask cooled to 17.5C and the second heated to 38.3C which to me wasn't very much

I really don't know what else, are my guesses correct?

So, as part of the experiment you measured the temperature difference for both?

Hypothetical situation: suppose your table was made out of a giant block of steel that was very cold (maybe cooled in a salt+ice slurry). What happens?

@chipbuster The temperature of the table would affect the reaction. Is that it?

@MaxW Yes I measured the temperature throughout the reactions

Think about the validity of your observations in the scenario @chipbuster proposes.

By the way: If you think you arrived at the answer you can post a self-answer here.

@bpedit what do you mean?

Would your temperature measurements be the same if you did your experiment as @chipbuster described? Wouldn't it be most meaningful if the temperatures you measured were only due to the reaction itself? Temperature changes during a reaction provide important information about the nature of that reaction.

@bpedit Unless you're heating the Erlenmeyer flask while it is in contact with the floor/desk, it would not act like a heat sink. Unless you're maintaining temperature for extended periods of time....Also, there's so little information that the OP gives...

@QuantumCAPUCCINO Any temperature difference between the table top and flask provides opportunity for heat transfer. Styrofoam, being a poor conductor of heat, minimizes this. You can debate whether the amount of transfer would be significant, but that's beside the point of this being good science. It eliminates, at least minimizes, the table as a variable in the temperature measurement.

@bpedit The air (conductivity 0.024) will be at the same temperature as the table. The table is probably made of wood (conductivity 0.12 ~ 0.30) or epoxy (conductivity 0.35). Styrofoam, on the other hand, has a conductivity of 0.03. So on conductivity alone the table is around 4~10 times larger. However, there's surface area. Assume that the Erlenmeyer is a cone with radius $r$ and height $2r$. The bottom will be touching the table, but the rest of the cone is in contact with air. The outside of the cone excluding the bottom is around 2.5 times the surface area of the bottom.

Lets now include the inside of the flask. Assuming that the inside of the flask is a smaller cone of radius $0.6r$ and height $1.2r$ gives us a surface area that is around 1.8 times the bottom of the flask. So...around 4.3 times the surface area of the flask is in contact with a colder or hotter "surface". I mean sure, its a good idea..but I think it borders on tunnel vision. If the teacher is so interested in the effect of the table, why not use an oven or a cooler?

@QuantumCAPUCCINO I don't understand the use of an oven or cooler. We're trying to minimize external influences not exercerbate them. I agree, as I've mentioned, the role of the table is minimal. The teacher here is making a point about the methods of science. Primary among those is the reduction or elimination of extraneous variables. I taught Chemistry for 24 years. While I might not have done anything as prissy as this teacher, I respect the point she is trying to make (although not the zero she threatens).

Our table is a phenolic resin countertop, if that gives more information.

Comments are not for extended discussion; this conversation has been moved to chat.

Spill of chemicals, not much on table but may sometimes contaminate some other places, and sometimes toxic/ corrosive chemicals do bad on us. How much cleanness is being expected, is depend upon the type of work done on this lab. For very-basic level pure chemistry I've never seen that much protection anywhere; however in a biology laboratory a mild contamination may spoil everything.

You provided the additional information in the comments that your experiment was about measuring the change in temperature that occurs upon mixing. This should give you a hint as to whether the substances mix with water in an endo or exothermic manner.
When measuring temperature changes it is crucial not to influence the temperature of the reaction mixture in any way. Naturally, that would for example mean not to keep the flask in your hand (body temperature is typically more than room temperature). But some error sources are not as easily visible. In your case, the table top was an error source. Whatever the material, it likely conducts heat better than air — meaning that if your flask was on the table a significant portion of the heat generated (or used) would be conducted away (or onto site) by the table resulting in a lower change in temperature. That is why these experiments are typically performed under maximally insulating conditions. One of the best and widely available insulators is polystyrene because of its high air content (air is a good insulator). Hence, keeping polystyrene under a flask reduces the error caused by the table conducting heat.

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


text:
Tried charging a penny/dime battery. Yellow goo bubbled. What did I create?
When I was a young child, I applied 9 volts across an electric cell I made using a penny, a dime, and a salted piece of paper napkin wet with saliva.
Immediately, an unknown mustard-yellow liquid began forming and bubbling from in between them. It discolored the coins.
What reaction took place? What was the yellow stuff?

Which coin was connected to which electrode?

Vapid -- Sorry, I do not recall.

You created an electrolytic cell. The reason why I have asked you in the comments about electrodes is to determine which coin served as a cathode, and which as an anode because this is crucial for the electrochemistry that took place in your experiment. Anyway, let's first discuss what happens if you electrolyze a solution of sodium chloride with copper electrodes and allow the solutions near the electrodes to mix (I was doing it this as a kid I remember). On the anode (the electrode connected to '$+$' on the battery) copper gets oxidized to its $+1$ oxidation state:$$\ce{Cu->Cu+ +e-}$$ (I will skip the reactions involving chlorine and all other reactions that are of no interest to us). On the other electrode (cathode, '$-$' on the battery) hydrogen ions that come from water dissociation: $$\ce{H2O->H+ + OH-}$$ are reduced to gaseous hydrogen:$$\ce{2H+ +2e- ->H2}$$ leaving out the $OH^-$ ions. When the solutions surrounding each electrode are mixed, Copper(I) hydroxide forms:$$\ce{Cu+ +OH- ->CuOH}$$ which instantaneously decompose to copper(I) oxide:$$\ce{2CuOH->Cu2O +H2O}$$ which is insoluble and yellow-orange in color. You can see this here (but it is better to do it yourself!). I believe that the 'yellow goo' you obtained in your experiment is the copper(I) oxide. This suggests that the penny served as an anode and the dime as a cathode (although both coins contain copper, so maybe it doesn't really matter which coin was which electrode).

Some of the color can also come from nickel and copper chloro- and aqua- complexes, they have a yellow-green coloration.

Hahaha, I had to translate for myself what a penny and what a dime is. Always the same for me; Fahrenheit, miles, ...
A penny is a copper-plated zinc coin. A dime is a nickel-plated copper coin. Knowing this, I would assume that the yellow color arose from the nickel content. If it would be blue, it would have more likely been copper.
The goo. Such a structure often occurs with proteins, polymers, and similar compounds. It's most likely that the bacterial farm on the coins you used, lead to the formation of this goo. It's impressive how many bacteria and what thick biofilms you can find on a coin. Additionally, your saliva also contains a multitude of living organisms and their remains.
Therefore, you should definitely reproduce this experiment! Clean the coins beforehand using ethanol, acetone, and water while scrubbing. Polish them with a linen cloth and not with sand paper. Otherwise, you would damage the plated metal film and surface the underlying metal (makes things now complicated). And don't spit in it yet! :-)


text:
Elements with multiple oxidation states
I downloaded a periodic table app which gives detailed information for each element. The app shows some elements, however, have two different oxidation states. Hydrogen, for instance, can have either a 1+ or 1- state. What do these two states signify? Why do these elements have more than one oxidation state?
Can someone explain this?

It means simply that the elements in question can have various charges, or oxidation numbers, when they are components of different molecules. Hydrogen is usually a +1, but in some cases (sodium hydride, or $\ce{NaH}$, for instance), it's a -1. See this Wikipedia page for some illumination and specific examples.

What's to explain? Some atoms (arguably, most of them) may exist in different oxidation states depending on how you treat them, much like some men may work at different jobs.

So when an elements denotes 1+/1- should I conclude that the element is predominately 1+ charge? Does that order show predominately charges first and less occurring charges secondary?

http://chemistry.stackexchange.com/questions/34106/why-do-atoms-form-either-a-positive-or-a-negative-charge-but-not-both

@DannyRodriguez Well, that depends on what the app designer was trying to do. There is no convention for listing common oxidation states first (often they are simply listed in ascending/descending order) and I personally would not jump to any such conclusion.

Except for helium and neon, multiple oxidation states are known for any elements. And for all other elements except for fluorine and argon, these different oxidation states can even feature different signs (although I’m not entirely sure all have been proven for the all of the others).

The oxidation state of an element in a compound is simply decided on the basis of electronegativity of other atoms. As the definition of oxidation state says

Oxidation state of an element in a  given compound is the charged acquired by its atom on the basis of electronegativity of  other atoms in the compound.

The basis of calculating oxidation number is that the more electronegative element acquires the negative charge and the less electronegative one acquires the positive charge. So in molecules in which hydrogen is less electronegetive it acquires a $+1$ charge while in molecules in which it is more electronegetive it will acquire a $+1$ charge ($1$ because it has the valency of $1$ and  usually oxidation number does not exceed valency).
Let's understand it with the help of examples:
In the $\ce{HCl}$ molecule, $\ce{H}$ is less electronegetive that's why charge of $+1$,for the same reason,$+1$ in $\ce{H2O}$.
In the compound $\ce{ZnH2}$, hydrogen is more electronegetive so it will have a charge of $-1$, for similar reason $-1$ in $\ce{NaH}$.
Usually metals (that make metal hydrides) are less electronegetive than hydrogen, That's why we also say hydrogen has an oxidation state of $-1$ in metal hydrides.

You see compounds like HCl in which the hydrogen is +1, and then there are compounds like NaH, where the hydrogen is -1.

The oxidation state, often called the oxidation number, is an
  indicator of the degree of oxidation (loss of electrons) of an atom in
  a chemical compound. Conceptually, the oxidation state, which may be
  positive, negative or zero, is the hypothetical charge that an atom
  would have if all bonds to atoms of different elements were $100 \% $
  ionic, with no covalent component. This is never exactly true for real
  bonds.

So an atom can gain $1,2,3$ or more electrons and also lose $1,2,3$ or more electrons thereby causing it to gain a certain degree of oxidation or reduction. This explains why an element can have multiple oxidation states.
Read More: https://en.wikipedia.org/wiki/Oxidation_state


text:
Where does computational chemistry overlap with organic chemistry?
I was wondering what research exists that combines computational chemistry and organic chemistry.
For the past three years, I've been using Amber and Gaussian (in hopes) to identify and understand trends in intramolecular interactions. This really just serves as an example of my past research which incorporated both physical chemistry and computational chemistry.
I would like to know what kind of research could be done that combines both organic chemistry with computational chemistry.
I've Googled computational research in organic chemistry, but haven't found much information. What are some examples of projects that have been done in organic chemistry that rely heavily on the computational side of things?

Drug screening is pretty lucrative and mostly organic chemistry. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3880464/ Or you could also work in creating force fields. There are many solvents whose force fields are not available, or accurate. Or in the spirit of this year's Nobel prize, you could look at the dynamics/simulation of molecular machines.

You could also try research in transition state/intermediate states. IRC, they have trouble determining them because of optimization problems (global vs. local). Since determining intermediates and transition states is important in Ochem, I suppose it's related, . Although computational chemistry is still not very good at this, you could research this since it is a rather important subject.

@QuantumCAPUCCINO Drug screening is also usually pretty computationally intensive, is it not? Worth looking into, but it might require more resources than you've got, Melanie.

the absolute coolest field of research right at the centre of chemistry is catalyst design. Using computational chemistry to design inorganic products with the end goal of doing fancy organic transformations, then scaling them up to the kilo level to make drugs and save the environment! and all along you continue to do computations, make predictions, perform experiments, and improve your understanding of the chemistry. imo the cat-chem/compchem/process chem area of chemistry is nested perfectly in the nexus of all scientific thought. It's like the perfect field.

honestly though, chemistry is all about getting wet. Having a good basis in compchem as an undergrad is ideal. Working full time on computational is painfully boring and bad for your back, plus calculations take long enough that you might as well be setting up reactions while you're waiting. The cool thing about comp-chem is that it takes an often empirical science and brings the scientific method front & centre. You go Theory>prediction>experiment>theory until you make better chemistry and save the world one kcal at a time!! The future is interdisciplinary. You've got to have both.

The first thing that comes to mind: read the backlist of Henry Rzepa's and (especially) Steven Bachrach's blogs.  Depending exactly on what resources you have available, there are lots of things they do on the computational side of organic chemistry that you could at first imitate and then use as a springboard to innovate.
In particular, I've found Bachrach's posts on reaction dynamics to be especially fascinating.  This is the phenomenon where the topography of the potential energy surface near a transition state is such that the reactions don't always follow the minimum-energy path due, effectively, to the inertia of the nuclei as they cross the TS.
I also have a personal interest in Bader's Quantum Theory of Atoms in Molecules, which uses the topology of the electron density to divide molecules up into atoms, where each atomic volume is enclosed by a 'zero-flux surface' in the electron density.  (There's some ongoing controversy over the theory, but it makes good sense to me, for whatever that's worth.)  Bader's book is a good place to start; or, you can search the journal literature for Richard Bader and a whole bunch of examples will surely turn up. I engaged with Dr. Rzepa last year in the comment thread of a post of his about electrides that touched on QTAIM, the electron localization function (ELF), non-covalent interactions (NCI), etc.
Finally, I've been very curious lately about 'charge-shift bonding', a relatively newly described class of bond where the nuclear framework tries to cram so many electrons into a small space that in the end Pauli repulsion drives them apart, causing what might otherwise appear to be a covalent bond to instead take the form of a resonance between two ionic structures.  (The electronic structure of $\ce{F_2},$ for example, is better described as a balanced resonance between $\ce{F- \!- F+}$ and $\ce{F+ \!- F-}$ overlaid atop a very weak covalent interaction, than as a purely covalent $\ce{F\!-F}$ species.) Search the literature for Sason Shaik's work for more information and examples.
Other things that I've come across that have seemed interesting include carbenes, cyclic alkynes, unusual Diels-Alder reactions, and fullerene systems.
I would also subscribe to, e.g., Computational Chemistry Highlights, CCL, and Jan Jensen's Computational Chemistry Daily -- the posts and/or questions that come across there might give you some inspiration.

I have complemented most of my organic chemistry ("wet chemistry") work with computational chemistry, and it's, in fact, quite common to do so today. I would say it pretty much depends on your interests. There are probably other members in the group of your professor. Maybe they face some problem where computational chemistry can assist in. One example is the possibility to simulate NMR, IR, UV-VIS or (V)CD spectra. Another is the calculation of ground states, intermediates and transition states of catalytic cycles in case the group is doing catalysis. Actually, I would say there definitely should be the possibility to apply computational chemistry in any chemistry group, the remaining question being how good the support will be and if your professor will appreciate what you are doing. There are many good books about (organic) computational chemistry out there. I particularly liked

Computational organic chemistry by Steven M. Bachrach
A Chemist’s Guide to Density functional theory by the authors Wolfram Koch, Max C. Holthausen
Exploring Chemistry with Electronic Structure Methods by James B. Foresman
Essentials of Computational Chemistry by Christopher J. Cramer
Introduction to
Computational Chemistry by Frank Jensen

more or less in that order. But there are lots of other good books of course. Another book for you, being interested in both physical and organic chemistry, is

Modern Physical Organic Chemistry by
Eric V. Anslyn,
Dennis A. Dougherty

which by the way has a nice treatment of some organic computational chemistry topics as well.

Ahh, yep, I realized after I went to bed that I forgot to mention spectroscopy. Frank Neese's pubs list and ORCA are a good place to start looking into that.

One thing that would be interesting to tie together is the relationship between chiral resolution and resolving agent structure.  Looking at the Dutch resolution on a molecular level would be cool.  Relating Fogassy parameters with what happens at the molecular level is another aspect of the same area of work.


text:
Why are the melting and boiling points of the group 1 metals lower than the group 2 metals in the same period?
I thought it was because the group 1 metals are smaller than the group 2 metals but the answer sheet says it's because there are more valence electrons and a stronger positive charge in group 2 metals. But why does it matter if there's more valence electrons and a stronger positive charge?

In the photo, you can see that for Group $2$ metals, the cation is doubly positively charged, making the attraction stronger.

Credits
The photo is from Figure 12.20, Section 12.5, Principles of General Chemistry (v. 1.0).

Is it doubly positively charged because there's more protons?

@Hamze partly. it is doubly positively charged because it is two electrons away from the stable octet state, so two electrons become delocalized.


text:
Dissociation of 4-aminobenzoic acid in aqueous solution
I believe that 4-aminobenzoic acid's structural formula is
$\ce{C7H7NO2}$
Hence, $$\ce{C7H7NO2 + H2O -> ?}$$
what are the dissociation products of the 4-aminobenzoic acid in aqueous solution ?

Try to draw a structure of 4-aminobenzoic acid.

You can't possibly deduce the reaction from the molecular formula.

4-aminobenzoic acid can be represented by the formula $\ce{H2N(C6H4)COOH}$, where the $\ce{C6H4}$ is the benzene ring minus two hydrogen atoms.
The group $\ce{RNH2}$ is basic, meaning the following happen:
$$\ce{RNH2 + H2O <<=> RNH3+ + OH-}$$
The group $\ce{R'COOH}$ is acidic, meaning the following happen:
$$\ce{R'COOH + H2O <<=> RCOO- + H3O+}$$
Combined:
$$\ce{H2N(C6H4)COOH + H2O <<=> ^+H3N(C6H4)COO- + H2O}$$

So, what is the water doing there?

@orthocresol catalyst of proton transfer?


text:
Why are higher-energy bonds preferred over lower-energy ones?
My chemistry book states that for example the bond S-O is preferred over the bond S-S because the first has a bond energy that is almost twice as the second one. Since the energy released by the formation of the bonds is equal to the energy needed to let the reaction occurr,why does stronger bond are preferred over lower ones?
I mean, if a S atom needs let's say 170 kJ/mol to start a reaction with another S, shouldn't this be more likely to happen respect to a bond S-O that needs 340 kJ/mol?

more energy released means more energy required to break the bond and if more energy is required to break the bond the bond is stable.I let you conclude from there.

When a mole of $\ce{S-O}$ bond is formed, $340$ kJ of heat is released.
This $340$ is not how much energy is needed to start a reaction. That energy (activation energy) cannot be derived from the bond dissociation energy.

Oh thank you I didn't understand very well. I was very confused with the bonds energy.

The energy in forming a bond is heat released, and so must be absorbed when breaking a bond to form separate atoms. To see if  one type of molecule will react with another we compare the heats of formation (approx. sum of bond strengths) to obtain a heat of reaction. If negative the reaction is exothermic then heat is released on reaction so products have lower energy and are more stable than reactants or if not then the reaction is endothermic which is the other way round.
The heat of reaction does not tell us whether the reaction is going to be fast or slow. This depends on the activation energy, which is a potential energy barrier between reactants and products, and has to be overcome if reaction is to occur. The more frequently this happens the faster the reaction. The activation barrier exists only very briefly, and is caused by rearranging the atoms in the reactive molecules, after they have collided with one another, and as new molecules are formed.


text:
Calculating the molar iodine concentration of a CT contrast agent
I want to calculate the molar concentration of Iodine in a CT contrast agent.
Name: Isovue 300 (Iopamidol - $\ce{C17H22I3N3O8}$)
Molecular weight: $777.09\ \mathrm{g \ mol^{-1}}$
The part that confuses me is this:
"Each $\mathrm{mL}$ of Isovue-300 (Iopamidol Injection $61%$) provides $612 \ \mathrm{mg}$ iopamidol with $1 \ \mathrm{mg}$ tromethamine and $0.39\ \mathrm{mg}$ edetate calcium disodium. The solution contains approximately $0.043\  \mathrm{mg}$ ($0.002\ \mathrm{mEq}$) sodium and $300\ \mathrm{mg}$ organically bound iodine per $\mathrm{mL}$." (source: https://www.drugs.com/pro/isovue.html)
Each molecule of Isovue contain 3 $\ce{I}$ atoms. So if I calculate the molar concentration (300 mg/ml / 777.09 g/mol * 1000 = 386.1 mM), do I need to correct for this?
I am sorry for this basic question, but I cannot find the answer on the manufacturer's website or elsewhere.
Thank you all in advance

I highly recommend not looking for literature from non-authoritative sources such as drugs.com: that's a paid-advertising site and is full of partial/misinformation.
For your compound, see the paperwork from the people who make it - Bracco Imaging.
Each $\mathrm{mL}$ of Isovue-300 contains more than what you list (namely, water, sodium hydroxide, and hydrochloric acid). That means your approach is close.
The wording about the 300 $\mathrm{mg}$ of organically-bound $\ce{I}$ per $\mathrm{mL}$ of solution makes sense when you consider the atomic weight of $\ce{I}$ (about 127 $\mathrm{mg}\,\cdot\mathrm{mmol}^{-1}$) and that 3 atoms of iodine comes out to 381 $\mathrm{mg}$. Thus, for that slightly dilute solution, the 300 $\mathrm{mg}$ makes sense. It's not an additional 300 - it's what's in the $\mathrm{mL}$ as-is.
So, the molarity of $\ce{I}$ is (I abbreviate iopamidol as "iop" below to save space):
$$\left({612\,\mathrm{mg\,iop}\over 1\,\mathrm{mL}}\right)\cdot
\left({3\,\mathrm{mmol}\,\ce{I}\over 777.09\,\mathrm{mg\,iop}}\right)\cdot
\left({1\,\mathrm{mol}\,\ce{I}\over 1000\,\mathrm{mmol}\,\ce{I}}\right) \cdot
\left({1000\,\mathrm{mL}\over 1\,\mathrm{L}}\right)
= 2.36\,\mathrm{M}$$


text:
Discrepancies between specific heat of vaporization of water and enthalpy of formation of water and steam
According to Wikipedia, the enthalpy of formation of water is $-285.8~\mathrm{kJ/mol}$ while the enthalpy of formation of steam is $-241.818~\mathrm{kJ/mol}$, implying the following:
$$\ce{H2O(l) -> H2O(g)}\qquad (\Delta H=44.0~\mathrm{kJ/mol)}$$

Let us derive the enthalpy of the above reaction using the specific heat capacity of water and the specific heat of vaporization of water instead.
$1~\mathrm{mol}$ of water weighs $0.018~\mathrm{kg}$.
To raise $1~\mathrm{mol}$ of water from $25~\mathrm{^\circ C}$ to its boiling temperature requires $(0.018~\mathrm{kg}) \times (4200~\mathrm{J~kg^{-1}~K^{-1}}) \times (75~^\circ\mathrm{C}) = 5.67~\mathrm{kJ}$.
To turn that amount of water to steam requires $(0.018~\mathrm{kg}) \times (2258~\mathrm{kJ~kg^{-1}}) = 40.644~\mathrm{kJ}$.
Adding these two terms give $46.314~\mathrm{kJ}$, implying the following:
$$\ce{H2O(l) -> H2O(g)}\qquad (\Delta H=46.314~\mathrm{kJ/mol)}$$

Is the discrepancies between the two thermochemical equations simply due to measurement inaccuracies?

The enthalpy change is a function of temperature. It could be that the data has been reported at different temperatures (e.g at 298K and 373K). In that case $\Delta _r H (\text{at 298 K}) \neq  \Delta _r H (\text{at 373 K}) $

@getafix Could you resolve the discrepancies?

You forgot to include the (hypothetical) enthalpy change for cooling the vapor back down to 25 C.

According to “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL, the standard molar enthalpy of formation at a temperature of $T=298.15\ \mathrm K$ is
$\Delta_\mathrm fH_\mathrm m^\circ=-285.8\ \mathrm{kJ\ mol^{-1}}$ for liquid water,
and
$\Delta_\mathrm fH_\mathrm m^\circ=-241.8\ \mathrm{kJ\ mol^{-1}}$ for gaseous water.
(These values are in agreement with the values that you have found on Wikipedia.)
Both values refer to a temperature of $T=298.15\ \mathrm K$; therefore, you do not have to raise the temperature of the liquid water to its boiling point. Nevertheless, you may do this; however, then you would have to cool down the steam back to $T=298.15\ \mathrm K$ later.
Instead, you can directly use the enthalpy of vaporization at the given temperature. Note that the enthalpy of vaporization depends on temperature, and that your value for the specifc enthalpy of vaporization of $\Delta_\text{vap}h=2258\ \mathrm{kJ\ kg^{-1}}$ refers to the boiling point at normal pressure, i.e. to a temperature of approximately $T=100\ \mathrm{^\circ C}$.
According to REFPROP – NIST Standard Reference Database 23, Version 9.0, the molar enthalpy of vaporization at a temperature of $T=298.15\ \mathrm K$ is
$\Delta_\text{vap}H_\mathrm m=43.987\ \mathrm{kJ\ mol^{-1}}$,
which is in good agreement with the difference of the values for the molar enthalpy of formation:
$-241.8\ \mathrm{kJ\ mol^{-1}}-\left(-285.8\ \mathrm{kJ\ mol^{-1}}\right)=44.0\ \mathrm{kJ\ mol^{-1}}$

If I raise the water from $298.15\ \mathrm{K}$ to the boiling point, then convert to steam using the $2258\ \mathrm{kJ\ kg^{-1}}$, and then use the specific heat capacity of steam to cool it to $298.15\ \mathrm{K}$, why do I obtain a different result?

@DHMO I am not sure what you tried. Anyway, note that the specific heat capacity of liquid water and steam also depend on temperature. Thus, for a precise calculation, you cannot use constant values for the heat capacity; you would have to integrate the values over the considered temperature changes. Therefore, for precise engineering calculations, the enthalpy balance based on values taken from so-called steam tables is usually preferred.

If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha. The corresponding results for a temperature of $T=298.15\ \mathrm K$ at equilibrium can be obtained using the input “water boiling at 298.15 K”.

Suggestion: Calculate the enthalpy change for the process $\ce{H2O} (\mathrm{l; 298~K}) \ce{-> H2O} (\mathrm{l; 373~K}) \ce{-> H2O} (\mathrm{g; 373~K}) \ce{-> H2O} (\mathrm{g; 298~K})$ for comparison. (I expect $\Delta \approx 0$ when compared to the above value.)

@Jan yes, $H_\mathrm m = 1.889 \to 7.521 \to 48.190 \to 45.876$, i.e. $\Delta H_\mathrm m =43.987$, which is the value that you can find in my answer (all values in $\mathrm{kJ/mol}$, obtained from REFPROP steam tables)

As I expected: the question’s ‘missing’ enthalpy derives from forgetting to cool the steam back down to room temperature.


text:
Rearrangement of carbocation
During rearrangement of carbocation, we have three choices: methyl shift, hydride shift and phenyl shift. If we have a situation where all the three are possible, then what will be the preference order?

Related: Determination of products in reactions involving carbocation rearrangement? and Precedence of 1,2 carbocation rearrangement

You should look at stabilization in the transition state. Also, this seems like a homework question. Is it?

No, it's not. We don't get such questions in home work, it's something which was confusing me.

Voting to leave open per OP's comment. But, @user375072, the question would benefit from adding more detail about what you are and are not confident about as regards these different reaction pathways.

In all cases we study the stability of the carbocation resulting from the hydride/methyl/phenyl shift. That shift will be more favored whose resulting carbocation is the most stable.


text:
Fitting of interatomic potentials to a training set
I have read several times in the literature the following:
"The fittings were accomplished using a combination of the singular value decomposition method for linear parameters and the Simplex method for nonlinear parameters."
What I understand from it is that the non-linear parameters are optimized iteratively and for each choice new optimized linear parameters have to be computed. In other words, the full optimization of all parameters therefore entails a linear optimization nested in a non-linear one.
I'm planing to implement this on my own code, although How can I justify this procedure?, is there any reference or book that you could recommend me for reading??
Lets take, for instance, the Buckingham potential
$\Phi _{{12}}(r)=A\exp \left(-Br\right)-{\frac  {C}{r^{6}}}$

for non-linear, dependent parameters you may consider global optimization techniques (genetic algorithms, monte carlo basin hopping, particle-swarm, etc.)

It has been suggested that this question might be appropriate on [scicomp.SE].  I have no problem with moving it there, but I will give you the choice.


text:
How does sulfuric acid react to heating?
How would $\ce{H2SO4}$ react to heat? I assume that it would form toxic sulfur (di/tri)oxide, but I can't find any information on this.
I'm concerned about fire hazards specifically, so assume very intense heating and very low concentration.

"very low concentration"? Then that's slightly acidic water and there's  not much difference what acid is there.

Sulfuric acid, $\ce{H2SO4}$, having an enthalpy of formation of $-814\ \mathrm{kJ/mol}$, is quite stable and won't decompose easily.
According to A Kinetic Study of the Decomposition of Spent Sulfuric Acids at High Temperature, Dominique Schwartz, Roger Gadiou, Jean-François Brilhac, Gilles Prado, and Ginès Martinez:

The decomposition of $\ce{H2SO4}$ to $\ce{H2O}$ and $\ce{SO3}$ is predominant between $400$ and $700\ \mathrm K$. The formation of a small amount of gaseous sulfuric acid can be observed. Above $673\ \mathrm K$, the equilibrium constant of the reaction R1 becomes higher than 1 and increases rapidly.
$$\ce{H2SO4 <=> H2O + SO3}\tag{R1}$$
The second process is the reduction of sulfur trioxide to $\ce{SO2}$. This endothermic reaction needs a high temperature to take place, the equilibrium constant of the reaction R2 being higher than 1 above $1050\ \mathrm K$.
$$\ce{SO3 <=> SO2 + \dfrac12O2}\tag{R2}$$

So, you only have to worry when the temperature reaches $400\ \mathrm K$, or $127\ \mathrm{^\circ C}$ (false precision).

@YdobEmos Usually available sulfuric acid is dissolved in water (except sulfuric acid in which case there is little water), so the temperature would be limited by the water.

In aqueous solution, you also have to worry about the water boiling and spitting acid at you. A speck of hot sulphuric acid on a mucus membrane is not a pleasant experience.

First, you don't need to worry about fire: sulphuric acid is non-flammable.
Around $1000$ kelvin, this reaction would occur:
$$\ce{2H2SO4 <=> 2SO2 + 2H2O + O2}$$
This is industrially used for the production of hydrogen gas, and the total reaction cycle is called the sulphur-iodine cycle.


text:
What exactly is a mole?
Upon coming back to University I've started studying basic chemistry again. I've always had trouble with the concept of "moles". According to my study book,

1 mole of any element or compound is equal to its molecular weight in
  grams.

I understand that part but later in the paragraph it also says:

One mole of any substance always contains exactly the same number of solute particles, that is, $6.02 \times 10^{23}$ (Avogadro's number). So whether you weigh out 1 mole of glucose (180 g) or 1 mole of water (18 g) or 1 mole of methane (16 g), in each case you will have $6.02 \times 10^{23}$ molecules of that substance.

If 1 mole equals the molecular weight of a compound, then how come 1 mole of glucose doesn't equate to 1 molecule of glucose? Basically, why is 1 mole of glucose $6.02\times10^{23}$ molecules instead of 1 molecule?

related http://chemistry.stackexchange.com/questions/53751/why-convert-to-moles-for-stoichiometry-problem-sets

one mole of glucose can not be equal to one molecule of glucose because one mole glucose weigh 180 gram while one molecule of glucose weigh 180u. Read the definition again.

because the molecular weight of a compound is the weigth of one mole of compound. Also, could you imagine if a single molecule of O2 weighted 32g? how would you breath it?

Why are there 12 eggs in a dozen eggs, and not 1?

I can't believe nobody's pointed out that the definition is actually wrong.  "a mole" is never equal to any weight.  The definition should be "1 mole of any element or compound has a mass equal to its molecular weight in grams."

A type of rodent, maybe?

Wanted to share that $2^{79} = 6.04*10^{23}$, and thus remembering "Six point oh-two times ten to the power of twenty three" is far more difficult but not much more accurate than remembering "Two to the power of seventy nine".

The first thing to realize is that "mole" is not a mass unit. It is simply a quantity - a number - like dozen or gross or score. Just as a dozen eggs is 12 eggs, a mole of glucose is $6.02 \times 10^{23}$ glucose molecules, and a mole of carbon atoms is $6.02 \times 10^{23}$ carbon atoms. "Moles" are only associated with mass because individual objects have mass, and thus a mole of objects also has a certain mass.
So why $6.02 \times 10^{23}$? What's so special about Avogadro's number? Well, nothing, really, it just makes calculations work out nicely. Avogadro's number is defined as the number of $\ce{^{12}C}$ atoms which weigh 12 g. So it's effectively a ratio: how many times larger is a gram than an atomic mass unit? If you have one atom of $\ce{^{12}C}$, it weighs 12 amu. If you have $6.02 \times 10^{23}$ of them, they weigh 12 g. If you have one molecule of glucose, it weighs 180 amu (or thereabouts). If you have $6.02 \times 10^{23}$ of them, they weigh 180 g (or thereabouts). -- This is just like if one egg weighs 60 g (on average), then a dozen of them weigh 720 g (on average), and if one cup of flour weighs 120 g (on average), then a dozen cups of flour weigh 1440 g (on average). The only difference is that dozen is defined forwards ("a dozen is twelve"), whereas a mole is defined "backwards" ("a dozen is the number of 60 g eggs that are in a collection of eggs that weighs 720 g.")
This convenience definition - pegging the value of Avogadro's number directly to the difference in scale between the amu and the gram - is what is probably throwing you. Moles are not a mass unit, but the definition is intimately tied to mass units. The equivalence in numbers at the atomic scale (amu) and at the macroscopic scale (grams) can also result in chemists playing fast and loose with terminology, quickly working back and forth from atomic to macroscopic scale, without a necessarily clear distinction between the two.

Note that the SI definitions are to be revised with Avogadro’s number to be turned into a fixed-value constant. That would obsolete the ‘$12~\mathrm{g}\ \ce{^{12}C}$’ definition. Timeframe: Soon™.

This was a brilliant explanation, thank you! I have a better grasp on the concept now.

This also means you can have a mole of moles. https://what-if.xkcd.com/4/

Following up on Jan's remark, you can read about the proposed SI changes in the Wikipedia article Proposed redefinition of SI base units

The second paragraph is not quite right. First the $\ce{^12C}$ atom is defined to have an atomic mass of exactly $12.00\bar{0}$ amu. Then for a $12.00\bar{0}$ gram sample of $\ce{^12C}$ the experimentally counted number of carbon atoms is Avogadro's number.

Please note that the atomic mass unit (amu) is the obsolete unit that referred to the relative atomic mass of oxygen which was taken as 16 (and unfortunately physicists used the atomic mass of the nuclide $\ce{^16O}$ whereas chemists used the average atomic mass of natural oxygen). The current unit (since 1961), which is defined as 1/12 of the mass of a neutral atom of the nuclide $\ce{^12C}$ in the ground state at rest, is the unified atomic mass unit (symbol: u).

Perhaps a hardware analogy might help.  Let's say you're in the business of putting things together using nuts, bolts, and washers. For the sake of argument, let's say that whenever you join something you always need one bolt, two washers, and one nut.  Oh, and you're going to be bolting together a lot of stuff.
So, you're going to need a lot of nuts, bolts, and washers, but that's okay, the hardware store sells them by the pound.
The bolts have a mass of 20 carats, the nuts are 5 carats and the washers are 2 carats.  What's a carat?  Doesn't matter really because you don't know how much stuff you're going to be joining - just "a lot".
So, you go to the hardware store and you buy 20 pounds of bolts, 5 pounds of nuts and 4 pounds of washers.
Now, you have no idea how many nuts, bolts and washers you have, but you do know that you have the same number of nuts as you have bolts and you have twice as many washers. So long as you convert the mass of the item in carats to the weight you buy in pounds, you can always make sure you get the items in the proportions you want.
Down the track a bit, you find out that there are 6,000 carats to the pound.  That means 20 pounds of bolts is actually 6,000 bolts, and 5 pounds of nuts is also 6,000 nuts, and 4 pounds of washers is 12,000 washers.
So, in this analogy, a lot (aka a "mole") of items is 6,000 or 6 x 103
If you relocate to a country with a metric system and their hardware stores sell the same bolts, nuts, and washers but by the kilogramme (heresy!), are you in the least bit phased?  Nah.  You buy 20 kg of bolts, 5 kg of nuts and 4 kg of washers.  Now you have 2.2 x 6,000 lots of joiner uppers. So, you're still in business.
The idea with atoms is basically the same, when you convert the mass in amu (or the dalton as I believe is preferred now) to grammes, then the mole is NA or Avogadro's number 6.02 x 1023.
The precise definition of exactly what is an amu (Da) is a fine point, important, but tangential to the concept of the mole which is what you're struggling with.  Interestingly, and equally tangentially, Avogadro's number may well be used to redefine what a kilogramme is:
https://www.youtube.com/watch?v=ZMByI4s-D-Y

Analogies should be banned as answers

Atoms combine on a particle by particle basis. That is, by numbers of particles.
Laboratory and industrial operations rely on measuring masses.
The mole is the bridge between these. It allows us to "count out" a particular number of atoms by massing the substance.

The first point is that it is a definition. In practical terms it enables us to have a convenient unit to express concentrations as $\pu{mol dm^{-3}}$ without having to use huge numbers.  More importantly it allows us to know how many atoms molecules etc. are in a given mass of substance.
A mole is defined as the amount of substance, n, that contains as many objects (atoms, molecules, ions for example) as there are atoms in 12 grammes of carbon-12. This number has been determined by experiment and is approximately $6.02214~10^{23}$. This is Avogadro's number.
If a sample has N atoms or molecules then the amount of substance it contains is $n=N/N_A$ where $N_A$ is Avogadro's constant and is $6.02214~10^{23} \pu { mol^{-1}}$. Thus $\pu {1 mol}$ contains $6.02214~10^{23}$ atoms , molecules or ions etc.
It follows that
The mass of one mole of a substance equals its relative molecular mass expressed in grams,
thus  18 g of water (for simplicity using O = 16;  H = 1 instead of exact masses) or  78 g of benzene contains  Avogadro's number of molecules, similarly 32 g of sulphur, 200g of mercury,  all have Avogadro's number of atom.

A mole is a quantity of discrete objects, not a dimension for measuring one thing. In the same way that a dozen objects is 12 or a score of objects is 20, a mole of objects is about $6.02214 \times 10^{23}$.
Technically, we shouldn't say "1 mole of glucose". We should instead say "1 mole of glucose molecules". The same goes for any other object: "1 mole of water" should be "1 mole of water molecules", and so on.  If we want to speak of raw elements, then we'd be talking about a mole of atoms of that element, rather than a mole of molecules, but the principle is the same. By convention, we usually omit that "molecules" part, but if we really want to be strict about things we should say it.
Avogadro's Number was chosen so that the mass of a mole of molecules (of the same thing) happens to equal $x$ grams, where $x$ is the molecular mass of that thing. Originally, it was defined so that 1 mole of hydrogen (which, remember, is really "1 mole of hydrogen atoms") weighs 1 gram. Nowadays, we define it so that one mole of Carbon-12 (again, "1 mole of Carbon-12 atoms") weighs 12 grams, but the proportions still work out, more or less. The folks behind SI are working on pinning this to a mathematical constant.

Since I rely on my chemistry textbook,so I am gonna provide you the knowledge about one mole I have gained through it.
Mole just indicate the amount of substance,it is SI unit of measurement of amount of substance.Three definitions can be given to a mole;

A mole is defined as that amount of substance which has mass equal to the gram atomic mass if substance is atomic or gram molecular mass if the substance is molecular.

For example,$H_2O$ have molecular mass of $18u$ thus its gram molecular mass is $18gram$. So you can say that $18$ gram of $H_2O$ is equivalent to $1$ mole  of $H_2O$.

2.A mole is defined as that amount of substance which contains Avogadro's number $(6.022140857×10^{23})$ number of atoms if the substance is atomic or Avogadro's number of molecules if the substance is molecular.

For example, $6.022140857×10^{23}$ molecules of $H_2O$ will form $1$ mole of $H_2O$.

3.In case of gases, a mole is defined as that amount of the gas which has a volume of $22.4$ litres at S.T.P ($101.3kPa$ and  $273.15 K$).

For example, At S.T.P $22.4$ litres of $CO_2$ gas will be said $1$ mole of $CO_2$.

Please note that you are using the old definition of STP ($T=273.15\ \mathrm K$ and $p=1\ \mathrm{atm}=101,325\ \mathrm{Pa}$) and the corresponding molar volume of an ideal gas of about $V_\mathrm m=22.4\ \mathrm{l/mol}$. This is acceptable since you explicitly mention the temperature and the pressure. However, the value of the standard pressure was changed in 1982 to $p=1\ \mathrm{bar}=100,000\ \mathrm{Pa}$. At this pressure, the molar volume of an ideal gas actually is $V_\mathrm m=22.710,947(13)\ \mathrm{l/mol}$.

YEah a agree,i suppose it was SATP right.Should i edit it.

No, I guess, you do not have to edit that. It’s ok since you explicitly define what STP is supposed to mean in your answer.

Okay Thanks,i will keep it in mind while writing answers in future

Mole is a measure of number. "Mole"-concept helps us to weigh or count definite-numbers of atoms (excluding the minute errors) just using a macroscopic pan-balance. If I know it is glucose, and take 180 gram of it ( since C6 H12 O6 has molecular weight 180); the sample contain 1 moles or avogadro numbers (N.A) or 6.023 * 10 ^ 23 molecules of glucose. Similarly, 18 gram H2O must contain 1 mole or 6.023 * (10 ^ 23) water molecules.
But how?
Let start with some macroscopic objects.
If we take 1 truck full of bricks and another truck full of brick-chips; the second truck would contain many times more brick chips than bricks in first one. Simple.
But if we took 1 truck brick (total mass of brick very heavy); and 1 bucket brick-chips (total mass of brick-chips much lighter than truckful bricks)... then? Perhaps the both will contain same (at least almost) number of particles.
I.e. if we take heavy objects in massive-amount, and lightweight objects in lighter amount; the number of objects of both cases, would tend to same.
In Quantitative terms:
 Say each brick-chip= 1 gram. 1 brick= 1000 gram.

Now, if we take 1234 * 1 gram brick-chips and 1234 * 1000 gram bricks; we'll get the same number of unit-particles in both case.
 Or if we take   X  *  1 gram brick-chips at a place,
 and  X * 1000 gram bricks at another place;
 in both case we would get the same number.

Molecular scale
1 atom of hydrogen (H) weigh 1 a.m.u. 1 molecule of Glucose ( C6 H12 O6 ) weighs 180 a.m.u.  1 molecule of H20  weighs 180 a.m.u.
  Now, 1 gram = 6.023 * (10 ^ 23) Dalton or N.A a.m.u.    (  *  )

So just like our above mentioned brick example;
1 gram of hydrogen (H) atom (or N.A amu H) , or 180 gram glucose (i.e.  N.A * 180 amu glucose ), or 18 gram water ( N.A * 18 amu water) would contain the same number of particle (Which is here 6.023 * (10 ^ 23) pieces or N.A. pieces due to relationship between gram and amu). This way 1 mole of any substance would contain NA number of molecules
Avogadro number (N.A.) in this example working the same way as X in previous brick example.
Here is a simplified diagram comparing real-life example and chemistry-example.

Advantage
We're given
1 C6 H12 O6 +  6 O2 = 6 CO2 + 6 H2O
The numbers at left side of compound's formula, or the stoichiometric coefficient;  is smallest possible number of molecules to complete a reaction.
we're given certain amount of glucose, and asked for amount of CO2 evolved after complete combustion in O2 ?
We can calculate that freshly from a.m.u. Larger calculation.
But mole concept help a lot since we can determine quantities of reactants and products directly in gram, from molecular formula and stoichiometric coefficients, without using any unit conversion between gram and a.m.u.
If we  could multiply the whole reaction
(180 amu + 6 * 32 amu = 6 * 44 amu + 6 * 18 amu for respective compound)
with NA (could run NA numbers of such reactions at a time)
i. e.
180 amu * NA + 6 * 32 amu * NA = 6 * 44 amu * NA + 6 * 18 amu * NA (respective compound)
or
1 mole glucose + 6 mole O2 = 6 mole CO2 + 6 mole H2O.
or
180 gram + 6 * 32 gram = 6 * 44 gram + 6 * 18 gram  (respective compound).
once we knew the fact instead molecule we could use mole; and we could easily write reaction for a mole parallel reactions. Then we convert between moles to gram using the chemical formula, we can easily determine the required or obtained amount of certain reactant or product. For a certain given value of 1 reactant or product
for say 2 H2 + O2 = 2 H2O. From  this reaction we easily conclude 2 mole of H and 1 mole of O2 forms 2  moles of H2O; or 2 * 2 g H2 and 32 g O2 makes 2 * 18 g H20. Now  using unitary method we could find an asked amount of reactant or  product from some given  amount of reactant or product.
references:

(  *  )  : http://chemistry.bd.psu.edu/jircitano/mole.html
Google Unit-conversion Result
Avogadro constant in wikipedia
Atomic Mass unit in Wikipedia
Mole (unit) in Wikipedia
Atomic mass in wikipedia
Molecular mass in wikipedia
Molar mass in Wikipedia
What we've been taught in basic chemistry classes


text:
Why is thionyl chloride preferred for preparing alkyl chlorides from alcohols?
Why is thionyl chloride ($\ce{SOCl2}$) preferred for preparing alkyl chlorides from alcohols, as opposed to say $\ce{PCl5}$ or $\ce{PCl3}$ ?

A follow-up question containing another reason can be found at If SOCl2 reacts with alcohols via SNi, why doesn't POCl3?.

Thionyl chloride is preferred for preparing alkyl chlorides from alcohols because the by-products formed in the reaction are $\ce{SO2}$ and $\ce{HCl}$ which are in gaseous form and escape into the atmosphere leaving behind pure alkyl chlorides.
$$\ce{CH3CH2-OH  + SOCl2   →  CH3CH2-Cl  +  SO2 ↑ + HCl ↑}$$

A comment about the possible side-products when using phosphorus pentachloride would be useful too.

add pyridine to your reaction vessel and see the jump in yield!

Preparation of  alkyl chlorides from alcohols by thionyl chloride is preferred over other methods mentioned because of the following reasons :
$$\ce{R-OH  + PCl5  →  R-Cl  +  POCl3 + HCl ↑}$$
$$\ce{3R-OH  + PCl3  →  3R-Cl  +  H3PO3}$$
$$\ce{R-OH  + SOCl2   →  R-Cl  +  SO2 ↑ + HCl ↑}$$

Both the byproducts formed in the last reaction i.e., $\ce{SO2}$  and  $\ce{HCl}$ are gases and can easily escape out into the atmosphere, so the equilibrium shifts more to the right side and hence, more products are obtained.
Moreover as both the gases easily escape out leaving behind pure alkyl halide in the last reaction, the purification process is minimized and thus the alkyl halide obtained can be easily purified.


text:
Identifying an unknown white oxidizing powder
I am working with a unknown white oxidizing powder that melts at around 500 degrees Fahrenheit. It turns my finger green when it comes into contact with my finger for a short while. It leaves a very soft almost slimy feeling even after washing it off with normal hand soap.
When melted it becomes a dark purple color And will turn into a solid very quickly when in contact with something cool. It has on the box it came in a diamond shaped yellow sign with the word OXIDIZER then the number 5.1 which means it is strongly corrosive.
It also seems to corrode rubies and sapphires when in this liquid/powder for about 40-45 minutes in 500 degrees Fahrenheit. So I would like to know if anybody has any idea what I am dealing with? Because the company that send this powder to my employer this powder doesn't seem so happy about giving up their secret and says it comes from the dead sea.

A picture of the box/substance might help. Does the box have any more information, and where did you get it from?

Ask them for the SDS, which they are obliged to provide

Do you have any kind of mass spec data?

Please wear gloves and avoid contact with skin.

Would it dissolve in water?

It may be a mixture of two or more compounds. From the description you gave it clearly contains something strongly alkaline, possibly sodium hydroxide. The melting point is about right, and it would indeed attack a ruby or sapphire. The purple color, however, does not fit so it suggests that there is a second compound involved (some oxidizer).


text:
What makes trans fats more harmful than saturated ones?
Why are by many trans fats considered to be worse for you than fully saturated fats?
I am asking specifically which of their structural features accounts for this condemnation. They won't form clumps any larger nor will their melting temperature be higher than that of saturated fats, unless I'm mistaken, so it must be due to the double bonds' reactivity in trans configuration. But is it different from that in cis one?
It's sometimes suggested that trans fats increase low-density lipoprotein (LDL) and decrease high-density lipoprotein (HDL) - could that solely result from the properties of the trans configuration of the molecules?

I'm sure i've seen this question a dozen times... Anyway: Trans-fats are thermodynamically more stable, and thus less useful synthetically. That's why evolution has mostly gone for the cis-fatty acids.

Related: When disqualifying trans fat, are we qualifying cis-fat as healthy?

First of all, it's important to distinguish between the types of fatty acids that may form the cell's membrane:

Note that both cis- and trans-fatty acids are unsaturated, the difference being the conformation around the double bond. A trans-fatty acid is more-or-less a straight line, whereas the cis-fatty acid has a noticeable "kink" at the point of the double bond.
Now, the suggested molecular mechanism behind the health risks associated with trans-fatty acids has to do with a property of the cell's membrane called membrane fluidity. This property affects the cell's ability to send and receive signals, and it greatly affects the function of many proteins whose structure and role are dependent on the membrane's state. The membrane has to be fluid enough in order to serve its purpose properly.
(If you don't have time to read a lengthy answer, you can skip to the last paragraph now. Otherwise, keep going...)
One of the main functions of the cell's membrane is to separate the inside of the cell from the outside world, in order to control the cell's internal environment, help it avoid damages from the surrounding environment and deal with harmful substances or unfavorable living conditions. However, the cell needs to communicate with the outside world through different signaling mechanisms which involve sending and receiving chemical signals, as well as adapt itself to different living conditions.
In order to do that, the cell has to have a membrane with a dynamic structure, i.e. one that will be able to adjust itself to the surrounding environment. Evolution decreed that this may be best achieved by a lipid bilayer.
This image shows a portion of a cell membrane, which is made of many fatty acids in a formation that is considered to be the most stable energetically, due to the hydrophobic/hydrophilic interactions of the fatty acids among themselves and with the surrounding water. This image also shows how cholesterol is embedded here and there within the bilayer.
The measure of a membrane's fluidity results from the type of fatty acids that form it and the membrane's cholesterol content.
To put it simply, the more straight fatty acids in the membrane, the more rigid (i.e. less fluid) it will be. The more "interruptions" to the hydrophobic interactions between the fatty acids' tails (by "kinks" in the tails or cholesterol), the more fluid the membrane will be.

The above picture shows two distinct states of the membrane. On the right-hand side the membrane contains mostly straight (i.e. saturated and/or trans) fatty acids, which results in a more rigid and viscous membrane. This state makes it more difficult for the cell to communicate with its surroundings (chemical signals cannot enter the cell easily, proteins in the membrane are less mobile, etc.).
On the left-hand side, the membrane contains more unsaturated fatty acids of the cis conformation. The "kink" that characterizes cis-fatty acids is the main reason this membrane is more fluid - it makes it harder for the fatty acids' tails to stick together due to hydrophobic attraction.
Cholesterol serves the same purpose as those "kinks". by disrupting the attraction between fatty acids' hydrophobic tails. This is a very important role of cholesterol in virtually all of the cells in the body, which makes cholesterol a very important molecule in the body.
To summarize, trans-fatty acids are considered unhealthy because when consumed, they will mostly remain intact (i.e. will not be metabolized; as this form of fatty acids is not found in nature, the body's enzymes have a limited ability to catabolize them), and then they can penetrate cellular membranes, form rigid hydrophobic bonds with the other fatty acids and decrease the membrane's overall fluidity.
See Diet and HD - article from Stanford university.
Other references may be found in most introductory biochemistry texts or online (be sure to read only scientific texts from reliable websites such as the one given above).

So the difference is that saturated acids, while also forming rigid structures, will not do so on such a large scale because they are easier to metabolize (owing to their common occurrence in the human diet)?


text:
How do electrolytes allow electrolysis to occur?
From what I know, when two electrodes are places in a electrolyte (say $\ce{H2SO4}$, which decomposes into $\ce{H}^{+}$ and $\ce{SO4}^{2-}$ in aqueous solution), the negative ion is attracted to the positive electrode and is oxidized while the cation is attracted to the negative electrode and is reduced.
This transfer of electrons from the negative electrode to the cation and from the anion to the positive electrode "completes" the circuit and electricity is conducted through the wire.
What I don't understand is how the electrolyte allows for the electrolysis of water, since the electrons are used in the redox reaction of the ions in the electrolyte in which water isn't even involved.

Hydrolysis and electrolysis of water are quite different, nearly unrelated things. Which one are you interested in?

@IvanNeretin I meant electrolysis (sorry)

You mean to ask that how does an aqueous electrolytic solution also contain H+ and OH- ions?

OK, then what would be the ultimate result of electrolytic oxidation of $\ce{SO4^2-}$?

@EdmDroid no I mean to ask how adding an electrolyte to allows electrolysis to occur since the e- are used to redox the ions in the electrolyte instead of breaking the bonds between the atoms of h2o

@Ivan Neretin SO4?

But there is no such compound, so it might exist only momentarily, and then it must decay somehow.

In the electrolysis of $\ce{H2O}$, a small amount of sulphuric acid is added. Why? Because pure $\ce{H2O}$ doesn't ionize much on its own ($K_w=10^{-14}$ approximately). So traces of an electrolyte ($\ce{H2SO4}$ in this case) is added so that it can help break one $\ce{O-H}$ bond and $\ce{H}^{+}$ released forms a coordinate bond with one $\ce{H2O}$ to form $\ce{H3O}^{+}$ ion. So now the ions present in the solution are: $\ce{H3O}^{+}$, $\ce{OH}^{-}$ and $\ce{SO4}^{2-}$. Of these, the oxidation potential of $\ce{OH}^{-}$ is higher than that of $\ce{SO4}^{2-}$ so at the anode $\ce{OH}^{-}$ gets oxidized to $\ce{O2}$. In this case only one cation goes to the cathode so it gets reduced to $\ce{H2}$. In case you use an electrolyte that gives a cation $\ce{M}^{n+}$ other than$\ce{H}^{+}$ then $\ce{H2}$ would be formed only if the reduction potential of $\ce{M}^{n+}$ is lower than that of $\ce{H3O}^{+}$. So, if the electrolyte used is $\ce{ZnSO4}$ then $\ce{H2}$ will be formed but if $\ce{CuSO4}$ is used then $\ce{H2}$ is not formed, but $\ce{Cu}$ is deposited at the cathode.
So the ionization of water doesn't have anything to do with the electrons used in the redox process. The ionization depends simply on the electrolyte used.

Why we need ion movement (electrolyte) in first place? Why water molecules don't oxidized/reduced at the anode/cathode respectively when no electrolyte is used?

@ Anton It's just much slower in the absence of extra electrolyte because the concentrations of hydrogen and hydroxide ions are so small, see this answer.


text:
Breaking Down PCB's in lakes
I was wondering if there is any natural process which breaks down PCBs (Polychlorinated biphenyl).
This is an extremely harmful contaminant that was pumped into various lakes by chemical plants up until about the 1950s. It is very stable, which is why it was used a coolant in these chemical plants. However this toxic substance has greatly impacted lake systems, as the concentration of the PCB is increased in each trophic level of the lake ecosystem (due to biomagnification). Anyways, I was wondering if there was any method to break down the PCBs.
Since they are a chlorine-containing compound, would I have to use something like a sodium or potassium based compound to safely break them down chemically. What do you guys think? Thanks!

Have a look at advanced oxidation processes, such as Fenton and photo-Fenton reactions, as well as photochemical degradation processes using titanium dioxide.

Thanks so much. This is a good idea. I will look into this and post results!

The problem is complex. PCBs would be easy to chemically decompose if you could isolate them. But some chemical that you'd put in a lake or river would have to be at high concentration and be more toxic than the PCBs. If you try to dig up the sediment in the Hudson River for 100 miles that is just an impossible task. That frankly is why we need the EPA. The only real answer is to prevent such pollution in the first place.

UV irradiation is supposed to be effective, unfortunately dissolved oxygen is a problem there.

But yea, the general problem is that the PCB are mostly in the sediment, not in the water.

According to this abstract$\mathrm{^{[1]}}$, both aerobic and anaerobic processes play a role in the natural degradation of PCB's.

Studies have identified two distinct biological processes capable of
  biotransforming polychlorinated biphenyls (PCBs): aerobic oxidative
  processes and anaerobic reductive processes. It is now known that
  these two complementary activities are occurring naturally in the
  environment. Anaerobic PCB dechlorination, responsible for the
  conversion of highly chlorinated PCBs to lightly chlorinated
  ortho-enriched congeners, has been documented extensively in the
  Hudson River and has been observed at many other sites throughout the
  world. The products from this anaerobic process are readily degradable
  by a wide range of aerobic bacteria, and it has now been shown that
  this process is occurring in surficial sediments in the Hudson River.
  The widespread anaerobic dechlorination of PCBs that has been observed
  in many river and marine sediments results in reduction of both the
  potential risk from and potential exposure to PCBs. The reductions in
  potential risk include reduced dioxinlike toxicity and reduced
  carcinogenicity. The reduced PCB exposure realized upon dechlorination
  is manifested by reduced bioaccumulation in the food chain and by the
  increased anaerobic degradability of these products.

1) Aerobic and anaerobic PCB biodegradation in the environment, Environ Health Perspect. 1995 Jun; 103(Suppl 5): 97–99


text:
Which measurement should I use?
In lab, I gathered everything and my partner weighed everything, I didn't realize til now that there is a problem. She measured the copper and the crucible separately, she got 0.89 g for the copper and 11.63 g for the crucible. Those two numbers combined equal 12.52 g. However, when we put the copper in the crucible it weighed 12.54 g.
Which number should I use for the calculation, 12.52 g or 12.54 g?

You use error bars and calculate error propagation.

Given that a 2-place balance is no more accurate than $\pu{+/- 0.01 g}$, you can say that the two measurements were within $\pu{0.01 g}$ of the average, $\pu{12.53g}$, then just use this average for your calculations.


text:
What role does glycerin or sugar play in bubble formula?
From: http://bubbleblowers.com/homemade.html

Exploratorium Bubble Formula
  from the Exploratorium web site
  2/3 cup Joy dishwashing soap
  1 gallon water
  2 to 3 tablespoons of glycerin (available in pharmacies or chemical supply houses)
Cyndi's Bubble Recipes from the Nathan's Wish web site 1/2 cup of
  dishwashing liquid (Dawn or Joy) 2 cups of water 2 teaspoons of sugar

What role does glycerin or sugar play in bubble formula?  Are sugar or glycerin better than each other in some way w.r.t this formula?

From:
https://www.scientificamerican.com/article/bring-science-home-best-bubbles/

The solution with only water and detergent probably made smaller,
  shorter-lived bubbles compared to the solutions with glycerin or corn
  syrup. Bubbles burst when the layer of water molecules between the
  detergent molecules evaporates. But Glycerin and corn syrup form weak
  bonds with the water molecules and slow down the evaporation process,
  thus improving the life span and durability of the bubble. Glycerin
  makes stronger, longer-lasting bubbles, but corn syrup is often
  substituted in bubble solutions because it is cheaper. Can you find
  other substances around the house that can be added to water to make a
  bubble solution? If you try products such as shampoo or liquid hand
  soap, you can check their ingredients to see what might be helping to
  make the bubbles form.


text:
How to Determine An Element's Colour
Different elements have different colours. Some elements may have similar colours like Sodium (Na) and Potassium (K), so that's why i'm asking "is there a way to calculate or even simply deduce the actual colour (not spectral lines) of an element based on its properties?"

Consider the energy required to elevate electrons from one "shell" to another, and the frequency of visible light that has this amount of energy.  The substance is more likely to absorb wavelengths of light matching the energy difference of that band gap, and reflect or transmit others.  The colors NOT absorbed can reach your eye as an observer, so (assuming a white light source) you see white (which is all the colors together) minus the colors absorbed by the material (matching the band gap).  For example, gold (Au) absorbs the blues and greens and leaves just the colors adding up to what you now think of as "gold."  If your light source is not white, what your eyes see will also be missing any colors your light source was missing.
The crystal structure can also make a difference, e.g. compare the colors of carbon (C) as graphite vs. diamond.
Useful links: See also this same question on ResearchGate, WebExhibits, and Quora.


text:
Which hydrogen halides react with group 14 oxides?
$$\ce{6HF +SiO2->H2SiF6 +2H2O}$$
$\ce{HF}$ reacts with $\ce{SiO2}$ to form $\ce{H2SiF6}$.

Do other hydrogen halides like $\ce{HCl}$ and $\ce{HBr}$ react with group 14 oxides in the same way?

Are you sure it's even useful to ask for the reactivities for all of group 14? The reactivities of lead and carbon are extremely different.

@ringo But still there must be some sort of periodic trend.

All reactions taken from this site.
First some maths: 4 hydrogen halides can react with 5 Group 14 monoxides and dioxides. So, there should be 40 different chemical reaction combination possible. But not all reactions are feasible.
HF

With $\ce{SiO2}$ (already mentioned)

$$\ce{GeO2 + 4HF → GeF4 + 2H2O}$$

$$\ce{SnO + 2HF ->[333 K] SnF2 + H2O}$$

HCl

$$\ce{GeO2 + 4HCl(g) ->[450-500 C] GeCl4 + 2H2O}$$

$$\ce{GeO2 + 4HCl(l) ->[170-180 C] GeCl4 + 2H2O}$$

$$\ce{SnO + 3HCl → H[SnCl3] + H2O}$$
$$\ce{SnO2 + 6HCl → H2[SnCl6] + 2H2O}$$
$$\ce{PbO + 2HCl → PbCl2 + H2O}$$
$$\ce{PbO2 + 4HCl → PbCl2 + Cl2 + 2H2O}$$

$\ce{HBr}$ and $\ce{HI}$ do not show such type of reaction.

Other posters have already commented on the thermodynamics of the reaction, but I will just point on the fluoride has a very high affinity for silicon, more so that oxygen. This is driving the thermodynamics of your reaction.
Fluoride is commonly used to remove silicon-based protecting groups for alcohols because it will displace the oxygen when other halides will not.


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Trends in the melting points of alcohols, aldehydes and ketones
Is there a general trend in the melting points of alcohols, aldehydes and ketones? Usually with the boiling point, it increases as the number of hydrocarbons increase, but what happens in the case of melting point?

Welcome to Chemistry StackExchange. Normally its best to try find the answer yourself first then ask a question based on this data that you still find puzzling.

Well, the trend in both melting and boiling point somewhat remain same. But the trend may alter due to reasons like hydrogen bonding, symmetrical structure leading to better packing in crystal lattice, increased stability due to some reasons, structure, etc.


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Effects of Changing Avogadro's Constant
The Avogadro project suggests that we redefine the Avogadro constant to be equal to our best known estimate, $N_\mathrm{A} = 6.02214179 \times 10^{23}$, and redefine the kilogram based on the Avogadro constant.
What happens if we define the Avogadro constant to be exactly $6 \times 10^{23}$ and reassign atomic weights correspondingly?
Would any unforeseen complications arise from here?
Example:
Under current definitions, $N_\mathrm{A}$ atoms of carbon-12 weigh 12 g. If we redefine the constant to have the value $N_\mathrm{A}^\prime = 6.0\times 10^{23}$, then $N_\mathrm{A}^\prime$ atoms of carbon-12 would weigh $\pu{11.955879 g}$, which we would have to assign as the new atomic mass of carbon-12.

Not exactly what you're asking about (it's about SI redefinition of the kilogram based on Planck constant rather than Avogadro), but a good read: http://www.bipm.org/en/measurement-units/new-si/faqs.html

It is an interesting question. But you should also ask whether there are any benefits at all of defining the avogadro constant as a round number. Other physical constants have not aimed for that and I can't think of any practical benefits of doing so.

That would mess with almost every other constant that we use in thermodynamics... Every standard energy table would need to be recalculated.

Why stop there? Let's just ditch the concept of moles and only refer to exaparticles (EP). By definition one EP of carbon-12 contains 10^24 atoms, and it weighs 19 grams. Let's also get rid of Coulombs and count charge and current in EPs. Volts should be replaced by Joules per EP.

@CurtF. - Well... the SI prefix "exa" means $10^{18}$. The SI prefix for $10^{24}$ is yotta. So $10^{24}$  particles would be yottaparticles.

You're right @MaxW -- oops.  Viva la yottaparticles (YPs)!  We can pronounce it "yips" to be cool.

As of May 20th 2019, the Avogadro constant will be set to $6.02214076 \times 10^{23}\ \rm{mol}^{–1}$, i.e. the Avogadro constant times one mole will be an integer, but not the one the OP asked about. In preparing for the switch of definition, the BIPM looked carefully at two ways of determining the constant, and then set it to the best estimate from experimental values, truncated to nine significant figures.
This way, there will be minimal practical changes of units even though the definitions have changed. Also changed is what is defined vs. what is measured. For example, the molar mass of 12-C is now based on measurement, and could change (certainly in terms of how many significant figures are known) as measurement techniques improve.
Setting the Avogadro constant to a dimensionless number as the OP asked about would make lots of relationships between quantities and constant dimensionally incorrect, such as $$ R = k_\mathrm{B} N_\mathrm{A}$$
with $k_\mathrm{B}$ the Boltzmann constant, R the universal gas constant and $N_\mathrm{A}$ the Avogadro constant.

Would any unforeseen complications arise from here?

Setting Avogadro's constant to $6 \times 10^{23}\ \rm{mol}^{–1}$ (I added the proper units) would have lots of complications, but they would not be unforeseen, as @Zhe discusses in the comments.