Why is Aluminum's 2nd ionization energy higher than Silicon's? According to the periodic table trend for I.E. Silicon should have a higher 2nd ionization energy than Aluminum so I'm confused.

Let us look at the electronic configurations of aluminum and silicon and their respective cations.

$\ce{Al}: [\ce{Ne}]\mathrm{3s^2\, 3p^1}$

$\ce{Si}: [\ce{Ne}]\mathrm{3s^2\, 3p^2}$

$\ce{Al^2+}: [\ce{Ne}]\mathrm{3s^1}$ It involves removing from $\mathrm{3s}$ sub shell

$\ce{Si^2+}: [\ce{Ne}]\mathrm{3s^2}$ It involves removing from $\mathrm{3p}$ sub shell

A $\mathrm{3p^1}$ electron has lower binding energy than that of $\mathrm{3s^2}$ electron.

It is due to more screening effect (as the predecessor sub-shells are higher in $\ce{Si}$) and higher stability attained by having a completely filled $\mathrm{3s}$ sub-shell


Using MestReNova: An Introduction Can anyone help me with a quick question regarding MestReNova 9.0. I have been fiddling with this thing all day and before you ask yes, I have consulted the users guide to no avail.

I want to do three things:

1)Rescale the chemical shift axis to focus only on the relevant sections of my spectra.

2)Set a threshold so that only peaks above a certain intensity will automatically label.

3)how on earth do you get the parameter box to fit onto the page? Ive managed to get it up but it is about three times larger than my spectra and rescaling using the little green buttons leaves it the same size but just cuts loads out.

Any help is much appreciated!

Testing this on version 9 will have to wait until tomorrow, but on version 8 (which I have in front of me, but which has limited differences to version 9 by memory)

1)Rescale the chemical shift axis to focus only on the relevant sections of my spectra.

Do a manual zoom.

->View >Zoom >Manual Zoom (or just type M, or click the manual zoom button)

Type in the values you want

2)Set a threshold so that only peaks above a certain intensity will automatically label.

Do Pick Picking via a Manual Threshold

->Analysis >Peak Picking >Manual Threshold (or just type K)

Left click and hold to define a box within which you will have your peaks picked

3)how on earth do you get the parameter box to fit onto the page? Ive managed to get it up but it is about three times larger than my spectra and rescaling using the little green buttons leaves it the same size but just cuts loads out.

You need to resize the text to a smaller font size

First select the parameters box.
Left click the parameters (which should have the green selected boxes shown) and choose properties. Make sure under the geometry tab, the Auto box is selected for size.
Then from the text toolbar (by default is usually at the bottom of the screen, but if not shown go to >View>Toolbars>Text) change the font size to something small enough to fit on the screen.

Try that for your version - if no luck I'll check on one of my other processing stations tomorrow which has version 9 installed. Let me know how you go.


what's the name of this org. compound. RR'-C=NNH2 What's the name of this compound? It's a semi-product of Wolff-Kishner reduction.

The complete reaction is:

The (now deleted) answer by NotCorey has already explained that the compound that is shown in the picture is a hydrazone that is derived from acetone (propan-2-one).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) for a hydrazone is generated substitutively as ‘ylidene’ derivatives of hydrazine rather than by functional class nomenclature as in previous recommendations.

  P-68.3.1.2.2 Hydrazones

  Compounds having the general structure $\ce{RCH=N-NH2}$ or $\ce{RR'C=N-NH2}$ are called ‘hydrazones’ and are named in two ways:

  (1) substitutively as derivatives of the parent hydride ‘hydrazine’, $\ce{H2N-NH2}$;

  (2) by functional class nomenclature using the class name ‘hydrazone’.

  Method (1) generates preferred IUPAC names.

Therefore, the PIN for the compound that is given in the question is (propan-2-ylidene)hydrazine.


Does the compound KFO₃ exist? As the compound potassium chlorate exists, is there something as $\ce{KFO3}$?
All the other halogenates are mentioned somewhere on the web but what about this?
If no, then why?

It has been already mentioned in the comments and the other answer, that fluorine is more electronegative than oxygen. Therefore it is highly unlikely that this compound exists.

To illustrate the problem at hand, take a look at similar, known compound with a fluorine oxygen bond: hypofluorous acid, $\ce{HFO}$, which is actually better described as hydroxylfluoride, $\ce{HOF}$. According to Holleman and Wiberg's[1,2] textbook the synthesis of this compound is already difficult. Low pressure fluorine gas is passed over ice at about $-40~^\circ\mathrm{C}$. Obtained is a mixture of $\ce{HF}$, $\ce{O2}$, $\ce{H2O2}$, $\ce{H2O}$ and also $\ce{HOF}$. Fractional freezing until about $-183~^\circ\mathrm{C}$ leads to a white substance, that turns pale yellow at about $-117~^\circ\mathrm{C}$, and boils at about $10-20~^\circ\mathrm{C}$. It decomposes via a radical pathway at about $25~^\circ\mathrm{C}$ and $100~\mathrm{mbar}$ with a half life of about 30 minutes to oxygen and hydrogen fluoride: $$\ce{2HOF -> 2HF + O2}.$$

The electronic structure of the molecule is most likely $$\ce{HO^{\delta+}\bond{-}F^{\delta-}}.$$

Salts, $\ce{M^{+}OF^{-}}$, that are derived from this compound are not known. Covalent compounds, however, are known, i.e. $\ce{O3ClOF}$ or $\ce{O2NOF}$.

It is easy to imagine, that a compound with an even higher oxygen load cannot be stable. Therefore we can conclude that a halogenate like $\ce{HFO3}$, let alone its salt $\ce{M^{+}[FO3]^{-}}$, cannot exist.

Notes and References

Unfortunately I only have access to the German version of this textbook. I tried to excerpt and translate the important parts to the best of my knowledge.
Holleman, A., Wiberg, N., Wiberg, E., et al. (2008). Lehrbuch der Anorganischen Chemie. Berlin, Boston: De Gruyter. Kapitel XII 4.2. Sauerstoffsäure des Fluors. S. 465.  (in german)
Translation of the 101st edition available as Egon Wiberg and Nils Wiberg (2001) Holleman, Wiberg. Inorganic Chemistry. Academic Press.


How does the human body deal with free radicals it creates itself? I know that free radicals are created all over your body all the time as a byproduct of its chemical processes, (or maybe I do not and that is false). However I am confused about the distinction between free radicals formed from UV photons that will affect DNA base pairings, and the ones that the human body creates on a consistent basis.

How does the body deal with the free radicals it created by itself?

And how are those free radicals different from the ones that cause oxidative damage (e.g., from UV rays)?

Some enzyme-catalyzed reactions produce free radicals as short-lived intermediates, but these are likely to be contained within the enzyme active site until they can react further to form product. Example: ribonucleotide reductase - which catalyzes the formation of deoxyribonucleotides and enables the production of DNA.

I can't think of any case where a biochemical process "intentionally" forms a free radical and then allows it to escape and cause damage elsewhere in the body. But biology is wonderfully varied and I would not want to say "never".


Linear combination of eigenfunctions In the textbook "Molecular Quantum Mechanics" by Atkins and Friedman [1, p. 11], it says in chapter 1:

  … a general function can be expanded in terms of all the eigenfunctions of an operator, a so-called complete set of functions.

Can ANY function be written as a linear combination of eigenfunctions of ANY operator? I think not.

What are the criteria here for which functions can be written as linear combinations of which operators?

References

Atkins, P. W.; Friedman, R. Molecular Quantum Mechanics, 5th ed.; Oxford University Press: Oxford ; New York, 2011. ISBN 978-0-19-954142-3.

The complete, formal mathematical answer is a bit complex. It might be better to get that discussion from Math Stack Exchange.

An answer to the world of Chemistry is more straightforward: Yes, for any physically relevant operator and any relevant function, you always can. That’s a fundamental property of quantum mechanics: (functional) states are superpositions of eigenstates.


Semi-Infinite Potential Square Well: Negative Potential I am having trouble with a problem involving a Semi-infinite potential square well:

I have written down some notes that I added to the post. I have problems understanding the physical situation. See the figure in the notes. When the potential is just shifted upwards, so that the $V(x)=0$ region is $0<x<a$, I have no trouble understanding the solution. However, I am just really confused about the negative potential.

As you will see in the notes, I have interpreted Region I as a bound state (e.g. an electron bound to a proton). Region II, I believe, is the free state (e.g. "bound" electron in energy level $n=\infty$), since $V(x) = 0$.  Is this correct?

I face some troubles with the boundary conditions, as one of my wave-functions simply disappear!

Answering your quantum mechanics question seems to become a hobby of mine :). First let me say that there is nothing wrong with shifting the potential up so that $V_{0} = 0$. This choice would indeed simplify the problem as it gets rid of negative energy values and thus avoids the following point of confusion: when $E < 0$ the wave function of a particle in region $\text{II}$ is not a plane wave $\mathrm{e}^{\mathrm{i} k x}$ ($k \in \mathbb{R}$) like that of a free particle moving in vacuum but a decaying exponential $\mathrm{e}^{-\kappa x}$ (with $\kappa \in \mathbb{R}$). So, the choice $V_{0} = 0$ is advisable. Shifting the potential back later is always possible and might be easier.

Nevertheless, of course the problem is solvable irrespective of the choice of $V_{0}$. I think your problem in this case is that you enforced an incorrect boundary condition. To avoid confusion I will start from the beginning:

I will only consider the case $E < 0$ in the following, so $E$ is always a negative number between $-V_{0}$ and $0$. For region $\text{II}$ you have to solve the Schroedinger Equation:

\begin{align}
  - \frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= E \psi_{\text{II}}(x) \\
  \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= - \frac{2mE}{\hbar^{2}} \psi_{\text{II}}(x) \ .
\end{align}

Ok, here comes the first "problem" with the choice of $V_{0}$: In your solution you chose $\frac{2mE}{\hbar^{2}} = k^{2}$ as you would naturally do for a free particle. But since $E$ is negative $k$ will be a complex number (and not a real number as it would be for a free particle) and so this differential equation does not have an oscillatory solution $\psi_{\text{II}}(x) = A \sin(kx) + B \cos(kx)$ with $k \in \mathbb{R}$. Rather you get a decaying exponential function by doing the following: $- \frac{2mE}{\hbar^{2}}$ will be a positive number so you simply set $- \frac{2mE}{\hbar^{2}} = \kappa^{2}$ to get

\begin{align}
  \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= \kappa^{2} \psi_{\text{II}}(x) \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .
\end{align}

Now, this differential equation has a solution of the form

\begin{align}
  \psi_{\text{II}}(x) = A_{\text{II}} \mathrm{e}^{\kappa x} + B_{\text{II}} \mathrm{e}^{-\kappa x} \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .
\end{align}

At this point you can enforce the boundary condition that the wavefunction $\psi_{\text{II}}$ must be square integrable over the whole interval $a < x < \infty$ (Note that you don't have to consider the interval $-\infty < x < \infty$ but only the interval on which $\psi_{\text{II}}$ is defined). Since $\lim_{x \to \infty} \mathrm{e}^{\kappa x} = \infty$ this mean that the prefactor $A_{\text{II}}$ must be equal to zero, so

\begin{align}
  \psi_{\text{II}}(x) = B_{\text{II}} \mathrm{e}^{-\kappa x} \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .
\end{align}

So much for region $\text{II}$. Let's move on to region $\text{I}$: Here you have to solve the Schroedinger Equation:

\begin{align}
  - \frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} - V_{0} \psi_{\text{I}}(x)  &= E \psi_{\text{I}}(x) \\
  \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} &= - \frac{2m(E + V_{0})}{\hbar^{2}} \psi_{\text{I}}(x) \ .
\end{align}

Again, care must be taken: Since $-V_{0} < E < 0$ the term $(E + V_{0})$ is positive and so $- \frac{2m(E + V_{0})}{\hbar^{2}}$ is a negative number and you can set $- \frac{2m(E + V_{0})}{\hbar^{2}} = -k^{2}$ which leads to

\begin{align}
  \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} &= - k^{2} \psi_{\text{I}}(x) \qquad \quad \text{with} \qquad k \in \mathbb{R} \ .
\end{align}

This differential equation has a solution of the form:

\begin{align}
  \psi_{\text{I}}(x) = A_{\text{I}} \mathrm{e}^{\mathrm{i} k x} + B_{\text{I}} \mathrm{e}^{-\mathrm{i} k x} \qquad \quad \text{with} \qquad k \in \mathbb{R} \ .
\end{align}

(You could also formulate an equivalent solution in terms of $\sin$ and $\cos$ functions using Euler's formula if that is more to your liking. But I prefer plane waves ;))
This is the point where you made your main mistake: You discarded one of the terms in your $\psi_{\text{I}}$ wavefunction because it blew up for $x \to \infty$. But $\psi_{\text{I}}(x)$ is only defined on the interval $0 \leq x \leq a$ and within this interval both exponential terms are perfectly square-integrable and you can't discard any of them.

Finally, you just have to remember that $\psi_{\text{III}}(x) = 0 \ \forall \ x$ and enforce the boundary conditions given by the system:

\begin{align}
  \psi_{\text{I}}(0) &= \psi_{\text{III}}(0) \quad &&\Rightarrow \quad A_{\text{I}} \mathrm{e}^{0} + B_{\text{I}} \mathrm{e}^{0} = 0 \quad &&\Rightarrow \quad B_{\text{I}} = -A_{\text{I}} \\
  \psi_{\text{I}}(a) &= \psi_{\text{II}}(a) \quad &&\Rightarrow \quad A_{\text{I}} \mathrm{e}^{\mathrm{i} k a} - A_{\text{I}} \mathrm{e}^{-\mathrm{i} k a} = B_{\text{II}} \mathrm{e}^{-\kappa a} && \\
  & &&\hphantom{\Rightarrow} \ \quad A_{\text{I}} \underbrace{\bigl(\mathrm{e}^{\mathrm{i} k a} - \mathrm{e}^{-\mathrm{i} k a} \bigr)}_{= \, 2 \mathrm{i} \sin(k a)} = B_{\text{II}} \mathrm{e}^{-\kappa a} \quad &&\Rightarrow \quad B_{\text{II}} = 2 \mathrm{i} \sin(k a) \mathrm{e}^{\kappa a} A_{\text{I}}
\end{align}

Edit:

Concerning your comment: You can indeed enforce another boundary condition. The probability
flux is required to be continuous across any surface, since otherwise the surface would
contain sources or sinks. This entails the requirement that both $\psi$ and $\nabla \psi$ must be continuous. The continuity condition $\psi_{\text{I}}(a) = \psi_{\text{II}}(a)$ has already been enforced but the one for the gradient is still left:

\begin{align}
  \frac{\mathrm{d} \psi_{\text{I}}(x)}{\mathrm{d} x} \bigg|_{x = a} &= \frac{\mathrm{d} \psi_{\text{II}}(x)}{\mathrm{d} x} \bigg|_{x = a} \ .
\end{align}

From the above discussion we know that:

\begin{align}
  \psi_{\text{I}}(x) &= A_{\text{I}} \underbrace{\left(\mathrm{e}^{\mathrm{i} k x} - \mathrm{e}^{-\mathrm{i} k x} \right)}_{= \, 2 \mathrm{i} \sin(k x)} = 2 \mathrm{i} A_\text{I} \sin (kx) = C_\text{I} \sin(kx)
\end{align}

and

\begin{align}
  \psi_{\text{II}}(x) &= \underbrace{2 \mathrm{i} A_{\text{I}} \sin(k a)}_{= \, \psi_{\text{I}}(x=a)} \, \mathrm{e}^{\kappa a} \mathrm{e}^{- \kappa x} = \psi_{\text{I}}(x \! = \! a) \, \mathrm{e}^{- \kappa (x - a)}
\end{align}

with $C_\text{I} = 2 \mathrm{i} A_{\text{I}}$, so

\begin{align}
  \frac{\mathrm{d} \psi_{\text{I}}(x)}{\mathrm{d} x} \bigg|_{x = a} &= \frac{\mathrm{d} \psi_{\text{II}}(x)}{\mathrm{d} x} \bigg|_{x = a} \\
  k C_\text{I} \cos(ka) &= - \kappa C_\text{I} \sin(ka) \\
  \kappa &= - \frac{k}{\tan(k a)}  \ .
\end{align}

While this equation looks innocent enough it is not easily solved for $E$ and you will need the assistance of something like Mathematica for that.


Numerical related to crystallisation of water below its freezing point Question:

  There is $16\ \mathrm g$ of pure water in a container at temperature $-20\ \mathrm{^\circ C}$. A small piece of ice is added to start crystallization. Amount of water in container when temperature reaches to $0\ \mathrm{^\circ C}$ is: (assume specific heat of water below $0\ \mathrm{^\circ C}$ is $1\ \mathrm{cal\ g^{-1}\ ^\circ C^{-1}}$.)

My attempt:

I have read this: Liquid water below freezing temperature. So, I understand that the small piece of ice is only added to provide a nucleation site to promote the freezing of water.

But, all physical processes must follow the Law of Calorimetry, right? So, for water below $0$ degrees to freeze, it must first accept heat from somewhere, to reach zero degrees. But, the container is at an even lower temperature, so it won't provide water with any heat (negative temperature gradient). We have also not been given the mass of ice added, so we can't calculate the amount of heat the water will gain from the ice (the latter is at a high temperature than water). The system is isolated, so there is no other possible heat source...

I am apparently at a dead end in this problem. I hope I have put everything in detail and correctly. What is the logical mistake in my thinking? And what is the correct way to approach this problem? Thank you!

In the question Crystallization begins after you add a small piece of ice
.
 So it has to begin that's been given in the question.

I don't know thermodynamically whether it's plausible or not, but it's a given in the question a logical alternative way to approach the question (according to me) is

 I formatted it as a spoiler just in case you want to think a bit more or anybody else does

   The temperature has increased because latent heat of fusion is being released i.e ice being formed from water is  releasing heat.

PS: I may be wrong


Reaction for the corrosion of iron Construct the equation for the reaction between iron(II) ions, hydrogen ions ($\ce{H+}$), and oxygen to form iron(III) ions and water.

The answer is $$\ce{4Fe^{2+}(aq) + 4H+(aq) + O2(aq)->4Fe^{3+}(aq) + 2H2O(l)}$$ Can you explain how to get this answer? Thanks.

Rusting of iron is an electrochemical phenomenon.
A certain site on the iron object acts as an anode.
$$\ce{(Fe(s) -> Fe^2+(aq) + 2e ) × 2}$$
Electrons released at this spot go to another site on the object and reduce oxygen in presence of $\ce{H+}$ ions which are either obtained from water out from acidic substances in water:
$$\ce{H2O -> H+ + OH-}$$
$$\ce{CO + H2O -> H2CO3}$$
$$\ce{H2CO3 -> 2H+ + CO3^2- }$$
This site behaves as cathode and reaction is
$$\ce{4H+ + O2 + 4e -> 2H2O}$$
Therefore the overall reaction is :
$$\ce{2Fe(s) + O2 + 4e -> 2Fe^2+(aq) + 2H2O}$$
Now the ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in form of hydrated ferric oxides.
$$\ce{Fe^2+(aq) + 2H2O + 1/2O2 -> Fe2O3 + 4H+(aq)}$$


Raise the pH of a solution and prevent build up of carbonates I need to raise the pH of a solution so the easiest approach is to add alkali hydroxides (NaOH, KOH, etc). But as most of you know, the CO$\ce{_2}$ in the air can react with these hydroxides to form carbonates which usually  precipitate out the other metal ion dissolved in the original solution. So what else are my options?

How about $\ce{NH4OH}$? This does not for carbonates as readily, but it does form complexes with many metals. It depends on your intended use.

It might be easier to keep things away from the air:

Use a closed container.
Float an inert liquid (e.g. paraffin oil) atop your solution, or just a layer of pure water, which might take a while to mix.
Use an inert gas over the container to displace air.


How can I clean solid sulfur and its oxides from test tube? I was experimenting with sulfur few years ago. The test tube I used for that purpose has still sulfur stuck onto it. I was searching for a way to dissolve it.

I found an idea to try warm acetone. Sounded like cheap solution, but doesn't work - or it will take ages to dissolve.

I'm not a chemist, so I don't have any strange chemicals at home.

Since you stated:

  I'm not a chemist, so I don't have any strange chemicals at home.

you will neither have facilities for their proper storage and  waste management.

Consequently, the only reasonable solution is: Get some new test tubes (about 0.20 Euro for a 160 x 16 mm tube), but don't risk your own health or pollute the environment by messing around with carcenogenic ($\ce{CrO3}$ in $\ce{H2SO4}$) or neurotoxic ($\ce{CS2}$) agents.


Solubility of Amides I was looking into how the solubility of amides compares to other compounds e.g. acids and amines, and came across these two statements on the wikipedia page for amides:
https://en.wikipedia.org/wiki/Amide#Hydrogen_bonding_and_solubility
"The presence of a C=O dipole and, to a lesser extent a N–C dipole, allows amides to act as H-bond acceptors. In primary and secondary amides, the presence of N–H dipoles allows amides to function as H-bond donors as well."
"Typically amides are less soluble than comparable amines and carboxylic acids since these compounds can both donate and accept hydrogen bonds."
I understand that amides can act as both H bond donors and acceptors, so I'm wondering how the second statement can make sense?
Additionally, would it be correct to say amides have greater solubility in water since they can form more H bonds in total compared to amines and acids?

Urea ($\ce{H2N-CO-NH2}$) is the quintessential amide. It is extremely soluble in water, and you could say it is a hydrogen donor and acceptor in water.
However, if you replace one of the $\ce{-NH2}$ groups with an alkane or other organic group, solubility is reduced, depending on the size of the group.
Amides generally have higher boiling points and melting points than related alcohols and acids - because of their unique ability to be internal hydrogen donors and acceptors. This donor-acceptor capability makes their solid state more stable and less able to be solvated to a greater degree in water. You could think of the internal bonds being like a dimerization, where the organic chain effect is doubled, while the double $\ce{-CO-NH2}$ interaction gains little benefit from solution in water.


Periodic trends: why is effect of protons greater than electrons? Why is it that adding protons has a greater effect than electron-electron repulsion on periodic trends like atomic radius and ionization energy (assuming # of shells constant)? It seems that if protons and electrons have the same charge and electrons are actually closer to the electrons in question (e.g. for ionization energy), that their effect would be the same?

This is because the attraction is not between each proton and it's corresponding electron, but between each electron and the nucleus. Since the number of protons increases the charge of the nucleus, but the charge of the electron stays constant, the attraction is greater. The attraction of the electron to the nucleus depends on the charge of the nucleus, and it is independent of the number of electrons in "orbit"

A great analogy for this is the solar system. (Gravity works very similarly to electrostatic force.) It won't matter if suddenly there is a mother earth orbiting in the outskirts of the solar system, Earth would be unaffected by that change, and we would still orbit the sun at the sae distance. However, if the sun suddenly becomes noticeably heavier, the planets will each be pulled in with a stronger force, which reduces the orbit radius. This will happen equally to each individual planet. It doesn't matter how many planets we have, The pull of the earth is only affected by the mass of the sun and the mass of earth.

This applies to the atom if you think of the nucleus as the sun and each electron as a planet. The force attracting the electron to the nucleus is only affected by the charge of the nucleus as each electron carries the same charge, and the charge is not the collective  charge of all electrons, but the individual charge. Hence all each electron perceives is the increase in the charge of the nucleus that is increasing its attraction to the nucleus.


If melting points and freezing points are the same, what state is pure equilibrium? If melting points and freezing points are the same, what state is pure equilibrium?

I understand melting points and freezing points are the same; adding or removing temperature of melting points for specific variables, like water at $\pu{0^\circ C}$ will melt or freeze them. Then, I have to ask the question, what is in between freezing and melting? What is that state? What is the behavior of the atoms if you somehow managed a perfect system where there wasn't extra heat being added nor removed and equilibrium was exact.

At the melting point there is a change of state. This means that solid is transformed in liquid or viceversa. For this process you need energy which is supplied or withdrawn in the form of heat. The whole process occurs at constant temperature, which is precisely the melting point.

Adding or removing temperature has no sense. It is heat what  is added or removed. The temperature is just a result of the state. When you have a mixture of solid and liquid in equilibrium, this temperature is defined, is the same in the whole system and is the melting point.

Since it is state change, both solid and liquid are present and the temperature will not change while both are there in any quantity and if the system is in equilibrium. The quantities of solid and liquid will not vary until there is some heat flow. However, on the microscopic level, the molecules are still moving and some will migrate from the solid to the liquid while others (the same amount) will go from the liquid to the solid. This results in that, even if the system is in equilibrium, the small crystals will dissolve while the bigger ones will grow bigger.


α,β-unsaturated carbonyl compounds and alkyl addition In my book it is suggested that in order to add alkyl groups in α,β-unsaturated carbonyl compounds we use organometallic reagents such as organolithium reagents.
In the previous chapters it was suggested that the addition of alkyl groups in carbonyl compounds with saturated chains it is conducted by using secondary amines that produce enamines.

I would like to know if it is possible to add alkyl groups in α,β-unsaturated carbonyl compounds by using secondary amines.

If I understand the question correctly (I do not have the book), it goes like this:

(Ref. J. Am. Chem. Soc.  85, 2, 207-222)

In short I would say that it is not possible to perform the reaction in question in the same way with the Stork enamine alkylation simply because—due to the α,β unsaturation of the ketone—the enamine can not be produced with pyrrolidine. Also, I performed a Reaxys search and found no relevant results. (Of course, I might have missed something.) However, I did find a couple of interesting reactions between cyclohexenone and pyrrolidine which I submit below for completion:

And the following L-Proline organocatalytic reaction taken from: Tetrahedron Letters, 2010, vol. 51, #38, p. 5086 - 5090. Note that compounds 1 and 2 in the scheme were isolated.

However it must be noted that generally, using a different α, β unsaturated ketone and a different amine to the ones used above, the reaction in question is possible as demonstrated by user55119 in his answer here: Why enamine formation does not occur with α,β unsaturated ketones?


Radicals vs atoms having unpaired electrons In organic chemistry course, I read that a radical is an atom, molecule or ion which have unpaired valence electrons.

So does there any difference between chlorine atom or cholrine radical? Also sodium atom contains one valence electron which is unpaired, then does sodium atom is also considered a radical?
Please clarify it.

Yes, a chlorine atom is very much the same as chlorine radical.

With sodium, the two things are technically similar too, but you'll hardly ever encounter a mention of "sodium radical".

Why is that? Well, I guess there is a difference between Na and Cl. A chlorine radical exists in its own right (if only briefly), and is an important intermediate in many reactions. A sodium radical, on the other hand, is pretty much not a thing. Single sodium atoms just don't wander around alone. You may see reactions involving metallic sodium, or ionic salts of sodium, or maybe (not sure about that) some compounds which can pass for covalent, or (unlikely) even $\rm\bf Na^{\pmb-}$... but that would be all.

So it goes.


Should the measurement uncertainty be included in the random uncertainty? So, the other day I was carrying out a titration and got the titres $33.4cm^3$ and $33.3cm^3$. I calculated the random uncertainty to be $0.05cm^3$. Now, the burette itself had an uncertainty of and I'm wondering whether I should add these two uncertainties together to find the uncertainty for the titres. Also, if I have to use the average of the titres to calculate the percentage uncertainty or whether I need to calculate the percentage uncertainty individually for the titres.

Yes - to calculate the total uncertainty when adding or subtracting measurements, it's the room-sum-squared of all the uncertainties.

$$
e_{tot} = \sqrt{(e_{meas})^2 + (e_{rand})^2}
$$
The more generalized form of this will be the root-sum-squared of all sources of error and uncertainty.
$$
e_{tot} = \sqrt{\sum{(e_n)^2}}
$$

To get the quantity "$e_{meas}$" you apply the root-sum-squared formula of all the uncertainties of your measurement devices used to transfer liquids (transfer beaker, graduated cylinder, burette, etc...).

$$
e_{meas} = \sqrt{(e_{beaker})^2 + (e_{burette})^2+(e_{grad.cyl})^2+(e_{burette})^2...etc}
$$

=========

To answer whether you apply the burette uncertainty to your measurement, the answer is also yes. Assuming that the burette has an uncertainty of $±0.01 cm^3$, the actual reading of your measurements will be: $33.3cm^3 ± 0.01 mL$ and $33.4cm^3 ± 0.01 mL$. The two uncertainties in the measurement will be applied to the $e_{tot}$ equation above.


Limiting reactant and percent yield If 2.30 mol of sodium reacts with 1.95 mol of water, how many moles of sodium hydroxide are produced?

Would it be 2.3 mol NaOH?

The first step is to figure out a reaction equation.

[I'm assuming elemental] sodium reacts with water violently because it releases a lot of energy when its only valence electron gets ripped apart. It creates \ce{NaOH} and \ce{H2}.

$\ce{Na + H2O -> NaOH + H2}$

We need to balance this.

$\ce{2Na + 2H2O -> 2NaOH + H2}$

$\ce{Na}$ and $\ce{H2O}$ are consumed at equal rates, so the $1.95$ moles of $\ce{H2O}$ are going to be used up completely (leaving $2.30 - 1.95 = 0.35$ moles of $\ce{Na}$ not reacted).

$\ce{NaOH}$ is formed at the same rate of consumption of the other two, so it's going to have $1.95$ moles produced.

--

Mathematically...

$\frac{2.30mol_\ce{Na}}{2mol_\ce{Na}} = 1.15$

and

$\frac{1.95mol_\ce{water}}{2mol_\ce{water}} = 0.975$

Effectively, water burns out faster in the reaction than sodium. So, we have to base our calculations off of moles of water.

For every 2 moles of water used, 2 moles of sodium hydroxide are formed.

$1.95mol_\ce{water} * \frac{2mol_\ce{NaOH}}{2mol_\ce{water}} = 1.95mol_\ce{NaOH}$


How to calculate the cation exchange resin capacity? If I have a cation exchange membrane and need to remove 2 g/L of sodium, how much resin would I need.

I used the amount of substance of $\ce{Na}$ to calculate the amount of substance of $\ce{SO3H-}$ needed for complete sodium removal. But this will not give me the correct mass since the resin is a polymer with that functional group attached. However I haven't been able to find any kind of complete structures for a cation resin.

First of all, you will need more sulfonic groups than sodium ions since sodium doesn't have a very good affinity for sulfonic groups.

Anyway, you can find some specs here — figure 2. This source is using sulfonated polystyrene with various crosslink percentages and expresses the capacity in $\mathrm{mmol}_\text{sulfonic groups}/\mathrm{g}_\text{resin}$ ($1\ \mathrm{meq} = 1\ \mathrm{mmol}$).


Equivalence point of titration Why is $\ce{[HIn] = [In-]}$ at equivalence point of a titration? I know that at equivalence point moles of known solution is equal to the moles of unknown solution, but I'm not able to relate that fact with the above conclusion. Perhaps I'm don't know something.

Ideally, the concentration of indicator is much, much lower than the concentration of the material to be titrated and of that used to titrate. Therefore, less than a drop of standard solution is needed to make the indicator entirely cross the equivalent point.

For most titrations, the equivalence point of the indicator is not the end point. It’s just somewhere near. And the only thing you can actually see and record is said indicator’s colour change. So it’s the only thing you can resort to.

The important choice in a titration is choosing the correct indicator. If you are titrating a strong acid with a strong base (so the titration’s equivalence point is approximately $7$), you want an indicator whose equivalence point is just behind that (at ca. $8$), so that the tiniest of drops added additionally will cause the colour change.


How would one favor the formation of p-TsCl over o-TsCl in the chlorosuflonation of toluene? I found an old synthesis of $\ce{o-TsCl}$ that purportedly gives a yield of $85\%$ here and was wondering how, still using chlorosulfonic acid, I could favor the formation of $\ce{p-TsCl}$ instead.

Chlorosulphonation using $\ce{ClSO3H}$ is an irreversible, electrophilic aromatic substitution reaction.
In the book Chlorosulfonic Acid, On taking Toluene as the aromatic substrate, it is seen to yield two products -

$para$-tosylchloride i.e. the thermodynamic product (because it is devoid of the steric hindrance offered by the methyl group)

$ortho$-tosylchloride i.e. the kinetic product (because of more activated ortho position due to proximity with methyl $(+I)$ group, which is favourable for the electrophile to attack on despite facing steric hindrance)

At low temperatures like $0-5°C$, the kinetic product is favoured and thus the ortho-isomer is formed. This reaction is exploited in the manufacture of the non-fattening sweetener $Saccharin$.
At higher temperatures like $75-80°C$, the thermodynamic product is favoured i.e. the para-isomer.
According to Chlorosulfonic Acid, The yield of para-tosyl chloride can be increased by

heating, or

addition of ammonium chloride or other catalysts

However, Clayden instead goes on to suggest on Pages 486 and 566, that the preference of ortho-isomer in chlorosulphonation and para-isomer in sulphonation is due to the fact that the latter is reversible unlike the former.

If the reaction is irreversible (as in chlorosulphonation), the alpha-product is usually formed. (with respect to naphthalene, here analogous to ortho position on toluene)
Sulphonation at low temperatures give the alpha-product by kinetic control and sulphonation at high temperatures gives the beta-product (corresponding to para-position on toluene) by thermodynamic control. The beta product is formed more slowly but it is more stable as there is less steric hindrance.... (as the case in toluene)

...which suggests that chlorosulphonation may not be temperature-dependent at all and shall always yield the ortho-isomer, which I am a bit doubtful about. (help needed)
However, the thermodynamic-kinetic explanation seems very satisfactory to me.


Solubility of ortho- and para-nitrophenol in benzene Which one between o-nitrophenol and p-nitrophenol is more soluble in benzene? And what is the reason? If the solvent had been polar, hydrogen bonding would have provided a lot of explanation, but what about non polar solvents?

It turns out that the opposite of what happens in polar solvents takes place when a non-polar solvent is used. At the same temperature, o-nitrophenol is more soluble in benzene than it's m and p isomers.
Sidgwick et al.1 did a study of this and obtained the following results. (Note that the solvent they used was toluene and not benzene, but they are similar enough that the same principles would apply.)

EDIT: The original version of this answer wrongly interpreted the graph and stated that o-nitrophenol would be less soluble in benzene than the other isomers. That was an error on my part. Thank you to Mathew Mahindaratne for pointing it out!
References:

Nevil Vincent Sidgwick, William James Spurrell, Thomas Ellis Davies, “CXXXII.—The Solubility of the Nitrophenols and Other Isomeric Disubstitution Products of Benzene,” J. Chem. Soc., Trans. 1915, 107, 1202–1213. doi:10.1039/CT9150701202.


Dissociation constant for Indicators So I came across a question which asked to calculate the $\mathrm{pH}$ at which an indicator of given $\mathrm{K_{a}}$ would change colour. The solution was given in the following way
Since when the color change occurs $\ce{[HIn]=[In^{-}]}$,
As $\mathrm{K_a}= \left(\frac{[\ce{H^+} ][\ce{In^-}]}{[\ce{HIn}]}\right)$,
At $\mathrm{K_a}=[\ce{H+}]$ the colour change would occur.
And consequently from the $\ce{H+}$ concentration the $\mathrm{pH}$ was calculated, the solution imposes that during the colour change the equilibrium is attained, but how can that be since to observe colour change you add an acid or base which disturbs the initial equilibrium for which the $\mathrm{K_{a}}$ was defined and hence the now won't hold true.

First of all, the indicator does not change colour instantaneously at $\mathrm{pH = pK_a}$. The indicator is a different colour than it's conjugate base, and exists in an equilibrium with it in solution. For the colour change to be noticed, the concentration of the base has to be at least 10 times more than that of the conjugate acid, or vice versa. This is why there is a range of colour change(look at these pH charts), with the upper and lower bounds having a $\mathrm{pH}$ difference of $2$. As an example, here's a simplified reaction for Phenolphthalein ($\ce{Ph}$ is the phenolphthalein anion)
$$\ce{\underset{\text{colourless}}{H2Ph} <=> 2H+ + \underset{\text{violet}}{Ph^2-}}$$
If you take a simpler, weak monobasic acid indicator $\ce{HIn <=> H+ + In-}$, the $\mathrm{K_a}$ is
$$\mathrm{K_a} = \frac{\ce{[H+][In^-]}}{\ce{[HIn]}}$$
The relative concentration of the indicator and it's conjugate base is only determined by the concentration of the $\ce{H+}$ ion, as $\mathrm{K_a}$ is a constant.
Answering your second query, the author takes the condition of equilibrium, simply because the system is in equilibrium, that is $\mathrm{Q_a = K_a}$. It is assumed that the reaction is instantaneous or that titration is performed slowly enough so that the system is in equilibrium at every single instant.


Does temperature change pressure in isochoric system? Consider the following equilibrium reaction:

$$\ce{2 NOCl (g) <=> 2 NO (g) + Cl2 (g)} \qquad  \Delta H > 0$$

My textbook has a statement that:

  If you decrease the temperature when keeping volume constant, then reaction shifts to the reactants.

I know that since the enthalpy change $\Delta H$ is positive, we can think of the reaction as:

$$\ce{2 NOCl (g) + \text{heat} <=> 2 NO (g) + Cl2 (g)}$$

And according to Le Chatelier's principle we can say that if we decrease the temperature, then reaction favor the side have heat in it. This comes me so reasonable but there is one more thing we have to consider in my opinion: pressure.

While keeping volume constant, decreasing temperature will decrease also pressure. So reaction will favor the side that have more molecules in it. In this case, it is the opposite side of the previous one.

Is my reasoning true? If it is, how can I conclude the side determining job?

Since it's given that ΔH > 0 (endothermic reaction) ,the reaction will proceed backward (towards the reactant). However as you know ,
Partial Pressure Of Gas= Total Pressure x Mole Fraction;
hence , even if the pressure is affected , the mole fraction increases/decreases in a way so that partial pressure remains constant.
If you calculate Kp ( Pressure Equilibrium constant) you will know that it depends on the partial pressure of the gases involved . Since partial pressure remains same no effect takes place. Hence , pressure  will have no effect on the equilibrium.

Also note that equilibrium constant is only affected by the change in temperature.
Hope this helps.

EDIT:-
In the light of the comments I would like to clarify ,

Partial pressure of a particular gas is the fraction of the gas multiplied by the total pressure which means it exerts a fraction of the total pressure. When I increase the total net pressure exerted by all gases , the fraction of moles of each gas decreases (according to the mathematical equation) thereby contributing lesser to the total pressure .

You can think of it as a ratio:-

If 1 mole X Gas exerts 10 bar pressure when total pressure is 20 bar , 0.5 moles of X gas will exert the same 10 bar pressure when total pressure is increased to 40 from 20 , thereby keeping the ratio same.


How exactly do alcohol molecules dissolve in seawater? The book "Lehrbuch der anorganischen Chemie" by Arnold F. Hollemann on page 47[link here][1] states that dropping 0.8g Alcohol (M = 46g/mol) in a sea or ocean would result in every litre seawater containing 8 molecules alcohol.

This completely baffles me. How exactly does this happen?

[1]: https://books.google.de/bookshttps://books.google.de/books?id=eGkvSDAqY9gC&pg=PA47&lpg=PA47&ots=MFR61amS0x&focus=viewport&dq=ein mol alkohol in den weltmeeren&hl=de&output=html_text

Here is my calculation.

The ocean has a volume of 1.3 billion cubic kilometers or 1.3e+18 cubic meters or 1.3e+21 liters. 0.8/46=0.0173 moles of ethyl alcohol x 6.02e+23 molecules per mole = 0.1047e+23 molecules, which if dissolved evenly in 1.3e+21 liters would give 10.47/1.3 or approx. 8 molecules of ethanol per liter.


How to derive Pauli Exclusion Principle without assuming anti-symmetry? So, it appears that the statement of the Pauli Exclusion Principle is equivalent to the statement that fermions are anti-symmetric. That is, if you assume that fermions are anti-symmetric, then you can derive the Pauli Exclusion Principle. If you assume the Pauli Exclusion Principle, you can derive anti-symmetry in the wavefunction.

At some point, one of these must have been proved independently of the other. I can't seem to find such a proof anywhere. I think this confusion also comes from the fact that spin is a difficult thing to conceptualize, let alone measure. Does some experimental data to necessarily imply one of the two? I guess, some insight on how spin was discovered would probably help.

The Pauli Exclusion principle is determined from experimental facts alone.
First, its consequences were observed and then it was incorporated in the theory. See Kaplan´s article for a more erudite answer. Also, see Towler´s exposition. Although, he dwells in foundational issues.

Important: they are not equivalent unless you consider only totally symmtric/antisymmetric functions! The antisymmetry is assumed, it cannot be derived rigorously from particle independence nor tested with experiments.

As for getting to the conclusion that the wavefunction of fermions has to be antisymmetric here is a weak argument without requiring the Pauli Exclusion principle.
You could also say that since you cannot follow the trajectory of particles therefore they are indistinguishable -- at least, if they are of the same kind. That is, the permutation of any two of them
$$ \Psi(\pmb x_1, \pmb x_2, ...) \to \Psi(\pmb x_2, \pmb x_1, ...) $$
maintains the same physical description
$$ |\Psi(\pmb x_1, \pmb x_2, ...)|^2 = |\Psi(\pmb x_2, \pmb x_1, ...)|^2.$$
The permutation can be described as the action of an operator that satisfies $P^2 = 1$.
There are only two ways to acomplish this. Either the wavefunction is symmetric (the eigenvalue of $P$ is $-1$) under the exchange or antisymmetric (eigenvalue $-1$).
$$ \Psi(\pmb x_1, \pmb x_2, ...) =  \Psi(\pmb x_2, \pmb x_1, ...) \quad \text{  Symmetric: bosons}$$
$$  \Psi(\pmb x_1, \pmb x_2, ...) = -\Psi(\pmb x_2, \pmb x_1, ...)\quad \text{Antisymmetric: fermions}$$

Therefore, if $\pmb x_1 = \pmb x_2$ then $\Psi(\pmb x_1, \pmb x_1, ...) = -\Psi(\pmb x_1, \pmb x_1, ...) = 0$. That is, Pauli Exclusion principle.

The reason as to why this argument, that is the one that was elaborated to justify Pauli Exclusion principle and is seen nowadays in most textbooks, is weak is neatly explained in the reference mentioned above.


Why does pH decrease when diluting a base? Take a strong base such a sodium hydroxide
Adding water will result in adding a hydroxide ions as well as hydronium ions to the solution
However if the pH decreases this means that the hydronium ion concentration will increase , hence the hydroxide ion concentration decreased
Why is this so ? What about all the hydroxide ions that were added from the water
I understand the mathematics behind it , but it doesn't intuitively make sense

You are overcomplicating it. The concept of pH and pOH stems from the concentration of $\ce{OH-}$ or $\ce{H+}$ per solvent unit $\ce{H2O}$, like so $\ce{\frac{[OH-]}{[H2O]}}$ or $\ce{\frac{[H+]}{[H2O]}}$. There is also a logarithm involved, but lets keep it simpler than that. You are adding far more of the denominator than the numerator, hence the fraction is becoming smaller and the pH is going towards pH of pure water.


Can anyone explain the reaction mechanisms of these synthesis? Can anyone explain the reaction mechanism of the following synthesis?

$\ce{2NH4Cl + Ca(OH)2 -> 2NH3 + 2H2O + CaCl2}$

$\ce{(NH4)2SO4 + 2NaOH -> 2NH3 + 2H2O + Na2SO4}$

The ammonium chloride, calcium hydroxide, ammonium sulfate, and sodium hydroxide are all ionic compounds, and when dissolved in water they dissociate into $\ce{NH4+}$, $\ce{Cl-}$, $\ce{Ca^2+}$, $\ce{SO4^2-}$, and $\ce{Na+}$ ions. The $\ce{NH4+}$ ammonium ion can donate a proton to the $\ce{OH-}$ ion, creating $\ce{NH3}$ and $\ce{H2O}$. The other ions don't really do anything, so these spectator ions remain in solution but the charges are still balanced. If you removed the ammonia and water, which is easy with a vacuum, you'd be left with the $\ce{CaCl2}$ and $\ce{Na2SO4}$ salts. These reactions will have some equilibrium, with $\ce{H2O}$ donating a proton back to $\ce{NH3}$ to form $\ce{NH4+}$ and $\ce{OH-}$ again.


Do we have some "rule of thumb" to estimate the miscibility of metals? For example, for ordinary fluids, we have polar and apolar ones. The "rule of thumb" is that polar fluids mix with each other, and also apolar fluids mix with each other, but polar fluids don't mix with apolars.
Of course, it is not a very exact rule. For example, bensene mixes in water - but it does it badly. While the also polar ethanol mixes water very well.
What is the case with liquid metals? Does some analogous - possibly more complex - category structure exist?

Bottom line: there are no simple rules of thumb.
To demonstrate this, I will show a round robin of phase diagrams for fcc metals ($\ce{Ag}$, $\ce{Cu}$, $\ce{Au}$, $\ce{Ni}$) with no intermetallic compounds - just fcc solid and liquid.

$\ce{Ag}$-$\ce{Au}$, full mutual solubility, liquid and solid enthalpies of mixing pretty close:

$\ce{Ag}$-$\ce{Cu}$, with limited mutual solid solubility and a eutectic:

$\ce{Ag}$-$\ce{Ni}$, no solid solubility, a huge liquid miscibility gap:

$\ce{Au}$-$\ce{Cu}$, complete mutual solid solubility plus a eutectic:

$\ce{Au}$-$\ce{Ni}$, solid miscibility gap at lower temperatures, but complete solid miscibility above that, a slightly irregular liquidus:

$\ce{Cu}$-$\ce{Ni}$, solid miscibility gap at lower temperatures, pretty ideal liquidus:

Bottom line: even for nominally 'simple' combinations of near-ideal fcc metals coming up with a workable rule of thumb just doesn't work.


How to recover 15g gold from 50 gallons of 30% Acid? Someone mistakenly poured a solution containing $15\mathrm{g}$ of gold into a nearly full waste drum that is around $30$% $\ce{HCl}$ with a little nitric acid in there as well.

Is there a safe way of separating the acid from the gold via distillation?

You could distill the waste easily enough, but the distillation gases will be rather nasty. You will also need a condenser to cool everything down again or the environmental penalties (and corrosion damage to the exhaust system) will be a lot more than the gold value.

Once you have the waste boiled down (literally) into a small flask you could use any of the known gold-extraction methods to get it out in elemental form.

At current market prices you have about US$580 of gold in the drum. No idea how much you will spend in fuel and cooling water to recover it but it will be a substantial percentage. Why don't you just tell "Someone" that they can either pay for the gold they dumped or mind the still themselves over the weekend (and pay for the propane). This will also encourage others to pay more attention to what is in the beaker.


How do I find out if ions are oxidised/ reduced based on half cell potentials? Assuming standard conditions, answer the following questions.
  (yes/no)

  Is $\ce{Fe^{2+}(aq)}$ capable of reducing $\ce{Cr^{3+}(aq)}$ to $\ce{Cr^{2+}(aq)}$?
  Is $\ce{Sn^{2+}(aq)}$ capable of oxidizing $\ce{Fe^{2+}(aq)}$?
  Is $\ce{Cr}$ metal capable of reducing $\ce{Fe^{2+}(aq)}$?
  Is $\ce{VO^{2+}(aq)}$ capable of oxidizing $\ce{Fe^{2+}(aq)}$?
  Is $\ce{H2(g)}$ capable of reducing $\ce{Ag+ (aq)}$?
  Is $\ce{Cr^{3+}(aq)}$ capable of oxidizing $\ce{Fe^{2+}(aq)}$?

I have

yes
yes
no
yes
yes
no

What am I doing wrong?

Look at this table of half-reactions: http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm

I'll explain how it contradicts your answer to the first question; you can then rework your answers and update your question with justifications.

  Is $\ce{Fe^{2+}(aq)}$ capable of reducing $\ce{Cr^{3+}(aq)}$ to $\ce{Cr^{2+}(aq)}$?

One entry of the table says $$ \ce{Cr^{3+}(aq) + e- -> Cr^{2+} (aq)} \quad –0.424.$$

The $-0.424$ is what's called the standard potential, expressed in volts.  It is a measure of how favorable the half-reaction is (compared with a standard half-reaction).  The higher the number, the more favored the products will be.

Now you have to find a half-reaction involving $\ce{Fe^{2+}}$.  Since you want Fe to reduce something, it will itself be oxidized.  The table shows no half-reaction where $\ce{Fe^{2+}}$ is oxidized, but that's because by convention only reduction half-reactions are shown; to get the opposite, oxidation, half-reactions, just swap the terms and negate the potential.  That is, take this line from the table:
$$
\ce{Fe^{3+}(aq) + e^{–}  -> Fe^{2+}(aq) } \quad 0.771
$$
and turn it into
$$
 \ce{Fe^{2+}(aq) -> Fe^{3+}(aq) + e^{–}} \quad {-}0.771
$$

Now add this, term by term, to the Cr half-reaction. The result is
$$
\ce{Fe^{2+} + Cr^{3+} -> Fe^{3+} + Cr^{2+}} \quad {-}1.195
$$
If the potential in the final reaction is negative, the reaction is impossible (more precisely, equilibrium lies to the left --- in the case very far to the left, becase 1 volt goes a long way).

That's the basics.  There is a subtlety involved when the number of electrons exchanged is not the same in the two half-reactions; you'll find it in your textbook.  You could try answering the other 5 questions now and if you get stuck, tell us precisely where.


Determining the mass of sodium sulfate that forms when reacting sulfuric acid with sodium hydroxide I have come across simple problem in the chemistry book which I cannot solve. The question is below. I would appreciate if you have any solution to this.

Question:

  $3\ \mathrm{g}$ of $\ce{NaOH}$ are mixed with $4.9\ \mathrm{g}$ of $\ce{H2SO4}$. How much sodium sulfate $\ce{Na2SO4}$ will form?

(The answer is $3.55\ \mathrm{g}$)

Note: I have calculated it as $5.33\ \mathrm{g}$ of $\ce{Na2SO4}$ don’t know where I am mistaken. It doesn’t seem to be a typo in the question because I couldn’t solve other similar problems in other books wither. There is something I always miss.

Rather than working with funny fractions, let’s use decimals.

$$m(\ce{H2SO4}) = 4.9~\mathrm{g}\\
m(\ce{NaOH}) = 3~\mathrm{g}\\
M(\ce{H2SO4}) = 98.09~\mathrm{g \cdot mol^{-1}}\\
M(\ce{NaOH}) = 40.00~\mathrm{g \cdot mol^{-1}}\\
M(\ce{Na2SO4}) = 142.05~\mathrm{g \cdot mol^{-1}}\\
~%
\\
n(\ce{H2SO4}) = 50~\mathrm{mmol}\\
n(\ce{NaOH}) = 75~\mathrm{mmol}$$

Remember when doing stoichiometry to correctly consider your reaction equations. In this case:
$$\ce{H2SO4 + 2NaOH -> Na2SO4 + 2 H2O}$$

If we want to do this, we see that we don’t have enough $\ce{NaOH}$ — it is present only in substoichiometric amounts. We would need $100~\mathrm{mmol}~\ce{NaOH}$ for a full reaction with $\ce{H2SO4}$:

$$\frac{n(\ce{NaOH})}{n(\ce{H2SO4})} = \frac{2}{1}\\
n(\ce{NaOH}) = 2 n (\ce{H2SO4})\\
n(\ce{NaOH}) = 2 \times 50~\mathrm{mmol}= 100~\mathrm{mmol}$$

If we now assumed that all $\ce{NaOH}$ would participate in the formation of $\ce{Na2SO4}$, we would arrive at $37.5~\mathrm{mmol}~\ce{Na2SO4}$ or $5.4~\mathrm{g}\ \ce{Na2SO4}$. But that would leave free $\ce{H2SO4}$ lying around. Instead, consider a stepwise process:

$$\ce{H2SO4 + NaOH -> NaHSO4 + H2O}\\
\ce{NaHSO4 + NaOH -> Na2SO4 + H2O}$$

We see that we need to use the first $50~\mathrm{mmol}\ \ce{NaOH}$ to generate $50~\mathrm{mmol}\ \ce{NaHSO4}$; with $25~\mathrm{mmol}\ \ce{NaOH}$ remaining. These $25~\mathrm{mmol}\ \ce{NaOH}$ can react with the $\ce{NaHSO4}$ to generate $25~\mathrm{mmol}\ \ce{Na2SO4}$ and leave $25~\mathrm{mmol}$ unreacted $\ce{NaHSO4}$.

$$n(\ce{Na2SO4}) = 25~\mathrm{mmol}\\
m(\ce{Na2SO4}) = n \cdot M = 25~\mathrm{mmol} \times 142.05~\mathrm{g \cdot mol^{-1}} = 3.55~\mathrm{g}$$


The chlorofluorocarbon reaction in the lab? The putative reason for banning Freon and its class of chemical (chlorofluorocarbons) is that they will rise to the top of the atmosphere, to the ozone layer and "catalyze" the reduction of ozone to ordinary $\ce{O2}$.

The objection that the amount of chlorofluorocarbons would be tiny in comparison to the amount of ozone, is answered by the idea that the chlorofluorocarbons, do not do a simple reaction, but catalyze at high rate the conversion of $\ce{O3}$ into $\ce{O2}$. In other words, the idea is that a single molecule of chorofluorocarbons could convert millions of ozone molecules into $\ce{O2}$ every minute or something like that.

Unfortunately, this supposed reaction (which would seem to be unlikely on the face of it) seems to have never been duplicated in a laboratory, which I would expect would be easy to do.

Would it be possible to easily set up and measure this reaction in a laboratory? The goal would be to determine the rate of reaction. For example, given a certain molar concentration of chlorofluorocarbon in the reaction vessel, how fast is a given molar concentration of $\ce{O3}$ converted to $\ce{O2}$.

The catalytic stratospheric ozone destruction by CFC's and other compounds is almost certainly too complex to fully replicate in a laboratory with current technology.  The only laboratory of proper scale and conditions is the earth's atmosphere as a whole (no pun here, seriously ;).

Even the 1995 Nobel prize winners Crutzen, Rowland and Molina had made serious omissions from their original work in the 70's that are now known to be critical to the mechanisms of stratospheric ozone depletion.  While Crutzen is pretty much a pure theoretician, Molina and Rowland did much of their work in the laboratory and combined that work with atmospheric measurements.  The bottom line is that only pieces of the whole process can be controlled in a single or set of experiments.  One important factor, for example, that was omitted from their original work was the role played by condensed-phase surface process in mediating catalysis by active chlorine.  Specifically, without the different types of polar stratospheric clouds (i.e. water ice, nitric acid trihydrate,  liquid supercooled solutions of sulfuric acid, nitric acid and water, and other complex matrices) catalytic ozone destruction does not occur. So, although the net of their predictions in the 70's was correct, it was the real-world observations decades later, not laboratory confirmations, that sent them to Stockholm for their prize.

In 2010, an article was published called "Key compound of ozone destruction detected; Scientists disprove doubts in ozone hole chemistry.", Karlsruhe Institute of Technology (KIT).  ScienceDaily. ScienceDaily, 22 July 2010, excerpt available here.  The summary states:

  For the first time, scientists in Germany have successfully measured
  in the ozone layer the chlorine compound ClOOCl, which plays an
  important role in stratospheric ozone depletion. Doubts in the
  established models of polar ozone chemistry expressed by American
  researchers based on laboratory measurements are disproved by these
  new atmospheric observations.

This represents an even more recent case in which rigorous laboratory attempts to simulate stratospheric ozone depletion processes failed in the face of experiments and measurements made in-situ in the "real world" laboratory.

The article further points out:

  The ozone hole above the Antarctic and the destructive role of
  chlorofluorocarbons (CFC) and their decomposition products have become
  a synonym of both global environmental problems and their solution by
  concerted agreements worldwide. Scientific fundamental research into
  ozone chemistry of the atmosphere was the basis of international
  agreements, such as the Montreal Protocol of 1987, which has put
  limits on CFC production. The success of the political implementation
  of these scientific findings is reflected by the fact that the
  chlorine content of the atmosphere and, hence, the ozone destruction
  potential recently started to decrease slowly.

So, the upshot here is that it was primarily in-situ experiments and measurements that led to predictions of catalytic chlorine mediated ozone destruction being later confirmed by measurements. The same was true for formulation of a solution to the problem, which again in-situ measurements confirmed.

In short, laboratory experiments have repeatedly been outperformed by in-situ experimentation and observation in the case of stratospheric ozone depletion studies.  The same is true for many complex, global chemical and physical process; it simply is not always feasible to re-create a system of large scale and complexity in a laboratory.


Still taught to reverse oxidation half cells in electrochemistry? In a question, Oxidation of metals/halogens by oxygen gas in acidic aqueous solution, there was a point made that reduction half-cells should not be reversed.

I was taught ($\approx 1970$) in the following manner. Given the reduction half cells:

  $$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\
\ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V}
\end{align}
$$

To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:

  $$
\begin{align}
\ce{Ag(s) &→ Ag+ + e-  } &\quad E^\circ &= \pu{-0.799 V} \\
\ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V}
\end{align}
$$

Now you can write a balanced chemical reaction as such:

  $$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{4Ag(s) &→ 4Ag+ + 4e-  } &\quad E^\circ &= \pu{-0.799 V} \\
\ce{O2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l)} &\quad E_\text{Total}^\circ &= \pu{+1.23 - 0.799 = +0.43 V}
\end{align}
$$

and to get the cell potential the two half cells are added, one being positive and the other negative. If the EMF is positive the forward reaction is spontaneous, if negative the reverse reaction is spontaneous.

Is flipping the reduction reaction not the standard method taught these days to balance redox reactions?

I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and USA. The Ostwald school of thought, wrote the electrode potentials in accordance with the electrostatic sign of the electrode with reference to the standard hydrogen electrode. This was the so-called European convention. Keep in mind that electrostatic signs of the electrodes are invariant. It does not matter how you write them. A silver electrode in Ag+/Ag half-cell is always positively charged with respect to the hydrogen electrode under standard conditions.

The American school of thought, with its own influence such as Gibbs, Lewis and Randall, Ligane, Latimer, had a thermodynamic view of the signs relating electrode potentials to the Gibbs free energy. Latimer wrote a very famous book "Oxidation Potentials" which is available from Internet archives. Yes, under those conditions we can flip the signs back and forth while keeping Gibbs free energy in mind. The tug of war between European convention vs. American convention went on until the 50s-70s. Textbooks are usually 20 years behind current research...these issues lingered on.

Finally, electrochemists decided, including Allan J. Bard (from US who wrote a very influential electrochemistry book taught all over the world), that let us go back to the original notion- keep the electrostatic signs associated with the electrode potentials which are invariant with respect to the way we write them. This has "almost" terminated the sign convention issue of US vs. European signs of electrode potentials. The plus and minuses took almost a century to resolve. It is amazing to see the insight of those early scientists such as Ostwald, that world finally accepted their views.


Why does concentration of solutions of alkali metals in liquid ammonia change the magnetic properties? In "NCERT Chemistry for Class XI - Volume II", in chapter "s-block elements", under the topic "Chemical Properties of alkali metals", it is given:

(vi) Solutions in liquid ammonia: The alkali metals dissolve in liquid ammonia giving deep blue solutions which are conducting in nature.
$$\ce{M + $(x + y)$ NH3 -> [M(NH3)_x]+ + [e(NH3)_y]- }$$
The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution. The solutions are paramagnetic and on standing slowly liberate hydrogen resulting in the formation of amide.
$$\ce{M+(am) + e- + NH3(l) -> MNH2(am) + 1/2 H2(g)}$$
(where ‘am’ denotes solution in ammonia.) In concentrated solution, the blue colour changes to bronze colour and becomes diamagnetic.

In the last set of lines, it is given that in concentrated solution, the blue colour changes to bronze colour and the paramagnetic solution becomes diamagnetic.
How does the change in concentration affect the magnetic properties of the solution?
Wikipedia has nothing to offer to clarify my doubt.

I always suggest students to try Google Scholar (scholar.google.com) when a simple Google search fails. I just searched three keywords : alkali metals ammonia solutions and the third result is highly relevant.

When your book talks about "in concentrated solution", it means more alkali metal in liquid ammonia. This paper, which you should search in Google Scholar, is titled A Revised Model for Ammonia Solutions of Alkali Metals, Journal of the American Society, Year 1962, volume 84, pg 2264). It talks about the formation of dimers of metals of the type $\ce{M2}$, because as you increase the metal conc. in ammonia solution conductivity decreases.
The authors write, raising objections to older models as completely wrong show the following evidence:

  "We picture the » species as a quadrupolar ionic assembly of $\ce{e- + 2M+}$ in which there is little distortion of either the ammoniated electrons or ammoniated metal ions. Presumably the electrons and ions are held in a square or rhombic configuration."

In short the picture is not that simple an no wonder the authors of a grade 12 book never discuss more details. You can start from here and search more.


Reaction of Benzene with 2-methylpropene? I'm studying organic chemistry at the moment and it's really confusing. What could ever happen between benzene and an alkene? and why is $\ce{H2SO4}$ there?

Sulfuric acid protonates the alkene to give the highest substituted carbocation, tert-butyl.  That electrophilically adds to a benzene carbon atom, to give the obvious next cation.  A proton is eliminated to obtain tert-butylbenzene.  Akyl groups activate benzene rings toward ortho, para attack (mild electron donation via  hyperconjugation), thus 1,6-di-tert-butylbenzene, certainly if the addend is in excess.  Especially stable carbocations can dissociate and wander around the benzene ring under Friedel-Crafts conditions.  If you push it, you get 1,3,5-tri-tert-butylbenzene.

Steric hindrance strongly discourages further reaction.  You can go to the literature to find how many adjacent tert-butyls can be jammed onto a benzene ring - and how it is done.

With due respect, you are in deep trouble for passing the course if the Friedel-Crafts reaction throws you.   Either look at things with the LCAO model, look in a few more textbooks, or talk with folks who have grasped it.  You must understand it.  You cannot survive substituent effects if you cannot flip bonds.  This is easy stuff compared to what will follow.  Chess can be well-played by computer.  Organic synthesis is done in wetware, though computers help help.

http://www.chemtube3d.com/model/electrophilic_aromatic_substitution/friedel-crafts/alkylation/benzene%2Btbu/mechanism.png
Close enough


What are the isomers of a complex having a structure of M(AB)2C2? Our instructor have us answer an isomerism problem of a crystal structure that has a form of $\ce{M(AB)2C2}$. I keep getting 9 isomers, which includes the optically inactive isomer, and my peers and I have different answers.

Referring to the replies of Mr. Ivan Neretin, complexes that has a form of M(AB)2C2 posses 8 stereoisomers, 2 of which are optically inactive. The other 6 isomers can be interchanged to their respective enantiomers.


What volume does one mole of an ideal gas occupy? This has been bugging me for a while now...

Obviously, to calculate the volume/space occupied by a mole of (an ideal) gas, you'll have to specify temperature ($T$) and pressure ($P$), find the gas constant ($R$) value with the right units and plug them all in the ideal gas equation $$PV = nRT.$$

The problem? It seems to be some sort of common "wisdom" all over the Internet, that one mole of gas occupies $22.4$ liters of space. But the standard conditions (STP, NTP, or SATP) mentioned lack consistency over multiple sites/books. Common claims: A mole of gas occupies,

$\pu{22.4 L}$ at STP
$\pu{22.4 L}$ at NTP
$\pu{22.4 L}$ at SATP
$\pu{22.4 L}$ at both STP and NTP

Even Chem.SE is rife with the "fact" that a mole of ideal gas occupies $\pu{22.4 L}$, or some extension thereof.

Being so utterly frustrated with this situation, I decided to calculate the volumes occupied by a mole of ideal gas (based on the ideal gas equation) for each of the three standard conditions; namely: Standard Temperature and Pressure (STP), Normal Temperature and Pressure (NTP) and Standard Ambient Temperature and Pressure (SATP).

Knowing that,

STP: $\pu{0 ^\circ C}$ and $\pu{1 bar}$
NTP: $\pu{20 ^\circ C}$ and $\pu{1 atm}$
SATP: $\pu{25 ^\circ C}$ and $\pu{1 bar}$

And using the equation, $$V = \frac {nRT}{P},$$
where $n = \pu{1 mol}$, by default (since we're talking about one mole of gas).

I'll draw appropriate values of the gas constant $R$ from this Wikipedia table:

The volume occupied by a mole of gas should be:

At STP
\begin{align}
T &= \pu{273.0 K},&
P &= \pu{1 bar},&
R &= \pu{8.3144598 \times 10^-2 L bar K^-1 mol^-1}.
\end{align}
Plugging in all the values, I got
$$V = \pu{22.698475 L},$$
which to a reasonable approximation, gives
$$V = \pu{22.7 L}.$$
At NTP
\begin{align}
T &= \pu{293.0 K},&
P &= \pu{1 atm},&
R &= \pu{8.2057338 \times 10^-2 L atm K^-1 mol^-1}.
\end{align}
Plugging in all the values, I got
$$V = \pu{24.04280003 L},$$
which to a reasonable approximation, gives
$$V = \pu{24 L}.$$
At SATP
\begin{align}
T &= \pu{298.0 K},&
P &= \pu{1 bar},&
R &= \pu{8.3144598 \times 10^-2 L bar K^-1 mol^-1}.
\end{align}
Plugging in all the values, I got
$$V = \pu{24.7770902 L},$$
which to a reasonable approximation, gives
$$V = \pu{24.8 L}.$$

Nowhere does the magical "$\pu{22.4 L}$" figure in the three cases I've analyzed appear. Since I've seen the "one mole occupies $\pu{22.4 L}$ at STP/NTP" dictum so many times, I'm wondering if I've missed something.

My question(s):

Did I screw up with my calculations?
(If I didn't screw up) Why is it that the "one mole occupies $\pu{22.4 L}$" idea is so widespread, in spite of not being close (enough) to the values that I obtained?

The common saying is a hold over from when STP was defined to be $\pu{273.15 K}$ and $\pu{1 atm}$. However, IUPAC changed the definition in 1982 so that $\pu{1 atm}$ became $\pu{1 bar}$. I think the main issue is a lot of educators didn't get the memo and went right along either teaching STP as $\pu{1 atm}$ or continuing with the line they were taught ("$\pu{1  mol}$ of any gas under STP occupies $\pu{22.4 L}$") without realizing it didn't hold under the new conditions.

Just as a "proof" of this working for the old definition.
\begin{align}
V &=\frac{nRT}{P}\\
  &=\frac{\pu{1  mol} \times \pu{8.2057338 \times 10^-2 L * atm//K * mol}
    \times \pu{273.15 K}}{\pu{1 atm}}\\
  &=\pu{22.41396 L}\\
  &\approx \pu{22.4 L}
\end{align}


Do non-English speaking countries use the same element symbols? The question does sound pretty absurd, but hear me out first. The Periodic Table of the Elements, as I know it, is supposed to be a common standard adopted by the global scientific community.

However, I see a hitch when it comes to a total world-wide acceptance of such a standard.

Every Periodic Table of the Elements that I've seen in print uses the Latin Alphabet, and quite a few chemical symbols are assigned based on the Latin words for the elements ( Fe- Ferrum, Sn- Stannum, Au- Aurum, Cu- Cuprum, etc etc).

Now this appears to be a rather clever compromise reached by the English-speaking countries (the U.S.A, the U.K, Australia, etc) and countries that don't really use English, but use the Latin Alphabet (Germany, France, Spain, etc, etc). Obviously, using a "dead-language" to assign chemical symbols helps eliminate resentment that could rise by favoring one language (say, German) over another (say, English). Well, this appears to be the same logic behind the use of Latin in biological nomenclature too.

Now here's the catch;

How do countries that don't use the Latin Alphabet deal with this?

Alright, it seems credible to believe that Middle-Eastern countries (Arabic) and African countries would use the standard, Latin Alphabet Periodic Table that we use, without complaint. I guess the same goes for former European colonies too.

But what about:

1) Countries like Russia (Cyrillic) and Japan (Kanji), which have made significant contributions to the modern Periodic Table? Since they did contribute quite a bit to the Table, they probably "feel" they have the right to use their own customized Periodic Table using Cyrillic or Kanji characters.

2) Countries that vehemently oppose cooperation with Western countries, and would obviously dismiss the Latin Alphabet Periodic Table as an 'instrument of the West'. For example North Korea.

Both North Korea and South Korea use the periodic table. The names, however, are different. Here is a periodic table in North Korea

The title translates to "Mendeleev's atomic periodic table." The names of the elements are different for some.

In South Korea
Hydrogen is known as and pronounced as soo-saw.
Phosphorus is known as and pronounced as in.
Potassium is known as and pronounced as kallium.
etc.

But some are the same.

In South Korea
Magnesium is known as and pronounced as magnesium.
Argon is known as and pronounced as argon.
Fluorine is known and pronounced as fluorine.

I haven't taken classes in North Korea or visited any research facilities. But it is likely that they have further modified the names  to be "less western".

All of these are names, and do not replace the symbols used in the periodic table.


IUPAC name for citric acid I wonder what the name is of the following structure is:

I think it's something like 3-hyrodxypentanoic diacid
However, I wasn't able to find how to name a side chain if it contains a $\ce{COOH}$ group.

Maybe I'm completely wrong, I hope someone can tell me how to name this structure.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  7. acids
  (…)
  17. hydroxy compounds (includes alcohols and phenols)
  (…)

Thus, the carboxylic acid group has priority to be cited as a suffix.

Systematic names of carboxylic acids are generally formed using the suffix ‘oic acid’ or ‘carboxylic acid’ or the prefix ‘carboxy’.

The relevant rule in this case reads as follows:

  P-65.1.2.2.1 If an unbranched chain is linked to more than two carboxy groups, all
  carboxy groups are named from the parent hydride by substitutive use of the suffix
  ‘carboxylic acid’, preceded by the appropriate numerical prefix ‘tri’, ‘tetra’ etc. and
  appropriate locants.

Note that the suffix ‘carboxylic acid’ expresses the substituent group $\ce{-COOH}$. Thus, the name of unsubstituted tricarboxylic acid corresponding to the compound given in the question is ‘propane-1,2,3-tricarboxylic acid’.

Therefore, the preferred IUPAC name (PIN) of the complete compound given in the question is ‘2-hydroxypropane-1,2,3-tricarboxylic acid’.

However, according to Subsection P-65.1.1.2.3, the name ‘citric acid’ is also retained for general nomenclature.


Molarity to Percent, Mole Fraction, Molality I have a series of questions similar to the following. I want to make sure I do this correct before I do the others. I am in need of a hint.

Given molarity of 6M NaOH with d=1.4g/mL, calculate, percent, mole fraction, and molality.

So in the above question, I have the molarity, but how am I supposed to get liters of solution?

I figured it was 6M per 1L of water, but then what am I supposed to do with the density? Is that the density of the water with the NaOH?

Then in a sub part it gives me the mole fraction as 0.1 with a density of 1.3g/mL. Why does the question give me that number? Is it possible to figure out percent, molality, and molarity based on mole fraction?

Thanks for you help.

I figured it was 6M per 1L of water

That is incorrect you have 6 mol per 1L of water. M(stands for mol/L) not for mol!

How to tackle this problem?
Check water density, is 1 $g/cm^{2}$. So yours is the density of the solution. Then understand what the problem is asking what is percent? What is mole fraction? What is molality? Are you able to answer these questions? If not go back to the text book.
Then is only a matter of simple algebra and units matching. And remember if you have weight and molecular weight you can calculate the number of mole!

So suppose you have a liter of this solution what's the weight of NaOH?

   molecular weight ($\frac{g}{mol}$) of NaOH $\times$ number of moles

Now what's the Overall weight?

   Overall weight=density $\times$ volume (remember to use the right units)

Percent what? In chemistry there are a lot of percent, you mean percentage by weight (wt%) I imagine. Try to figure out from your self:

  $ = \frac{NaOH weight}{Overall weight}\times100 $

Mole fraction require mole of both, we start from the weight, so whats the weight of water?

   you got the overall weight and the weight of NaOH

Now calculate the moles of water, and it is simple to find the mole fraction, and molality. Make these steps in the opposite way and you will find the answer to your last question.


Significance of sign of standard reduction potential I am very confused with the signs of standard reduction potentials.
For example the standard reduction potential of Zinc is −0.76 V and that of copper is +0.34 V.

I cannot understand the  significance of the signs.

As far as I know,if the value is negative then the reaction is non-spontaneous(from the relation between Gibbs Free Energy change and Electrode potential).But if the reaction is non-spontaneous, then why are we classifying it as an electrochemical cell?

Should'nt it be classified as an electrolytic cell?
(Sorry if that sounded incorrect but my basics are weak)

There are several questions here so let me start by defining the different types of cells and standard reduction potentials.

Definition of Electrochemical (Galvanic) cell: A cell that converts chemical energy into electrical energy. In these cells a redox reaction creates electrons that can do work.

Definition of Electrolytic cell: A cell that converts electrical energy into chemical energy. In these cells the electrical energy source provides the electrons to perform a reaction.

Standard reduction potential: The tendency for a chemical species to be reduced and is measured in volts at STP. The more positive the potential is the more likely it will be reduced.

Now, here are your two reduction equations for $\ce{Zn}$ and $\ce{Cu}$:

$$\begin{alignat}{2}
\ce{Zn^2+(aq) + 2e- &-> Zn}\qquad&{-}0.76\ \mathrm V\\
\ce{Cu^2+(aq) + 2e- &-> Cu}\qquad&{+}0.34\ \mathrm V
\end{alignat}$$

The more positive the potential the more favorable the reaction as it is written will be. Remember that  $\Delta G = -nFE^\circ$ and that when $\Delta G$ is positive, the reaction is non-spontaneous, and when $\Delta G$ is negative, the reaction is spontaneous.  Positive values of $E^\circ$ will lead to negative values of $\Delta G$ and vice versa.

So, the reduction of $\ce{Cu^2+}$ to form $\ce{Cu}$ is more favorable than the reduction of $\ce{Zn^2+}$ to form $\ce{Zn}$. This means that  $\ce{Cu^2+}$ is a better oxidant than $\ce{Zn^2+}$. For an electrochemical cell, the cell potential can be calculated by the following equation:

$$E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}$$

For a working electrochemical cell we need $E^\circ_\text{cell}$ to be positive. I would use $\ce{Zn}$ as the anode (oxidation) and $\ce{Cu^2+}$ as the cathode to give an $E^\circ_\text{cell}$ of $+1.10\ \mathrm V$.


Why is my sodium acetate not resulting in any crystals? My son received a chemistry set for Christmas and we've been working through the experiments over the last few weeks.

One of them involves boiling white vinegar and adding baking soda to produce sodium acetate. After boiling this down it is supposed to result in some crystals forming on the pan. However, after trying it a couple times we haven't been successful in producing any crystals. What are we doing wrong?

Our process is:

Add 4 cups of distilled vinegar to a pan. Boil it on the stove.
Stir in 4 tablespoons of baking soda. Do this 1 tablespoon at a time, stirring gently to dissolve each spoonful before adding more.
Keep at a high boil until it is no longer fizzing.
Reduce to a medium heat and boil until only 1 cup remains. This usually takes about an hour.

At this point our booklet says that crystals should be forming around the outside of the pan. However, both times we have no crystals - only a liquid that looks like weak tea.

The solubility of sodium acetate is interesting (that means confusing!). Sodium acetate exists in two forms: anhydrous and trihydrate. When compounds exist in two forms like this, the solubilities, given in grams per 100 mL H20, are predictable: the hydrated form is much more soluble, because it is only partially the compound, and the other part is extra water, the solvent.

However, sodium acetate displays a different behavior:

https://en.wikipedia.org/wiki/Sodium_acetate

Sometimes different hydration stages of compounds have different temperature ranges of stability and solubility, but the behavior of sodium acetate is unique. The precipitate that forms so impressively upon cooling is probably the trihydrate, because the amount of sodium acetate in solution according to the recipe does not seem (in my quick estimate) to exceed the solubility of the anhydrous compound, and so the solution produced might be metastable. For some length of time. But, we may be able to initiate crystallization by scratching the glass container, or stirring it, or allowing a drop to dry completely to produce some solid, then using that as a seed for the rest of the solution. Or freezing it.

What I have done is to heat some (pure) sodium acetate to form a concentrated solution (calculated to be supersaturated at room temp), then cooled it: no ppt. Then I touched the surface of the solution with my fingertip and BOOM! Crystallization proceeded like a shot, producing a mass of crystals that filled the container and solidified the former liquid mass. (Did I say it was supersaturated?) I was surprised at the mass of crystals compared to the amount of material I had added to the water at the beginning, but did not think about it till now.

It turns out that the solubility curves of trihydrate and anhydrous sodium acetate are quite separated:

https://www.google.com/search?q=solubility+of+sodium+acetate&client=safari&rls=en&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjt-un3m4XlAhXrY98KHQExDiEQ_AUIESgB&biw=1004&bih=846#imgdii=qbOI-c2mtjJb3M:&imgrc=lRafmM5w0nsXhM:

The formation of trihydrate at low temperatures must be a difficult process, and the brown goo obtained here might signify some crystallization inhibitor is present. Maybe a slight excess of NaHCO3 or HCH3CO2 is holding things up - you could try adding a drop of the vinegar, or a few grains of NaHCO3 to see if that changed anything. But If you have a glass rod, scratch the side of the glass container - that works almost all the time.


What is the hybridization of the central carbon in buta‐1,2‐diene? State the hybridisation of asterisked carbon in $\ce{CH3-CH=\overset{\ast}{C}=CH2}.$

I am not sure between $\mathrm{sp^2}$ and $\mathrm{sp}$ hybridization. Because of the presence of double bonds I think it should be $\mathrm{sp^2},$ but because of the two π-bonds I think $\mathrm{sp}$.
Which one is correct?

You must be clear that the hybridization is decided by the number of σ-electrons, so for $\ce{CH4}$ the number of σ-electrons is four and so its hybridization is $\mathrm{sp^3}.$
In case of $\ce{CH3-CH=\overset{\ast}{C}=CH2}$ the $\ce{\overset{\ast}{C}}$ is having two σ- and two π-bonds, so its hybridization is $\mathrm{sp}.$
While deciding hybridization we need to only care about bonds, and that too only the σ-bonds. Remember that possible hybridization for $\ce{C}$ are $\mathrm{sp^3},$ $\mathrm{sp^2}$ and $\mathrm{sp}.$


What does A485 mean? And what does the values on the y-axis expresses? This question is from past exam paper. I don't want an answer to the multiple choice question below, I just want to understand what A485 means in the context of the question.

$A_{485}$ means the same thing as the text of the multiple choice question "absorbance of the solution at 485 nm".

"nm" is an abbreviation for nanometers.

In other words, the absorbance of light having a wavelength of 485 nanometers is plotted on the y-axis of the graphs.


In Ion Chromatography instrument, Does the specific conductivity (peak area) of an analyte depends upon background conductivity? In Ion Chromatography instrument,
Does the specific conductivity  (peak area) of an analyte depends upon background conductivity ?

Assume Specific Conductivity (peak area) of fluoride ion is 4 unit for 1 ppm fluoride standard with background conductivity of 15 uS/cm (Carbonate/ bicarbonate eluent with suppression.

What will happen if the background conductivity increases to 30 uS/cm or 50 uS/cm.

Pl do not consider the change in retention time.

https://www.metrohm.com/easyDB/Web-61006520_std_carbonate/orig

The heart of an ion chromatograph is the suppressor, whose sole purpose is to minimize the background conductivity and convert analytes into highly conductive ions for example fluoride ion becomes HF, chloride becomes HCl and so on. If your background conductivity is high, it means that the conversion of F(-), Cl(-) into corresponding acids is also not 100% efficient-peak height can potentially decrease as a result. The peak areas will also change. Although the background is higher, one can still make a calibration curve (it will be less sensitive than the one with low background).
Remember that peak area in chromatography is not an absolute measure like mass; it is only true for one particular condition and set-up.


Can I perform electroplating with sodium carbonate and copper? I want to avoid anything poisonous during electroplating. I can't avoid copper oxide though.

I have dissolved $\ce{Na2CO3}$ in water and I'm trying to electroplate aluminium with copper:

As you can see I'm doing it with no actual chemical equipment. Actually I'm trying to discover ways that can be done at home.

In the water, I can see copper hydroxide ($\ce{Cu(OH)2}$) and copper carbonate ($\ce{CuCO3}$), both formed from the copper anode. But the aluminium heatsink does not seem to have any copper on it. Will it take time or am I just wasting copper here?

If this is not gonna work, what is gonna work? Could I use vinegar?

You can try the electrodeposition in a solution prepared by mixing baking soda with vinegar and the copper sulfate. Wear glove and protection glasses. The aluminium must be cleaned before plating. Please let me know the result.

Acidic copper plating on aluminium will cause evolution of hydrogen, so less acidic solution is better. On the other hand you need to have the solution conductive and the baking soda could help.


Figuring out the theoretical mass of reactants I'm working on my chemistry lab report. The initial instructions were to test the hypothesis that the mass of products of a reaction will equal the mass of the reactants.
Basically we where given a equation, our job was to balance and figure out the theoretical mass of reactants, and the theoretical mass of products.
There we're a total of three questions. In the first one I got the following differential values, 0.2 and 0.1. This seemed to be the most logical for me since I knew the law of conservation of mass implied that mass cannot be created nor destroyed but could be transformed. I attributed this small differential value to this transformation theory.
However, the next two questions I attempted still gave a very large differential value (40-120), which rejects my initial hypothesis.
I believe I may have missed something in the second and third question, below are my attempted solutions.
Question 1:
Initial Equation: $\ce{CH4 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{CH4 + 2O2 -> CO2 + 2H2O}$
$$
\begin{array}{ccccccc}
\ce{CH4} & + & \ce{2O2} & \ce{->} & \ce{CO2} & + & \ce{2H2O}\\
1(12+4) & + & 2(32) & \ce{->} & 1(12+32) & + & 2(2+16)\\
1(16) & + & 2(32) & \ce{->} & 1(44) & + & 2(18)\\
16 & + & 64 & \ce{->} & 44 & + & 36\\
\end{array}
$$
Theoretical Calculation:
$$
\begin{array}{ccc}
\ce{CH4}=\pu{16g} & & \ce{H2O}=\pu{36g}\\
+ & \ce{->} & +\\
\ce{O2}=\pu{32g} & & \ce{CO2}=\pu{44g}\\
\end{array}
$$
Actual Calculation:
$
\ce{CO2}=\pu{43.8g}\\
\ce{H2O}=\pu{35.9g}\\
$
Differential Value(Theoretical Calculation - Actual Calculation):
$
\ce{CO2}=\pu{0.2g}\\
\ce{H2O}=\pu{0.1g}\\
$
Question 2:
Initial Equation: $\ce{C2H6 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}$
$$
\begin{array}{ccccccc}
\ce{2C2H6} & + & \ce{7O2} & \ce{->} & \ce{4CO2} & + & \ce{6H2O}\\
2(24+6) & + & 7(32) & \ce{->} & 4(12+32) & + & 6(2+16)\\
2(30) & + & 7(32) & \ce{->} & 4(44) & + & 6(18)\\
60 & + & 224 & \ce{->} & 176 & + & 108\\
\end{array}
$$
Theoretical Calculation:
$$
\begin{array}{ccc}
\ce{C2H6}=\pu{60g} & & \ce{H2O}=\pu{108g}\\
+ & \ce{->} & +\\
\ce{O2}=\pu{224g} & & \ce{CO2}=\pu{176g}\\
\end{array}
$$
Actual Calculation:
$
\ce{CO2}=\pu{78.59g}\\
\ce{H2O}=\pu{48.2g}\\
$
Differential Value(Theoretical Calculation - Actual Calculation):
$
\ce{CO2}=\pu{97.41g}\\
\ce{H2O}=\pu{48.2g}\\
$
Question 3:
Initial Equation: $\ce{C3H8 + O2 -> CO2 + H2O}$
Balanced Equation: $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$
$$
\begin{array}{ccccccc}
\ce{C3H8} & + & \ce{5O2} & \ce{->} & \ce{3CO2} & + & \ce{4H2O}\\
1(36+8) & + & 5(32) & \ce{->} & 3(12+32) & + & 4(2+16)\\
1(44) & + & 5(32) & \ce{->} & 3(44) & + & 4(18)\\
44 & + & 160 & \ce{->} & 132 & + & 72\\
\end{array}
$$
Theoretical Calculation:
$$
\begin{array}{ccc}
\ce{C3H8}=\pu{44g} & & \ce{H2O}=\pu{72g}\\
+ & \ce{->} & +\\
\ce{O2}=\pu{160g} & & \ce{CO2}=\pu{132g}\\
\end{array}
$$
Actual Calculation:
$
\ce{CO2}=\pu{82.5g}\\
\ce{H2O}=\pu{45.0g}\\
$
Differential Value(Theoretical Calculation - Actual Calculation):
$
\ce{CO2}=\pu{49.5g}\\
\ce{H2O}=\pu{27.0g}\\
$
Not sure if it matters, but we where given this program to enter our calculations to simulate the reactions.

Try to answer to these questions, don't cheat, check the answer passing over the yellow box.
I think this software is confusing you always start from how much reactants you have and then calculate how much products you will have, suppose you have $\pu{100g}$ of oxygen and $\pu{44g}$ of propane.
How many mole of propane and oxygen ($\ce{O2}$) do you have?

 Oxygen moles$=\frac{\pu{100g}}{\pu{32g/mol}}\sim \pu{3mol}$,  Propane moles$=\frac{\pu{44g}}{\pu{44.096g/mol}}\sim \pu{1 mol}$

If this is your balanced equation:
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$$
How many oxygen molecules do you need to react with one molecule of propane?

 You need 5 oxygen molecules for every propane molecule. So you need 5 moles of oxygen for every mole of propane.

Which is the ratio?

 $\frac{1}{5}$ if you have $x$ moles of oxygen you need $\frac{x}{5}$ moles of propane. e.g. if you have 5 moles of oxygen you need? 1 mole of propane.

But you have only 3 moles of oxygen to get the products, oxygen is a limiting reagent! So how many moles of propane will react:

 $$\pu{\frac{3}{5}mol} =\pu{0.6 mol} $$

So how many moles of the products will you have?

 $3\times0.6$ moles of $\ce{CO2}$ and $4\times0.6$ moles of water

And so how many grams?

 for $\ce{CO2}$, $\pu{1.8mol}\times \pu{44g/mol}$ for water, $\pu{2.4mol}\times \pu{18g/mol}$


What is the significance of molar mass What exactly is the significance of molar mass in chemistry? Why not just mass?

molar mass of a substance is the mass of one mole of a substance

This is not quite correct. If you said the speed of a car is the distance it travels in one hour, cars would go at speeds of 120 miles (race car) or 30 miles (in town). However, speed is defined as distance traveled per time. As a result, speed has the dimension of distance divided by time, and a typical unit (at least where I live) is miles per hour.

Molar mass is defined as the mass per amount of substance. The dimensions are mass divided by amount of substance, and typical units are g/mol.

  Why does molar mass have the same value as molecular mass?

If the molar mass is given in g/mol and the molecular mass is given in u (unified atomic mass unit) or Dalton, the number is the same. The reason is that both the mole (at least until May 2019) and the Dalton are based on the mass of the carbon-12 isotope. A mole of a substance is defined as the amount that has the same number of particles as present in 12 grams of carbon-12. The Dalton is define as 1/12th of the mass of a carbon-12 atom.

From these definitions, you will find that the molar mass of the carbon-12 isotope is 12 g/mol, exactly, and the molecular mass of carbon-12 is 12 Dalton, exactly.

  What will change in May 2019?

The mole will be defined as "One mole contains exactly 6.02214076×10^23 elementary entities." The molar mass of carbon-12 will be an experimentally determined quantity.


Is there any software to simulate a laboratory? I wanted to do some experiments with some chemicals but unfortunately I do not have access to them. Is there any software I can use to get the products?

I don't want to do very complicated things, I just want to gain some experience in basic chemistry, with simple chemicals like $\ce{H2O2}$, some salts, sulfuric acid, potassium permanganate, etc.

I tried to search on the net but most of the software I found is too complicated for me.

Do you know about any such software? Or do you have any alternatives?

There is a product called Yenka Chemistry which simulates basic chemicals/apparatus, and allows basic manipulations (for instance to show the pH change as more acid is added to base, or the amount of gas evolved).

The program evolved from another piece of software called Crocodile Clips, which was aimed at GCSE science in the United Kingdom. For this reason it is incredibly basic, mainly focusing on inorganic reactions (acid/base, titrations, reactivity of group 1 and 2 metals), with no real capacity for demonstrating organic reactions.


What is the mechanism of the Clemmensen reduction? I have learned that we can reduce a ketone to an alkane using the Clemmensen reduction, where we treat the ketone with amalgamated zinc in the presence of $\ce{HCl}$.

My book didn't mention the mechanisms for this type of reaction and I couldn't find an answer by searching Google.

How does this reaction actually take place? Are there carbocation rearrangements in this reaction?

As with many complex mechanisms, many variants can and indeed have been proposed. The most widely accepted is the one proposed by Rosanti (J. Org. Chem. 1991, 56, 4269. DOI: 10.1021/jo00013a036).

The key steps of the mechanism are given below, crucially showing that the alcohol is not an intermediate (hence why secondary alcohols are not able to undergo a Clemmensen reduction to afford the corresponding alkane).

A single electron reduction of the ketone takes place to give a radical anion. The exact species responsible for this  has been debated, but is thought to either be the zinc chloride radical (as shown) or the mercury chloride radical (not shown)
The radical anion picks up a hydrogen from water (the reaction is carried out in aqueous hydrochloric acid), this generates a hydroxyl radical, but also means that the alkoxide derived from the starting ketone is now simply an anion rather than a radical anion.
Homolytic cleavage of the C–O bond, the oxygen has now been completely removed from the substrate, simply leaving behind a secondary radical which must gain a hydrogen (again from water) in order to provide the desired product..

As shown below, several other steps also operate in order to quench out the various radical intermediates produced during the course of the reaction.

  Proposed mechanism of the Clemmensen reduction

  Ref: Comprehensive Organic Named Reactions, Wang.


Gaussian Error: Zero Point Energy Calculation on Fluoroform fails I have been performing isodesmic reaction calculations and one of my molecules is Fluoroform. It has already been optimized and its frequencies checked at the HF level of theory. However, when I went to calculate the energy at  B3LYP/6‐311+G(3df,2p) I received an error. I have isolated my error to the route section.

Below is my input file:

#T B3LYP/6‐311+G(3df,2p) Temperature =0

Optimization of Fluoroform

0 1
C
F     1    B1
F     1    B2   2    A1
F     1    B3   2    A2   3    D1
H     1    B4   2    A3   3    D2

B1  1.490000000
B2  1.490000000
B3  1.490000000
B4  1.090000000
A1  109.4712206
A2  109.4712206
A3  109.4712206
D1 -120.0000000
D2  120.0000000

Additionally, here is the error output Gaussian prints on termination:

QPErr --- A syntax error was detected in the input line.
 #T B3LYP/6â€311+G(3df,2p) Temperature =
           '
 Last state= "GCL"
 TCursr=      3523 LCursr=        10
 Error termination via Lnk1e in /cluster/software/gaussian/gaussian-09-D.01/g09/l1.exe

As you can see in the error message

#T B3LYP/6â€311+G(3df,2p)

there is some Unicode garbage in your input file; the dash between the 6 and 3 isn't an ASCII dash. Deleting and replacing it should fix the problem.


Why electronegativity instead of electropositivity When I learnt about polarity, I always come to the term electronegativity and always use the electronegativity chart. However, when I studied further, they have the word electroposivity.

So, I've been thinking why we use electronegavity more often than electropositivity? And why we have a electronegativity chart instead of electropositivity chart?

The first thing that should be highlighted here is that electropositivity is simply the opposite of electronegativity, any of the two can be used interchangeably with the necessary modifications to the sentence.

It is true though that electronegativity is more commonly used then electropositivity. For example, the Wikipedia article on the subject is electronegativity, and mentions electropositivity as its opposite. I believe there are two reasons for that, related to each other.

First: Consistency

It's very important in the scientific community, from reasons that go from facilitating students understanding to the writing of papers, that standards are used, and that everyone is talking about the same thing. It's easier if everyone is talking about the same thing, even if they are the complete opposite. It just makes us think faster. Why choose electronegativity then?

Second: How we explain chemistry

It's unecessary to explain why and how the study of the electrons "behaviour" in atoms and molecules are important to chemistry. Usually, when explaining a phenomenon, we talk about where the electrons are "going to". We say a reaction occurs because an electron is taken by an atom, or maybe because it's donated by one. We explain the polarity of the H-O bond by saying that the oxygen will attract the electrons of the covalent bond a lot more than hydrogen will, and not (usually) that the hydrogen doesn't attract the electrons as much as oxygen does.

Again, it's obvious that you can explain everything by means of electropositivity, but, in my experience, we usually explain things by saying which atom/group has the property that makes electrons go to them with higher value, rather than saying which atom/group has the property that makes electrons leave them with higher value.