text:
How do I determine the molecular vibrations of linear molecules?
I know how to do it for just about every other point group, but the $D_{\infty \mathrm h}$ and $C_{\infty \mathrm v}$ character tables aren't as straightforward. In particular, I'm interested in the vibrational modes of carbon dioxide, $\ce{CO2}$.
$$\begin{array}{c|cccc|cc} \hline
C_{\infty\mathrm{v}}  & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} &  & \\ \hline
\mathrm{A_1} \equiv \Sigma^+ & 1 & 1 & \cdots & 1 & z & x^2 + y^2, z^2 \\
\mathrm{A_2} \equiv \Sigma^- & 1 & 1 & \cdots & -1 & R_z & \\
\mathrm{E_1} \equiv \Pi      & 2 & 2 \cos\phi & \cdots & 0 & (x,y), (R_x,R_y) & (xz,yz) \\
\mathrm{E_2} \equiv \Delta   & 2 & 2 \cos 2\phi & \cdots & 0 & & (x^2-y^2,xy) \\
\mathrm{E_3} \equiv \Phi     & 2 & 2 \cos 3\phi & \cdots & 0 & & \\
\vdots & \vdots & \vdots & \ddots & \vdots &  & \\ \hline
\end{array}$$
$\,$
$$\small \begin{array}{c|cccccccc|cc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\mathrm{A_{1g}} \equiv \Sigma^+_{\mathrm{g}} & 1 & 1 & \cdots & 1 & 1 & 1 & \cdots & 1 &  & x^2 + y^2, z^2 \\
\mathrm{A_{2g}} \equiv \Sigma^-_{\mathrm{g}} & 1 & 1 & \cdots & -1 & 1 & 1 & \cdots & -1 & R_z & \\
\mathrm{E_{1g}} \equiv \Pi_{\mathrm{g}} & 2 & 2\cos\phi & \cdots & 0 & 2 & -2\cos\phi & \cdots & 0 & (R_x,R_y) & (xz,yz) \\
\mathrm{E_{2g}} \equiv \Delta_{\mathrm{g}} & 2 & 2\cos 2\phi & \cdots & 0 & 2 & 2\cos 2\phi & \cdots & 0 & & (x^2-y^2,xy) \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & & \\
\mathrm{A_{1u}} \equiv \Sigma^+_{\mathrm{u}} & 1 & 1 & \cdots & 1 & -1 & -1 & \cdots & -1 & z & \\
\mathrm{A_{2u}} \equiv \Sigma^-_{\mathrm{u}} & 1 & 1 & \cdots & -1 & -1 & -1 & \cdots & 1 & & \\
\mathrm{E_{1u}} \equiv \Pi_{\mathrm{u}} & 2 & 2\cos\phi & \cdots & 0 & -2 & 2\cos\phi & \cdots & 0 & (x,y) & \\
\mathrm{E_{2u}} \equiv \Delta_{\mathrm{u}} & 2 & 2\cos 2\phi & \cdots & 0 & -2 & -2\cos 2\phi & \cdots & 0 & & \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & & \\ \hline
\end{array}$$

Perhaps not the answer you are looking for, but the most straightforward way is just to learn what they are instead of figuring them out. It is quite logical, there are two stretching modes (symmetric $\Sigma_g^+$ + antisymmetric $\Sigma_u^+$) and two degenerate bending modes $\Pi_u$. You can visualise them here: http://www.chemtube3d.com/vibrationsCO2.htm

@orthocresol True, but that only gives an answer for the case where it is $D_{\infty h}$ with 3 atoms. This is the case that we see most often, but it is useful to have a general way to derive it rather than memorizing cases.

As Tyberius noted, the projection formula does not work for infinite order groups (this is because the Hermitian form on characters is defined to be G-invariant by averaging over all elements in a group. That is, it works because it is possible to hit all the elements of the group in some order).
How can one work in the infinite dimensional groups then? One property that still holds for these particular infinite groups (though not in general 1) is Maschke's Theorem, that every representation is the direct sum of irreducible representations. So we can still determine what irreps make up a rep by adding them up.
First we need to find the representation we wish to reduce. I find that the easiest way to do this is to start with just the atoms (instead of the vectors) and then include the vector contribution afterwards. For the sake of a new example, I will do the linear $\ce{I4^{2-}}$ anion instead of $\ce{CO2}$.
$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{atoms} & 4 & 4 & \cdots & 4 & 0 & 0 & \cdots & 0 &\\
\end{array}$$
Now, we multiply by the vector's contributions to the character.
$$\begin{array}{c|c}E&3\\\hline C_2&-1\\\hline \sigma&1\\\hline i&-3\\\hline C_n&1+2\cos(\frac{2\pi}{n})=1+2\cos\theta\\\hline S_n&-1+2\cos(\frac{2\pi}{n})=-1+2\cos\theta\end{array}$$
These come from taking the trace of the matrix representations of each symmetry operation. For example, a rotation about the z-axis by an angle $\theta$ is given by:
$$\left[\begin{array}{ccc}\cos\theta&\sin\theta&0\\-\sin\theta&\cos\theta&0\\0&0&1\end{array}\right]$$
$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{xyz} & 12 & 4+8\cos\theta & \cdots & 4 & 0 & 0 & \cdots & 0 &\\
\end{array}$$
Now, we are only interested in the vibrational modes so we must subtract off the translational and rotational degrees of freedom. We can check the character table for these (they are the ones which correspond to $z$, $(x,y)$ and $(R_x, R_y)$, $R_z$ doesn't count because we define the molecular axis to be the $z$-axis.
$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{xyz} & 12 & 4+8\cos\theta & \cdots & 4 & 0 & 0 & \cdots & 0 &\\
-\Gamma_\text{rot} & 2 & 2\cos\theta&\cdots&0 & 2 & -2\cos\theta&\cdots&0 \\
-\Gamma_\text{trans} & 3 & 1+2\cos\theta&\cdots& 1 & -3 & -1+2\cos\theta&\cdots&-1 \\
\hline\Gamma_\text{vib} & 7 & 3+4\cos\theta&\cdots& 3 & 1 & 1&\cdots&1 \\
\end{array}$$
So now we have the actual representation of the vibrations. How do we go about decomposing it? We want to eventually subtract off all of the irreps until we have the zero vector so we consider what effect each irrep has and try to move our remaining rep closer to zero. The first thing to note is that only $\Pi_g$ and $\Pi_u$ irreps have $2\cos\theta$. $\Pi_u$ adds $2\cos\theta$ under both proper and improper rotations, while $\Pi_g$ adds to proper and subtracts from improper. We note that in our rep, we only have $\cos\theta$ terms under proper rotations and so we need to subtract off enough $\Pi_u$'s until there are the same number of $\cos\theta$ (with different signs) under proper and improper rotations, at which point we should subtract off $\Pi_g$s until there are no $\cos\theta$ terms.
$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{vib} & 7 & 3+4\cos\theta&\cdots& 3 & 1 & 1&\cdots&1 \\
-\Pi_u & 2 & 2\cos\theta&\cdots&0 & -2 & 2\cos\theta&\cdots&1 \\
-\Pi_g & 2 & 2\cos\theta&\cdots&0 & 2 & -2\cos\theta&\cdots&1 \\
\hline\Gamma & 3 & 3 &\cdots& 3 & 1 & 1&\cdots&1 \\
\end{array}$$
Similar to the $\Pi_g/\Pi_u$, the $\Sigma_g^+$ shift adds 1 to all positions, while $\Sigma_u^+$ adds 1 to all positions before $i$ and subtracts from all after. We subtract enough $\Sigma_g^+$ until all the positions before and after $i$ have equal magnitude.
$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma & 3 & 3 &\cdots& 3 & 1 & 1&\cdots&1 \\
-2\Sigma_g^+ & 2 & 2 &\cdots&2 & 2 & 2&\cdots&2 \\
-\Sigma_u^+ & 1 & 1 &\cdots&1 & -1 & -1&\cdots&-1 \\
\hline \Gamma& 0 & 0 &\cdots&0 & 0 & 0&\cdots&0
\end{array}$$
Thus, we have found that the vibrational modes of $\ce{I_4^{2-}}$ to be $2\Sigma_g^+ +\Sigma_u^+ + \Pi_u + \Pi_g$.
We can check this with the formulas given in Group Theory And Chemistry by David M. Bishop as noted by Feodoran. One can also follow Tyberius's suggestion and work in $D_{2h}$ which will give the decomposition $2A_g+B_{1u}+B_{2u}+B_{3u}+B_{2g}+B_{3g}$.
It is important to note that correlation from the $D_{2h}$ point group to $D_{\infty h}$ works fine for vibrations because you cannot have terms higher than $\Pi$ and so there is no ambiguity. For things like electronic states, though, in which direct products must be decomposed, you will run into some problems. If the $D_{2h}$ result includes $A_g+B_{1g}$ does that correlate with a $\Sigma_g^+ +\Sigma_g^-$ or a $\Delta_g$ state in $D_{\infty h}$? Working in the parent infinite group may be more difficult without the reduction formula, but these problems are avoided.
The reason going to lower symmetry works is because all characters under the infinitely many proper and improper rotations are the same for each irrep (namely $2\cos n\theta$) and so we can represent the behavior of all of them by the behavior of one of them. Hence, we pick $\theta=\pi$ and ignore all the others and this produces the $D_{2h}$ point group (the two improper rotation axes are equivalent to $i$ and the two proper axes are equivalent to one $C_2$ about $z$). This is origin of the ambiguity. In $D_{\infty h}$, $\Sigma_g^+ + \Sigma_g^-$ are equal to $\Delta_g$ at $\theta=\pi$, but not in general.
Having now identified the origin of the ambiguity, one can tune how fast these ambiguities occur though. The problem occurs when the $\theta$ chosen to represent all $\theta$ has the property $1=\cos n\theta$ where $n$ represents the period of the ambiguity. So for $D_{2h}$, $\theta=\pi$ and $1=\cos2\theta$ and the period of the ambiguity is 2. That is, every other term will be ambiguously identical ($\Sigma, \Delta, \Gamma, \ldots$ are the same and $\Pi,\Phi,\ldots$ are the same). By dropping to $D_{4h}$ instead, $\theta=\pi/2$ and $1=\cos4\theta$. The ambiguity period is longer and we can distinguish $\Delta$ from $\Sigma$ by correlation from the $D_{4h}$ point group. We can thus select the point group to descend to based on how much resolving power we expect to need in the parent point group (though we should stick to even numbers to maintain inversion symmetry for $D_{\infty h}$). Of course we could also stay in the parent point group and never have a problem, but also have no reduction formula.

Instead of trying to use symmetry tables it is possible to calculate the vibrational normal modes directly using the equations of motion, rather as would be done for a double pendulum for example.
It is normally assumed that the vibrational motion of the atoms is harmonic, i.e. Hook's law applies, and we additionally assume that the only force constants are those between adjacent atoms; this is equivalent to assuming that the valence bond model describes the bonding. A molecular orbital model would consider forces constants between one atom and every other atom.
The starting point is to calculate the potential energy and, from this, the forces on the atoms. As carbon dioxide is symmetrical, there is only one force constant, k, but as the masses are different it is found that the mass-weighted force constants are needed in the calculation and these are different for each type of atom.
Stretching vibrations
The normal modes for a simple molecule can easily be sketched and as the molecule is linear there are $3N − 5$ or four modes in total, two are stretches and the other two are the degenerate bending motion in the plane of the page. One where the carbon atom moves down and the two oxygen atoms move up and then vice versa. The other bending normal mode is the similar motion but perpendicular to the plane of the page. Only the stretching modes are considered here.

Placing a vector $s_1,\,s_2,\,s_3$ on each atom and pointing in the same direction and along the long axis of the molecule, the potential energy V is the sum of terms for the stretching of each bond; therefore,
$$V(s_1,s_2,s_3)=\frac{k}{2}(s_1-s_2)^2+\frac{k}{2}(s_2-s_3)^2 $$
These equations are just for the stretching motion, a second calculation is needed with vectors perpendicular to the molecular axis for the bending modes.
The forces are the negative derivatives with respect to the displacements
$$-\frac{dV}{ds_1}=-k(s_1-s_2); \quad -\frac{dV}{ds_2}=k(s_1-s_2)-k(s_2-s_3);\quad -\frac{dV}{ds_3}=k(s_2-s_3)$$
With f as the force, placing these equations into matrix form gives
$$\begin{bmatrix}
f_1\\ f_2\\f_3
\end{bmatrix}
=
\begin{bmatrix}
-k & k & 0 \\
k & -2k & k\\
0 & k & -k
\end{bmatrix}
\begin{bmatrix}
s_1\\ s_2\\s_3
\end{bmatrix}$$
but, because the masses are different, we must change to mass weighted force constants, using the formula $K_{i,j}=k_{i,j}/\sqrt{m_im_j}$  where i and j are the atom indices; for example $K_{1,2} = k\sqrt{ m_Om_C}$, is the mass-weighted force constant between atoms 1 and 2, if $m_O$ is the oxygen mass and $m_C$ that of the carbon. The matrix of force constants becomes
$$\boldsymbol{K} =
\begin{bmatrix}
\displaystyle\frac{-k}{m_O} & \displaystyle\frac{k}{\sqrt{m_Om_C}} & 0 \\
\displaystyle\frac{k}{\sqrt{m_Om_C}} & \displaystyle\frac{-2k}{m_C} & \displaystyle\frac{k}{\sqrt{m_Om_C}}\\
0 & \displaystyle\frac{k}{\sqrt{m_Om_C}} &\displaystyle \frac{-k}{m_O}
\end{bmatrix}$$
which is solved as a secular determinant with eigenvalues $\lambda$,
$$ \lambda_1=-k/m_O; \quad \lambda_2= -k\frac{2m_O+m_C}{m_Om_C}; \quad \lambda_3 = 0 $$
The frequency of each vibration is $\omega^2 = −\lambda$  so the square of the normal mode frequencies are
$$\omega_1^2=k/m_O; \quad \omega_2^2 = k\frac{2m_O+m_C}{m_Om_C} ; \quad \omega_3=0$$
Using the measured stretching vibrational frequencies for $\ce{CO2}$, for example, which are 1337 & 2349 cm$^{-1}$, the force constants are 1680 & 1418 N/m respectively. Note that the frequency $\omega = 2\pi\nu c$ with $\nu$ in cm$^{-1}$.
The normalised eigenvector matrix x is, after some rearranging and with the total mass as $M=2m_O+m_C$
$$\boldsymbol {x}= \frac{1}{\sqrt{2M}}
\begin{bmatrix}
-\sqrt{M} & \sqrt{m_C} & \sqrt{2m_O} \\
0 & -2\sqrt{m_O} & \sqrt{2m_C}\\
\sqrt{M}& \sqrt{m_C} & \sqrt{2m_O}
\end{bmatrix}$$
Notice that the eigenvectors do not depend on the force constants; this matrix is used to produce the geometry that is related to the symmetry of the vibrations, and cannot depend on the value of the force constants. The normal modes depend only on the geometry because the ‘springs’ connecting the atoms can only vibrate in certain patterns governed by the geometry or symmetry of the molecule; the frequency and size of extension depend on the force constants.
The normal mode coordinates are calculated using $\boldsymbol{Q = x^Tq}$  where $\boldsymbol{x}^T$ is the transpose of the eigenvector matrix and q the vector of mass weighted coordinates with $q=s\sqrt{m}$.
$$\begin{bmatrix}
Q_1\\Q_2\\Q_3
\end{bmatrix}
=
\begin{bmatrix}
-\sqrt{M} & 0 & \sqrt{M} \\
\sqrt{m_C} & -2\sqrt{m_O} & \sqrt{m_C}\\
\sqrt{2m_O}& \sqrt{2m_C} & \sqrt{2m_O}
\end{bmatrix}
\begin{bmatrix}
s_1\sqrt{m_O}\\s_2\sqrt{m_C}\\s_3\sqrt{m_O}
\end{bmatrix}$$
and the individual modes are
$$Q_1=\sqrt{m_O/2}(-s_1+s_3); \quad Q_2 = \sqrt{\frac{m_om_C}{2M}}(s_1-2s_2+s_3);\\ \quad Q_3=(m_Os_1+m_Cs_2+m_Os_3)/\sqrt{M}$$
The symmetry of the normal mode can be seen from the displacements s. The last, $Q_3$, is just a translation of the molecule as all the s vectors point in the same direction and has frequency $\omega=0$. $Q_1$ would correspond to a symmetric stretch and therefore $Q_2$ to the assymetric stretch, both O atoms move in one direction and the carbom moves the opposite way to keep the centre of gravity fixed in space.
We can choose any Q to be zero to find the s displacements
(i) Suppose we choose $Q_1 = Q_2 = 0$, then all the atom displacements are the same, $s_1 = s_2 = s_3$ and are $Q_3/\sqrt{M}$. As each atom is moving in the same direction this cannot be a vibrational normal mode as the bonds are neither stretched nor compressed, but represents a translation, and clearly, this corresponds to the zero frequency eigenvalue $\omega_3 = 0$. Note that if $Q_3 = 0$ this must mean that the centre of mass does not change, no translation occurs, and so we expect to produce normal modes with this condition.
(ii) If $Q_1=Q_3=0$ then $\displaystyle s_1=s_3= \frac{m_C}{2m_OM}Q_2$ and $\displaystyle s_2= - \frac{2m_O}{2m_OM}Q_2$ In this case this is the asymmetric stretch with the oxygen atoms moving in the same direction and opposed to that of the carbon, and this is $\omega_2$. The relative motion is $s_1 = s_3 = 0.115Q_2$ to $s_2 = −0.308Q_2$ so the carbon atom moves further than the oxygen atoms do, which is not surprising, because the centre of mass has to be held constant and the C atom has to compensate for the motion of two O atoms.
(iii) Finally, if $Q_2 = Q_3 = 0$ then the equations are solved when $s_2 = 0$ and $s_1 = −s_3$, which is the symmetric stretch $\omega_1$ with displacements $\pm Q_1/\sqrt{2m_O}$ or $0.176Q_1$.
There is a very general method with which to do these types of calculations called the GFG matrix method or some similar name equivalent to this. As this is a matrix method it is easy to calculate equations using symbolic algebra programmes such as SymPy or to do the calculation numerically using for example Python. (Both SymPy & Python are free to use).
The secular determinant to be solved is
$$  | \boldsymbol{GFG} -\omega^2\boldsymbol{I} |=0$$
where I is a unit diagonal matrix, $\omega$ the normal mode frequencies and the matrices are G is a matrix zero everywhere except on the diagonal where it has values $1/\sqrt{m}$, i.e.
$$\boldsymbol{G} = diag \left [ 1/\sqrt{m_i}\right ] $$
and the F matrix is that of the force constants;
$$\boldsymbol{F}
=
\begin{bmatrix}
k_{11} & k_{12} & k_{13} & \cdots\\
k_{21} & k_{22} &\cdots & \cdots \\
\vdots & \vdots & & k_{nn}
\end{bmatrix}$$
and the product is just the mass weighted matrix of force constants
$$\boldsymbol{GFG}
=
\begin{bmatrix}
k_{11}/m_1 & k_{12}/\sqrt{m_1m_2} & k_{13}/\sqrt{m_1m_3} & \cdots\\
k_{21}/\sqrt{m_1m_2} & k_{22}/m_2 &\cdots & \cdots \\
\vdots & \vdots & & k_{nn}/\sqrt{m_n}
\end{bmatrix}$$
The eigenvectors x can be used to produce the normal mode displacements as $\boldsymbol{Q=x^Tq}$ where $\boldsymbol{q=G^{-1}s}$ so that each element is $q_i= s_i\sqrt{m_i}$ and the coordinate displacements via, $\boldsymbol{ s =GxQ}$.
Bending vibrations
In the case of bending vibrations in a molecule with masses $m_{1,2,3}$, such as HCN, the potential is written as
$$V= k_br_1r_2(\delta\theta)^2/2$$
where the bond lengths are $r_{1,2}$ and the bending force constant $k_b=k_\theta r_1r_2$ which keeps its units in N/m. For small angle bends $\displaystyle \delta\theta = \frac{(s_1 - s_2)}{r_1} -  \frac{(s_3 - s_2)}{r_2}$  (where vector s is perpendicular to the internuclear axis just as s was along the axis) and the mass weighted K matrix of force constants  becomes
$$K =
\begin{bmatrix}
\displaystyle \frac{k_3}{m_1}\left(\frac{r_2}{r_1}\right) & \displaystyle\frac{-k_3}{\sqrt{m_1m_2}}\left(1+\frac{r_2}{r_1}\right) & \displaystyle \frac{k_3}{\sqrt{m_1m_3}} \\
\displaystyle\frac{-k_3}{\sqrt{m_1m_2}}\left(1+\frac{r_2}{r_1}\right) & \displaystyle\frac{k_3}{m_2}(2+\frac{r_2}{r_1}+\frac{r_1}{r_2}) & \displaystyle \frac{-k_3}{\sqrt{m_2m_3}}\left(1+\frac{r_1}{r_2}\right)\\
\displaystyle \frac{k_3}{\sqrt{m_1m_3}} & \displaystyle\frac{-k_3}{\sqrt{m_2m_3}}\left(1+\frac{r_1}{r_2}\right) & \displaystyle\frac{k_3}{m_3}\left(\frac{r_1}{r_2} \right)
\end{bmatrix}$$
which means that the ratio of bond lengths must be known if they are different from 1.
The algebraic eigenvalues are immensely complex so it is necessary to calculate them numerically except for the case of $D_{\infty h}$ point group. In this case the force constants are zero, or $2k_3(2/m_2 +1/m_1)$ where $m_2$ is the central atom.
In the case of $\mathrm{CO}_2$, $m_1 = 16$, $m_2 = 12$ and the bond length $r=0.116$ nm. The only non zero (and doubly degenerate) frequency is $\omega^2 =2k_3(2/m_2 +1/m_1)$ . In $\mathrm{CO}_2$ the bending vibration has a frequency of $667$ cm$^{-1}$ and then $k_3 = 57$ N/m.
(source. The arguments presented here follow closely that in Beddard 'Applying Maths in the Chemical & Biomolecular Sciences, an example based approach' publ OUP, and where there is a more detailed description. Herzberg gives several examples (Vol II, chapter 2) and Wilson, Decius & Cross 'Molecular Vibrations describe the method in great detail.)

I actually didn't know how to do this either, but I've found how to work with these infinite symmetry point groups. You have noticed that the reduction formula would not work properly for these groups. So, what we can do to solve for $D_\mathrm {\infty h}$ is solve for the irreducible representation in a lower order symmetry group, say $D_\mathrm{2h}$, and then correlate the results to the higher order symmetry group using the tables below.
$$\begin{array}{c|cccccc}
C_\mathrm{\infty v} & \mathrm{A_1 = \Sigma^+} & \mathrm{A_2 = \Sigma^-} & \mathrm{E_1 = \Pi} & \mathrm{E_2 = \Delta} & \cdots \\ \hline
C_\mathrm{2v} & \mathrm{A_1} & \mathrm{A_2} & \mathrm{B_1 + B_2} & \mathrm{A_1 + A_2} & \cdots
\end{array}$$
$$\begin{array}{c|cccccc}
D_\mathrm{\infty h} & \Sigma_\mathrm{g}^+ & \Sigma_\mathrm{g}^- & \Pi_\mathrm{g} & \Delta_\mathrm{g} & \cdots & \Sigma_\mathrm{u}^+ & \Sigma_\mathrm{u}^- & \Pi_\mathrm{u} & \Delta_\mathrm{u} & \cdots \\ \hline
D_\mathrm{2h} & \mathrm{A_g} & \mathrm{B_{1g}} & \mathrm{B_{2g} + B_{3g}} & \mathrm{A_g + B_{1g}} & \cdots & \mathrm{B_{1u}} & \mathrm{A_u} & \mathrm{B_{2u} + B_{3u}} & \mathrm{A_u + B_{1u}} & \cdots
\end{array}$$
I got these tables, and a lot of information I didn't know about point groups, from this page from the University of Massachusetts Boston.
I also found a nice walk-through on Wikipedia, which explains how to do this for $\ce{CO2}$, which is identical in terms of symmetry grouping.

Again only part of an answer. As a reference I used the book Group Theory And Chemistry by David M. Bishop.
There it is stated (Chapter 9-8) that linear molecules are special cases, and they give generalized results for $N$ atomic molecules:

asymmetric molecules ($C_{\infty v}$) have:

$N-1$ longitudinal vibrations belonging to $\Sigma^+$
$N-2$ pairs of transverse vibrations belongig to $\Pi$

symmetric molecules ($D_{\infty h}$) have for even $N$:

$\frac{N}{2}$ longitudinal vibrations belonging to $\Sigma^+_g$
$\frac{N}{2}-1$ longitudinal vibrations belonging to $\Sigma^+_u$
$\frac{N}{2}-1$ pairs of transverse vibrations belonging to $\Pi_g$
$\frac{N}{2}-1$ pairs of transverse vibrations belonging to $\Pi_u$

symmetric molecules ($D_{\infty h}$) have for odd $N$:

$\frac{N-1}{2}$ longitudinal vibrations belonging to $\Sigma^+_g$
$\frac{N-1}{2}$ longitudinal vibrations belonging to $\Sigma^+_u$
$\frac{N-3}{2}$ pairs of transverse vibrations belonging to $\Pi_g$
$\frac{N-1}{2}$ pairs of transverse vibrations belonging to $\Pi_u$

I tried the $\ce{CO2}$ example, but didn't get very far: The representation we need to reduce is (Chapter 9-6 in the above book explains how to do this):
\begin{array}{c|cccccccc|cc}
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma^{red} & 9 & 3+6\cos\phi & \cdots & 3 & -3 & -1+2\cos\phi & \cdots & -1 \\
\end{array}
As already stated in Tyberius' answer, the problem is to apply the reduction formula. Here we need to sum over infinitely many $C_\infty^\phi$ axes. We can do this by integration, for example in the reduction formula for the IRREP $A_{1g}$ we get the following term:
\begin{equation}
    2\times 1\times\int\limits_0^{2\pi}(3+6\cos\phi)\mathrm{d}\phi = 6\pi
\end{equation}
In a similar way for the improper rotational axes $S_\infty^\phi$ we get $-4\pi$. What I don't know yet is how to treat the infinit many elements of the $\sigma_v$ and $C_2$ classes.

The reduction formula cannot be applied to infinite dimensional groups.


text:
How do we predict the properties of the materials we make through chemical reactions?
I find chemistry fascinating. I've taken a couple of classes and so am familiar with the rudiments of orbital theory, equilibria, acid/base chemistry, and a small bit of organic chemistry. So I can attempt to make predictions about which reactions might happen, but I'd like to know what the properties of the materials made are ahead of time. Is there a way to do this? I know that there are materials composed of the same elements in different structures that have vastly different properties (melting point, color, conductivity, etc.), so what can we actually do to predict these things (i.e. figure out the structure we'll get, and the properties these structures produce)?

Predicting properties is a massive can of worms. By and large chemists extrapolate. WE look at similar compounds and extrapolate how changes would effect properties of known compounds. // Some things like protein folding just defy calculation.

Well, you've described one of the goals of computational chemistry. It turns out that this problem is harder than intuition might predict. I recently read an article on the state of quantum computing that happens to summarize the issue for chemists nicely.

We do not have enough computing power to do this well.

Computational quantum chemical models have become more and more sophisticated to take advantage of increasing computing power, but there still is not enough computing power. Part of the issue is that at the center of what we are trying to do is a multibody problem, which means that any attempts to optimize the electronic state for one electron in a molecule perturbs the state for other electrons. Then we go and adjust those other electrons, which perturbs our first electrons. This is a certified Really Hard ProblemTM.
At the end of the day, we currently cannot determine exact solutions to multielectron wavefunctions. So approximations get made, and the approximations are getting better. In fact for smallish molecules, we are getting quite close to a variety of known experimental properties. However, for larger molecules, the algorithms still need trained and calibrated on real experimental data. As the molecules get bigger, or the materials being modeled become more complex, our ability to throw quantum chemical calculations at them diminishes. Modeling of proteins mostly uses molecular mechanics and statistical dynamics.

I think the problem is not really the computing power, but our models. Models that generate results that are quantitative often have a formal scaling of $N^6$ (CCSD) or $N^7$ (CCSD(T)). Even if our computational power will be 10 times stronger than what it is now, it would mean nothing for those models.

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text:
Organic molecules with fast non-radiative decay rates
What structural features in an organic molecule would promote fast non-radiative decay to a ground state, for instance in an aromatic π-conjugated system ?

Anything that has potential energy surfaces that cross or get close in energy. Offhand things like s-indacene and the indenofluorene family of compound, there are many others as well (azulene mentioned below). In general, if the $S_0$ and $S_1$ state for a $\pi$-conjugated organic come close in energy there is the possibility for all of that energy to be used on vibrations instead of luminescence. As far as predicive rules, that is a difficult thing to generalize. Perhaps you could argue that a low $S_0 \ce{->} S_1$ transition energy opens the possibility for non-radiative decay since these potential energy surfaces are not seperated by much energy, biradicaloid $\pi$ systems fall in this category.This is the same reason there is typically not emission from higher energy state, $S_n \ce{->} S_1$ transitions are usually quite close in energy and have crossings to get to $S_1$ efficiently through molecular vibrations. For further examples and explanations see, for example:
http://www.sciencedirect.com/science/article/pii/S0009261414008847?via%3Dihub
Or Josef Michl's excellent book "Excited States and Photochemistry of Organic Molecules" for more details on the topic.

Not an expert on this by any means but azulene was the example that we were taught (ultrafast S1->S0 internal conversion). Then again, I don't know what "structural features" would lead to an intersection between the PES's.

The cause fast non-radiative decay (excluding reaction, such as cis-trans isomerisation, as a non-radiative event ) is spin-orbit coupling which links singlet and triplet states and opens up a non-radiative pathway between them, in addition to fluorescence and internal conversion. The SO coupling 'breaks' the forbidden singlet-triplet selection rule. In internal conversion the operator is anharmonicity since no spin change is involved.
The second important factor, but only in the condensed phase,  is the gap between the initial and final electronic states, the smaller this is the the larger non-radiative rate constant. As this is an application of a Fermi Golden Rule it is quite general and applies to all molecules not just aromatics. This answer explains it in detail Non-radiative transitions or what features of rhodamine result in it being so highly fluorescent?
Additionally heavy atoms, as a substituent or a solvent, increase non-radiative rate constants (again by spin-orbit coupling) as do paramagnetic species such as molecular oxygen dissolved in solution.
Should also mention that dipole selection rules for fluorescence /phosphorescence via excited & ground state symmetry means an additional factor is present to determine if the excited state is long or short lived.

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text:
Do conjugated double bonds take priority over a +1 formal charge?
Do conjugated double bonds take priority over a +1 formal charge?
The first one is what I solved for (ignore the hexagon that has no double bonds attached to it), and the one below it is the answer from the solutions.
Why did they move the conjugated bonds around first? This ends up causing the formal charge to move in the opposite direction than mine. Does it matter for this problem?

Some may think that mesomeric structures be ‘real’, that molecules ‘resonate back and forth between them, like a pendulum’ or that there be any type of ordering within mesomeric structures.
All of that is not the case. Mesomeric structures are our crutches to understand what is only one state of the molecule. We draw them in a line interconnected by a series of mesomery arrows because we are uncapable of drawing multi-dimensionally. There is no ‘first’ or ‘later’, no ‘full single’ or ‘full double’ bond. You must understand that all those structures are equal and equally far from reality. Reality is much closer to a continuous wave function with electron deficiencies on the benzylic, ortho and para carbons, with varying (but constant for a given bond) bond orders somewhere between 1 and 2.
Your question in its original form is basically moot. However, a marking scheme of your exam may require you to draw five mesomeric structures rather than just four to get full marks. That is up to your teaching staff.

Given the symmetry of this benzyl cation, the "direction" does not matter. I have put direction in quotes because there actually is no such thing. Electron pushing arrows are a helpful tool (at the very least for accounting purposes), but it is not what is happening in this molecule (or most other cases, either). What you have here is a resonance structure, meaning that the true situation is not well modeled by a single Lewis structure, thus we use several, and the true situation is somewhere in between the structures with varying contributions from the individual ones. So it does not matter how one arrives at these structures, it is important that they be reasonable and do contribute.
One could argue that the answer given is more extensive than your attempt, but I would consider both equal. The reactivity of the molecule is not dominated by the aromatic ring, thus placing extra emphasis on it seems superfluous.

Our primary motive is to delocalise the charge. More delocalisation of charge implies more stability.
The book solution had moved the conjugated pie bonds inside the ring just to show that they are also in resonance and hence is counted towards the overall structure stabilisation. It doesn't at all matter in which direction your charge being delocalised. What matters is it should be delocalised to maximum extent. More delocalisation => more stability of the overall structure. And don't think the pie conjugation inside the ring gets delocalised first and then the positive charge. There is no such order. As I said what matters is delocalisation. You can delocalise it in any direction. So you are also correct and the book's answer is also correct.
The more contributing resonance structure you have, the more will be the stabilisation of the overall molecule. To know which resonance structure (canonical structure) is more contributing and which is least contributing read the article here: http://web.chem.ucla.edu/~harding/tutorials/resonance/imp_res_str.html


text:
How to assign priority groups of tartaric acid?
How do we assign priority groups of tartaric acid? Counting outwards from any of the chiral carbons we get the same atom.

First you have to rank the molecules around the two chiral centres of tartaric acid according to Cahn Ingold Prelog principle. (I have added the principle in below after this 3 points)
After ranking the molecules according to Cahn Ingold Prelog principle, then orient the molecule such that the lowest priority group is directed away from you. The three remaining groups then project toward you.

If you intend to find the configuration of the molecule (i.e. R or S), then let me add this point:

If the three groups projecting towards you are ordered from highest priority to lowest priority (i.e from rank 1 to rank 3) in clockwise manner, then the configuration is "R". If the three groups projecting towards you are ordered from highest priority to lowest priority (i.e from rank 1 to rank 3) in counterclockwise manner, then the configuration is "S". I mentioned "from rank 1 to rank 3" - because the 4th one is oriented away from us and is not considered while checking the order.

Cahn Ingold Prelog principle:
Rule to prioritize the groups attached to a chiral carbon:

Prioritize the four atoms, or groups of atoms, attached to the chiral center based on the atomic number of the atom that is bonded directly to the chiral center. The higher the atomic number, the higher the priority.

Like say you have carbon and lets the 4 atoms attached to it are: $\ce{H},\ce{OH},\ce{Cl},\ce{Br}$.
So here the carbon is chiral, as all the 4 atoms attached to it are different and now for ranking, we rank it on the basis of atomic number. Here as Br has highest atomic number, It would be ranked 1, then Cl would be ranked 2, then OH will be ranked 3 (oxygen will be considered as it is directly attached to the chiral carbon) and then hydrogen will be ranked last.

Atoms participating in double/triple bonds are considered to be bonded to an equivalent number of similar "phantom" atoms by single bonds. This is done just for the purpose of the rule. See this pic for better understanding:
If two or more of the atoms that are bonded directly to the chiral center are the same, then prioritize these groups based on the next set of atoms (i.e. atoms adjacent to the directly bonded atoms)

Lets take an example here. Say to the carbon I have $\ce{CH3},\ce{COOH},\ce{Cl},\ce{H}$ attached.
Here first we will compare the atomic number of the atoms which are directly attached to the chiral carbon i.e. C, Cl and H. From these as Cl has highest atomic number, hence it is ranked 1. Then, for the carbon, as for both the groups i.e $\ce{CH3}$'s and $\ce{COOH}$'s, it is the carbon atom which is directly connected to the chiral carbon. Hence, we need to look for the next set of atoms. $\ce{CH3}$ next set of atoms are H, H & H; while $\ce{COOH}$'s next set of atoms are O, H and O-C (see pic provided in rule 2). So from here easily we can say $\ce{COOH}$ is ranked 2, then $\ce{CH3}$ (i.e. rank 3) and then H.

Hi there, thank you for your answer. I was indeed asking about how to rank the molecules around the two chiral centre's of the tartaric acid. I am fine with the finding the configuration of a molecule once the groups have been ranked. Could you help me with ranking the groups please?

I have added the rules.

For simplicity, here is tartaric acid without any stereochemistry:
$$\ce{HOOC-CHOH-CHOH-COOH}$$
And here is a depiction that includes the stereochemical information:
 O=C-OH
   |
 H-C-OH
   |
HO-C-H
   |
 O=C-OH

(using the Fischer convention)
The chiral carbons are the central ones; we need to assign priorities to each of the atoms bound to them. For both, the discussion is similar, because they are chemically practically identical.

$\ce{O}$ from $\ce{OH}$.
$\ce{C}$ (carboxylic acid group)

$\ce{O}$ from the double bond
$\ce{O}$ from the $\ce{OH}$ group
$\ce{(O)}$ (a ghost oxygen) from the double bond.

$\ce{C}$ ($\ce{CHOH}$ group)

$\ce{O}$ from the $\ce{OH}$ group
$\ce{C}$ from the distal carboxy group
$\ce{H}$

$\ce{H}$

These lists are written out and the first point of difference is noted; here it is the higher priority of 2.2 (an oxygen atom) with respect to 3.2 (a carbon atom) — and this does not even require extensive knowledge on how to open double bonds.
Thus, the priorities of the groups on the two carbon atoms are:
 O=C-OH      b         .           O=C-OH
   |         |         :             |
 H-C-OH    d-C-a    ···c···        H-R-OH
   |         |         |     –>      |
HO-C-H    ···c···    a-C-d        HO-R-H
   |         :         |             |
 O=C-OH      ·         b           O=C-OH

And we realise that I have drawn (2R,3R)-tartaric acid.


text:
Why can two carbon atoms not form more than triple bond with each other?
Why two carbon atoms don't form more than triple bond with each other? Please tell me simply as I am a student of 10th class.

As you are in class 10, I believe that you haven't started learning about Molecular Orbital Theory. If you draw the MOT diagrams for $C_2$ molecule, you get a bond order of 2. This would mean that $C_2$ is stable.

Related: http://chemistry.stackexchange.com/questions/594/bonding-in-diatomic-c2-a-carbon-carbon-quadruple-bond

See: http://onlinelibrary.wiley.com/doi/10.1002/anie.201208206/abstract for a extensive discussion by Roald Hoffmann, Henry Rzepa and Sason Shaik in Angewandte Chemie about this topic.

Unfortunately, an answer to this requires a quick dip into orbital theory. You may have heard of s, p and d orbitals; carbon as a main group element of group 14 has the electronic configuration of $[\ce{He}]\,\mathrm{2s^2\,2p^2}$, meaning two of its electrons are in an s orbital and two in p orbitals in its atomic ground state. The process by which these orbitals form bonds is simply overlap: If you have two orbitals that can be moved together and these two orbitals overlap, you generate a bonding and an antibonding orbital from these two. (This is a mathematical operation known as linear combination; plugging two orbitals into a linear combination means that exactly two orbitals are mathematically required to come out of the linear combination.)
The bonding (or antibonding) orbitals that are generated by this process are labelled σ or π orbitals. Note that these are the Greek letters corresponding to s and p: a σ orbital is basically an s orbital streched in one direction and a π orbital is a p orbital streched in the same manner.
Due to a number of constraints that you will learn about soon enough™, a carbon–carbon triple bond $\ce{C#C}$ will consist of one σ bond and two π bonds as shown in the image below.

Figure 1: $\ce{H-C#C-H}$ showing the σ and π orbitals required to make the respective bonds. Image taken from jahschem.wikispaces.org.
See how two of the three possible p orbitals on each carbon (figure 1’s left half) overlap to form two π bonds; the third possible p orbital takes part in both the σ bond to the other carbon and that to the hydrogen — it has been linear combined with the s orbital to form an sp hybrid orbital outside the scope of this answer.
The main point of interest here is that it is, for reasons of symmetry, not possible for more than two p orbitals of a single atom to form π bonds to a single other atom at any one time. And carbon does not have any additional accessable orbitals that it could use for further bonds. Thus, it is not possible for carbon to form quadruple bonds.
The next orbital available would be a d orbital; the transition metals are those in which the d orbitals are valence orbitals and take part in bonding. Once you have a better understanding of orbitals, it may be worthwhile to look up quadruple bonds of transition metals such as in $\ce{[(H2O)Cr({\unicode{x3bc}-}OAc)4Cr(H2O)]}$ or dirhenium octachloride. The latter is particularly interesting because the fourth — δ, in analogy to d — bond can be cleaved by irradiation causing a structural change that clearly shows there must have been a quadruple bond previously. To the best of my knowledge, quadruple bonds have not been shown to exist between main group elements though.
And finally, $\ce{C2}$ does exist; however, explaining it requires orbital theories on a much higher level than in the scope of this answer.

When you say four bonds with each other, it means 1 hybrid orbital for sure which will be the sigma bond inbetween the two carbon atoms. Now inbetween two atoms you can never have more than 1 sigma bond , the reason is for sigma bond we need head one overlapping of hybrid orbitals. If hybrid orbital of one carbon atom is already in head on overlap with the hybrid orbital of the other carbon atom, the other 3 hybrid orbitals of the carbon atom cannot do any further head on overlapping because angle between the hybrid orbitals would be 109 degrees approx. So basically they cant reach the hybrid orbital of the other C atom. So you can't have more than 1 sigma bond.
As we need one hybrid orbital for sigma bond, we would be creating only one hybrid orbital then, hence sp hybridisation.
If we talk about four bonds inbetween the carbon atom, then the other three has to be pie bonds that will be from 3 pure p orbitals, but 3 pure p orbitals is not possible. Because one of the p orbital has already used up in sp hybridisation with the s orbital creating one sp hybrid orbital for sigma bonding. So we have only two pure p orbitals left with us. So how can you form 3 pie bonds when you have only 2 pure p orbital remaining with us. Hence its not possible to have 4 bonds inbetween two carbon atoms.

I think a picture in your answer would be helpful to illustrate why there is no more space for a third pi-bond. I only suggest this because I do not think the OP has yet covered, in class, the orientation of sigma and pi bonds in space.


text:
Ambiguous nature of aldehydic group in glucose

In aqueous solution, α-anomer and β-anomer of glucose
  remain in equilibrium with each other along with a small amount of
  glucose in open chain form ($0.02~\%$). The open chain form in the
  solution contains a free aldehydic group and hence gives several
  reactions involving free aldehyde.

Though little, the open chain form of glucose reacts with many reagents which are used to treat free aldehydic groups. Therefore, glucose in aqueous solution also reacts with those reagents.
$$\ce{glucose_{(aq)} + HCN -> gluco cyanohydrin} $$
$$\ce{glucose_{(aq)} + H2NOH -> gluco oxime} $$
$$\ce{glucose_{(aq)} + H2NNHPh -> gluco phenylhydrozone} $$
$$\ce{glucose_{(aq)} + 3H2NNHPh-> gluco osazone} $$
$$\ce{glucose_{(aq)} + Br2_{(aq)}-> gluconic acid} $$
Glucose also answers Tollen's reagent test, Fehling's solution test and Benedict's solution test.
However, as interesting it may seem, glucose is a two-face molecule. Here are some reactions which glucose does not like:
$$\ce{glucose_{(aq)} + NaHSO3 -> no reaction} $$
$$\ce{glucose_{(aq)} + {2,4-DNP} -> no reaction} $$
$$\ce{glucose_{(aq)} + {Schiff's reagent} -> no reaction} $$
Why is that glucose does not react with $\ce{NaHSO3}$ or $\ce{{2,4-DNP}}$ or Schiff's reagent while it answers many other reactions where the free aldehydic group is involved?
I am looking for a convincing answer. Most answers in web tell that some are strong and some aren't strong enough. If you give enough time, all the carbonyl reactions should work with glucose because there will always be $0.02~\%$ in open chain form. If some of it gets used up, then the equilibrium will shift and hence more open chain form of glucose will be formed.

glucose does react with brady's reagent to give an osazone,do make the correction

My textbook says it doesn't.  Can you cite some reference? I would expect it to react with DNP. Isn't the mechanism for reaction with phenylhydrazine with glucose same the reaction of DNP with glucose? So, if glucose can form hydrazones and osazones then it should react with Schiff's reagent, DNP, etc. too.

So you say that it will work. Agreed, but given enough "time" is the crunch factor here. A bit of math using the Arrhenius formulae, considering the extremely low collision frequency, would lead that time factor to be around $\ce{4.6 * 10^10}$ seconds

chemistry.stackexchange.com/questions/22269/does-glucose-react-with-bradys-reagent

@QuarkyLittleThing I have been to that question before asking this one. Unfortunately, I don't consider that answer as a credible source of reference/information.

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

im sorry, its with phenyl hydrazine

I do agree that DNP reacts with carbonyl compounds but glucose seems to show a weird behavior. The answerer of that question quoted text from the DNP wiki page (which says that DNP reacts with carbonyl compounds) and then wrote that glucose should, therefore, react with it. He/She could be someone just like me who expects glucose to react in such a manner. But as you see, glucose behaves differently, we should be careful before drawing such conclusions.

Keep in mind that just because you form an imine or a hydrazone does not mean that it will not ring close to form the analog of the hemiacetal. The ring version is probably not colored and quite possibly dominates the equilibrium in some of these cases.

The distal hydroxy groups are in a perfect distance to form pyranose (six-membered oxane) or furanose (five-membered oxolane) rings. Any equilibrium reaction that leaves the carbonyl group liberated for extended periods is prone to lose free aldehydes to the (unreactive) hemiacetal forms. Any reaction whose equilibrium is strongly product-sided or which proceeds somewhat irreversibly due to precipitation etc. will not be affected by that effect.
This explains why Fehling’s and Tollens’ solutions react with glucose: in both cases one of the products precipitates which drives the reaction to completion. None of the aldehyde that is used up in the test reaction will be regenerated. The same is true for the osazone formation which involves a redox step or the reaction with bromine which is also a redox reaction.
The reaction with sulphite to give sulphonic acids is an equilibrium: sulphite can attach but is equally quickly liberated again. However, sulphite is a rather soft nucleophile and an acceptable leaving group. The reaction is reversible and due to the formation of hemiacetals not much of the sulphonic acid will be observed. In general sulphite testing, the precipitation of the strongly hydrophilic sulphonate drives the reaction to completion but sugars themselves are already strongly hydrophilic so this driving force cannot exist.
A cyanohydrin, however, is a much more stable intermediate because cyanide is a much worse leaving group. Thus, here again we are driving a reaction to completion due to a practically irreversible step — and contrary to sulphite which is based on the same mechanism but is much more reversible.
The same difficulty occurs when attempting to generate imines: these are also prone to attack by the distal hydroxy functions giving O,N-acetals. While these may be stable products, they do not have the same features as the imines usually generated by the reaction of carbonyl compounds with amines and thus are not captured by the tests. This also prevents productive reaction of glucose with Schiff’s reagent — and additionally, Schiff’s test requires the addition of a sulphite which we learnt already is unfavourable.
The big question remains why some hydrazines react with glucose according to your list but some don’t. I have no satisfactory explanation why that would be the case. The only difference between 2,4-DNPH and phenyl hydrazine is the former’s electron-withdrawing groups that could lead to an overall slower reaction or less tendency to undergo the required redox step to form an oxazone. It is not a great explanation but maybe a starting point.

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text:
Rationale for titant standardization in alkalinity measurements [ISO 9963-1:1994]
I work in a environmental chemistry laboratory and I will start to do some alkalinity measurements in my project. My research group is pretty big on following any and all standardization possible, which most often mean compliance according to ISO-standards.
In the ISO-document Water quality - Determination of alkalinity - Part1: Determination of total and composite alkalinity (ISO 9963-1:1994), you need to prepare your titrant, in this case $0.1 \text{M HCl}$ and a standard solution $0.025 \text{M Na}_{2}\text{CO}_{3}$. Then you need to standardize your titrant using either a potentiometric detection (titrate a dilute standard solution to pH 4.5 and note the volume used) or visual endpoint detection (titrate a dilute standard solution with bromocresol green-methyl red indicator until color change). And lastly, a blank using water as the titrand instead of the standard solution.
Using these two acid volume consumption you then recalculate your $0.1 \text{M HCl}$ to it's actual concentration. And if you keep your stock acid solution for a longer period you should perform this standardization every week. Then you can perform the actual titration and determination of the alkalinity in the sample.
My question is, why is this necessary? Why does the ISO stipulate this standardization?
I understand that if I prepare an acid it won't be super precise since there are effects that influence the preparation such as my handling, age of chemical, pipetting, graduated cylinder error intervals etc. But the same would apply to my preparation of the sodium carbonate. In principle I am shifting the error effects from the acid preparation to the sodium carbonate preparation which I am standardizing against.

Hello and welcome to Chemistry.SE. If you have any questions about how the site works, feel free to take the short [tour] or visit the [help]. Best of luck with your question!

According to this listing, $\ce{Na2CO3}$ is on of the primary standards used in volumetric analysis, i.e. is easily available in a reproducible form and concentration, has a high molecular mass, is -- once the container is opened -- not (so much) hygroscopic; and is stable.
I speculate, a dilute solution of HCl (0.1 mol/L), if stored over weeks in a bottle recurrently opened, maybe stored on a window bench (seen so frequently in the undergrad's labs) subject to ventilation and underneath heating by the radiators does not have this resillience.
Using a cartrige to prepare a standard solution of HCl that you rinse into a volumetric flask and fill up with deionised water may be seen valid as alternative for a secondary standard by convention. (ISO standards are settled agreements; similar to ASTM standards, if you see things differently, you may contact them.)  Obviously, alteration of a single step within a whole chain of procedures may invalidate the protocol's fit into ISO; a less severe issue in a teaching lab than for a GMP-accredited lab.
Addendum
"How is the diluted solution of HCl obtained?" brought an other element into play that I did not consider earlier: the use of concentrated hydrochloric acid. If purchased and used freshly, it may be a batch of concentrated (37%) or even "fuming HCl" (about 40%), and both are hygroscopic.  Small errors about the concentration of the stem solution of this corrosive reagent may indeed yield larger systematic errors, down the road of repeated dilutions.  (Obviously not so important if the sole intend is to fill up again the bottle of 1 M or 2 M HCl for the extractive workups ...)

I think you might be correct that it has to do with stability and changes over time of the dilute acid. I would presume my dilute acid at preparation is as accurate as the standard solution I prepared but by the convention of the ISO-document it should always be standardized. I guess I never kept the acid around long enough to ever feel it would be warranted (except by virtue of being able to say the experiment was performed according to the ISO standard).

@NFA I just added a small, yet potentially significant part: concentrated HCl is hygroscopic.  At least sometimes in the preparative lab we don't care if dilution of the former will yield a 1.0 or 1.2 M HCl and hence pay less attention than the analytical chemists about the precise concentration of the concentrated stem solution.  Provided the intended 0.1 M HCl in your case, I'm not sure if such a dilute solution is still hygroscopic; if so, it could represent an additional source of error, too.


text:
Should an entire compound or just the redox-active element be written in redox pairs?
For oxidant/reductants and conjugate pairs, for example $\ce{MnO2 -> Mn^2+}$, do we write $\ce{Mn^4+}$ or $\ce{MnO2}$ as the oxidant? Websites conflicts what my text book says, because it appears that my text book would write the conjugate pair as $\ce{Mn^4+/Mn^2+}$ which implies that it considers $\ce{Mn^4+}$ is the oxidant not $\ce{MnO2}$.
I just realised my example wouldn't work for what I was trying to say since $\ce{MnO2}$ does not exist in ionic form. But using a compound that did exist in the ionic form, would I be saying that the oxidant is the entire compound or the ion that has a change in oxidation number?

You generally use whichever compound is actually present in solution.
For manganese(IV), it will practically always be $\ce{MnO2}$ so use that. Manganese(VII) will always be $\ce{MnO4-}$ so use that. Manganese(II) salts dissolved in water dissociate and coordination complexes of manganese(II) are very labile. Thus, it doesn’t make sense to assume any $\ce{[MnL_n]^2+}$ complex to predominate and you should simply use $\ce{Mn^2+}$.
It does, for example, make a difference if your iron redox pair is $\ce{Fe^2+/Fe^3+}$ or $\ce{[Fe(CN)6]^3-/[Fe(CN)6]^4-}$. Likewise, a reduction using tin(II) should better be written $\ce{Sn^2+/[SnCl6]^2-}$.
If you are in doubt whether something dissociates or whether it should be considered a single entity, the cleaner solution is probably to write the fully dissociated salt, so you can get away with using $\ce{Sn^2+/Sn^4+}$. However, as soon as molecules or oxyanions are involved you cannot resort to this; oxalate reductions have to be $\ce{C2O4^2-/CO2}$ and permanganate has to be $\ce{MnO4-/product}$.

Some data for different Co(III)/Co(II) and Fe(III)/Fe(II) couples: https://i.stack.imgur.com/nCYyg.jpg (Yeah, that's my handwriting haha)


text:
Basicity of boric acid
My book defines the basicity of the acid as the number of $\ce{H^+}$ ions furnished by one mole of the acid in solution. Well, I know that $\ce{B(OH)_3}$ is a weak Lewis acid and it accepts an $\ce{OH^-}$ to form an anion - thereby increasing the net $\ce{H^+}$ concentration in the solution.
So, what should exactly be its basicity? Should I just assume its basicity to be one as well - although it doesn't satisfy the definition?

Boric acid does not dissociate in aqueous solution, rather it is acidic due to its interaction with water molecules (i.e accepts $\ce{OH-}$) to form the tetrahydroxyborate ion:
$\ce{B(OH)3 + H2O -> [B(OH)4]- + H+}$
$\ce{B(OH)3}$ does not act as a proton donor but behaves as a Lewis acid i.e. it accepts a pair of electrons from $\ce{OH-}$ ion and in this process $\ce{H+}$ ions are furnished.
I understand monobasic means there should be 1 replaceable hydrogen atom in the molecule, but $\ce{B(OH)3}$ in itself actually doesn't contain any replaceable hydrogen atom; rather, on interaction with water (i.e after accepting a $\ce{OH-}$ ion) a hydrogen ion is liberated. But if you see the after effect, say adding a monobasic acid (which holds the definition correctly) like $\ce{HCl}$ to water, per molecule it liberates one $\ce{H+}$ ion, similarly each $\ce{B(OH)3}$ also liberates one $\ce{H+}$ after it accepts $\ce{OH-}$ from water. So only looking at the after effects/ final result, both $\ce{HCl}$ and $\ce{B(OH)3}$ liberate 1 hydrogen per molecule though the process through which it was liberated is different.
So considering the after effect we consider $\ce{B(OH)3}$ as an acid and as it finally liberates 1 electron per molecule (though after accepting $\ce{OH-}$), we still can consider it to be a monobasic Lewis acid. At least, Wikipedia calls it a monobasic Lewis acid.


text:
Relating pH and buffer capacity
So, here is the equation for a weak acid - strong base pH curve

And here is the expression gotten when the derivative of concentration of hydronium ions with respect to volume of base added is applied:

Additionally, here is the buffer capacity formula found:

Now, I want to be able to relate these two (buffer capacity versus pH). I've done a buffer capacity versus pH experiment wherein I measured how buffer capacity changes with pH. The graph I got reaches its peak when pH = pK value.
My challenge is, how do I relate the two equations for buffer capacity and pH. Additionally, is there any experiment you suggest I do to further the theoretical model of the pH curve (beyond just doing a simple acid-base titration for verification)?

I suggest you plot a logarithmic diagram of your system, i.e. pH = f(log Ci), where Ci is the concentration of each protolyte in the solution. The buffer capacity can then be plotted as
pH = f(log(β/ln10)).
If you have plotted your logarithmic diagram properly, you can directly plot the line for log(β/ln10).
I will give you an example:
Suppose we have a weak acid HA with the pka = 4.8.
The line for log(β/ln10) goes from (0; 0) along the line of log [H3O+] until it reaches the crossing with the line for log [A-]. Here it bends upwards 0.3 log C units above the crossing point. It then follows the line for log [A-] close to pH = pka. Here it bends down, passing pH = pka, 0.6 log C units below the line for log Ctot. Then the line for log(β/ln10) will follow the line for log [HA] until it reaches the crossing point with the line for
log [OH-]. Here the line for log(β/ln10) bends up again and passes the crossing point 0.3 units higher. Then the line for log(β/ln10) will follow the line for log [OH-] to pH = 14.
From this example, we can immediately identify a local maximum for the buffer capacity, i.e. at pH = pka. From the diagram, we can also identify two local minima, which will be at about pH 3 and close to pH 9. We can also directly estimate the maximum buffer capacity from the diagram.
Logarithmic diagrams are excellent tools for this kind of problems. If you understand the rules, you can plot the buffer capacity (log(β/ln10) directly in the diagram without any calculations.
I tried to copy in a logarithmic diagram of the example above, but it did not work.

This class of problems is solved quite simply. You start with Henderson - Hasselbalch $r = 10^{pH - pK}$ in which $r$ is the ratio if dissociated to undissociated acetate. It's easy to see that the fraction of the total acetate that is dissociated is $f_1$ = 1/(1+r). This is the charge on acetate ions. The buffering capacity of the acetate is then simply $\partial f_1\partial pH$. Water has a net charge of $Q_w = 10^{-pH} - 10^{pH - pK_w}$ The buffering capacity of water is then $\partial Q_w/\partial pH$ and the buffering of the mixture the sum of those two derivatives.
To see the relatioship bewtween buffering capacity and pH one simply calculates $mf_1 + Q_w$ in which $m$ is the number of moles of acetate in a liter of water and makes a plot vs pH.If you stay away from p < 3 (where you are 'titrating' water - not acetic acid) the curve should look like a stair step with the pK of acetic acid half way up the riser (charge $-m/2$). Differentiating this with respect to $pH$ gives the buffering capacity. Obviously, there will be a major bump at $pK_a$
The fact that the curve is so simple to compute means that if you hace software that will allow the fitting of user defined functions to data you can estimate $m$ and $pK$ from laboratory titration data.


text:
Does Br2/H2O oxidize all aldehydes including carbohydrates?
In many books, it is mentioned that aldehydes are easily oxidized even by mild oxidizing agents. However, I have never seen bromine water explicitly mentioned. But in the carbohydrates chapter, it is mentioned that glucose etc. are oxidized by bromine water.
So is it that only carbohydrates are oxidized by bromine water or all aldehydes?
Also, do they react with bromine water under the conditions of the bromine water test for unsaturation? Or do they react only under vigorous conditions?

Aldehydes, including aldoses, are oxidized to their respective carboxylic acids in the presence of $\ce{Br2}$ in $\ce{H2O}$. The reason this reaction is often discussed with carbohydrates is that it is useful for differentiating aldoses from from ketoses, which cannot be further oxidized. A solution of $\ce{Br2}$ and $\ce{H2O}$ will lose its characteristic brown color in the presence of aldoses, but will not in the presence of ketoses.
The general mechanism for the oxidation of an aldehyde under these conditions is:

Though ketones will be able to form a hydrate like aldehydes, no deprotonation can occur to eliminate the bromide ion as we see in aldehydes. Hence no further oxidation can occur.

How is the dihydroxy alkene formed? Shouldn’t an alcohol be created?

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


text:
What does the following sentence mean - "Li is least reactive but the strongest reducing agent among all alkali metals"
Isn't it contrasting? Is reactivity only concerned with kinetics of reaction ?
PS - I am a high school student, probably I might be missing small detail!

It means that the thermodynamics of reduction are more favorable but the kinetics are less so than with other metals.

Oh I see it. I guess I was right, thank you very much. Really grateful for your clarification....


text:
Environmental effects of terephthalic acid
What are the effects of terephthalic acid on the environment? Especially oceans
What would happen if the levels of terephthalic acid increased in the ocean?
What would be a way to reduce the impact of this?

Did you check the SDS of terephthalic acid for GHS hazard statements and pictograms? Is it a dangerous substance according to GHS?

No, how do i find them? @klaus Warzecha

I recommend Googling: "MSDS terephthalic acid"...

Here is an MSDS from Eastman Chemical company for Terephthalic acid. In a lot of cases, companies won't have a whole lot of information in their MSDS and some of the handling instructions are just generically generated, but this actually seems to have a lot of the information you are looking for. In particular, it notes that the $LC_{50}$ (the concentration at which $50\%$ of an exposed population dies)for fish is actually greater than the solubility of the substance in water. In general, it looks like it has a very low toxicity, though that doesn't necessarily rule out other ways it could be harmful.

According to "Health and Environmental Effects Profile for Terephthalic Acid, Environmental Criteria and Assessment Office, Office of Research and Development, U.S. Environmental Protection Agency, Cincinnati OH 45268, EPA Report Number EPA-600/X-84-109", terephthalic acid appears to be a remarkably benign compound with respect to long term health effects on mammals and microbes. Several species of bacteria were shown to thrive on a diet from which their only carbon source was terephthalic acid. Multiple studies showed that rats who were sustained on a diet of 5% terephthalic acid developed growths and stones in the bladder and kidney. Rats who's diets contained less than 1% terephthalic acid didn't exhibit any long term effects.
Regarding the environmental fate and transport of terephthalic acid, the above publication stated:

Pertinent data regarding the atmospheric fate and transport of tereph-
  thalic acid were not located in the searched literature. Terephthalic acid aerosols and fumes could conceivably be released during shipping accidents and from molten production processes, but there are no reports of such incidences.

Degradation of terephthalic acid by soil bacteria was shown by multiple studies to happen rapidly (near complete removal via several different studies within a few days) via both aerobic and anaerobic bacterial degradation.
The article states that, based on a single study in 1966, the half-life of terephthalic acid in aquatic media is estimated to be 1.1 years with respect to chemical destruction via hydroxyl radicals. However the net aquatic half life should be much shorter due to biodegradation.
With regard to global production of terephthalic acid, this Wikipedia articl states:

Virtually the entire world's supply of terephthalic acid and dimethyl
  terephthalate are consumed as precursors to polyethylene terephthalate
  (PET). World production in 1970 was around 1.75 million tonnes.[2] By
  2006, global purified terephthalic acid (PTA) demand had exceeded 30
  million tonnes.

Although this represents a significant increase in global production, the fact that terephthalic acid readily biodegrades rather than bioacumulates should result in minimal long term accumulation in the environment.
With the caveat that there is some dearth of studies on the topic, the above evidence seems to indicate that terephthalic acid is a relatively benign compound to both mammalian and bacterial life, and that it is fairly rapidly degraded in the environment.


text:
What is the difference between these two Electrochemical cell
can someone please tell me the difference between the 1st and 4th electrochemical cell in the picture, I understand that in the 4th cell Cu is 0.1M so does that make it nonspontaneous and would the cell voltage be different, and also someone know the cell voltage for the fourth cell.

Does the keyword "Nernst equation" ring a bell?

Ecell = Eocell - (RT/nF) ln Q
Where Eocell is standard cell potential
Standard reduction potential of Cu2+ = + 0.339 V
Standard reduction potential of Zn2+ = - 0.762 V
We know that more positive standard electrode potential implies more tendency to get reduced. Hence Cu2+ will be reduced. That means Zn will get oxidised.
Standard oxidation potential of Zn will be +0.762V
Now net reaction:
Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s)
Q = [Zn2+]/[Cu2+]
So difference in concentration in Cu2+ will make Q different and as Ecell = Eocell - (RT/nF) ln Q ,  Ecell will also change
Now your second question is to find Ecell for the fourth cell:
Ecell = Eocell - (RT/nF) ln Q
Ecell = Eocell - (2.303 RT/nF) log Q
(i.e just converted ln to log base 10)
2.303 RT/F at 298K = 0.059
Ecell = Eocell - (0.059/n) log Q
Where n is the total number of electrons exchange in the reaction.
Q = [Zn2+]/[Cu2+] = 1/0.1 = 10
Eocell = 0.339 V + 0.762 V = + 1.101 V
(Eo cell i.e. standard cell potential  = standard reduction potential + standard oxidation potential)
Zn(s) + Cu2+(aq) + 2e- --> Zn2+(aq) + Cu(s) + 2e-
In the net reaction you can note that net electron exchange is 2. Hence n = 2.
Now substituting all the values in the Ecell expression.
Ecell = Eocell - (0.059/n) log Q
Ecell = 1.101 - (0.059/2) log (10)
Ecell = 1.101 - 0.059/2
Ecell = 1.0715 V


text:
Do copper and ammonium hydroxide react?
According to the activity series positions of copper and ammonium, with ammonium being higher up on the series compared to copper, it would be predicted that ammonium replace copper in a single displacement reaction involving the two elements. Therefore, a reaction between copper and ammonium hydroxide would not be predicted to occur. Despite this, after placing a copper wire in a solution of ammonium hydroxide, the copper appears to react with the ammonium hydroxide, with the evidence for such being the case being that the solution turns from a clear liquid to a light blue-coloured liquid. Do these reactants indeed react? If they do, what are the products? (My suspicion is that the reactants yield copper hydroxide, which is light-blue coloured, ammonia, and hydrogen, as in $$\ce{Cu + 2NH4OH -> Cu(OH)2 + 2NH3 + H2}$$)

But ammonium is not a metal. It is not on the activity series in the first place.

Did you exclude air from the solution?  Air could oxidize the copper.

But the problem is that copper(II) hydroxide will further react with ammonia solution to form tetraammine complexes and other various copper-ammine complexes and thus you will not get a definite product and only get a mixture of compounds.

@NilayGhosh I object. You will get $\ce{Cu(NH3)42}$, and that's pretty definite.

No, I believe that copper(II) hydroxide will not form rather a soluble copper-ammonium complex will be formed.
This handbook discuss the corrosion of copper in ammonium hydroxide solution:

Copper and copper alloys suffers rapid attack by concentrated solution of ammonium hydroxide to form a soluble complex of copper-ammonium compound. In some cases, the corrosion of copper exposed to dilute solution of ammonium hydroxide is low. Copper specimen submerged at room temperature for one week in 0.01N ammonium hydroxide solution suffered a weight loss of 3.5mil/yr.

Also, this paper discuss the rates of Copper Dissolution in Aqueous Ammonium Hydroxide Solutions:

The rates of dissolution of polycrystalline metallic copper in aqueous ammonium hydroxide solutions have been determined under various experimental conditions. Over a considerable range of conditions the dissolution has been found to be autocatalytic. An empirical equation is presented which represents the autocatalytic rate data with an average deviation of $\ce{\pm4\%}$. The rate equation is discussed in terms of a mechanism which assumes that the rate controlling process is the removal of cuprous ion species from the copper solution interface by diffusion. Data are also given for an essentially zero-order rate of dissolution observed when the dissolving medium was concentrated with respect to oxygen or very dilute with respect to cupric ion species. The autocatalytic dissolution is well represented by a half-order rate equation.


text:
How will Back-Bonding affect molecular geometry?
While going through a textbook, I came across the following line:

$\ce{R3C-}$ is pyramidal while $\ce{(CN)3C-}$ is planar due to lack of back bonding from the lone pair of carbon atom into the $\pi^*$ orbital of the $\ce{CN}$ group.

I am having trouble understanding how exactly back-bonding occurs in the first species and how it causes the structure to be pyramidal while the latter is planar.

You might think of it as resonance instead of back-bonding...

related http://chemistry.stackexchange.com/questions/32354/is-the-conjugate-base-of-trinitromethane-aromatic-what-is-y-aromaticity


text:
Models for diastereoselectivity in nitronate additions
In the scheme below (taken from the referenced Chem. Eur. J. paper on the synthesis of Tamiflu), there is a step in which a nitronate anion (the anion derived from 5) is added into an electrophile 6, undergoing what is essentially a Michael addition.
This Michael addition sets the nitro stereocentre, as the nitronate can evidently react from one of two faces. This isn't entirely unusual, the same thing happens with enolates during aldol reactions, and the facial selectivity is explained using various models depending on whether the reaction goes via an open or closed TS (i.e. Zimmerman Traxler vs open chain models).
Clearly the nitronate anion is able to exist in several conformations, however if one draws the Newman projection, then the conformation leading to the product is the one in which H is perpendicular to $\ce{C=N}$ as if were the large group (not the most sterically favored since the ester and the alkyl chain are clearly larger).
I'm hoping someone might know the name of a model that explains the selectivity (in the same way that a Zimmerman Traxler TS rationalises a chelated aldol and the Felkin Ahn model may be used to rationalise the stereochemistry from addition into chiral aldehydes). The closest thing I can think of is the Houk model, however, this itself doesn't give the correct product and I can find no literature from Houk's group that mentions this class of reactions.

Ref: H. Ishikawa, T. Suzuki, H. Orita, T. Uchimaru, Y. Hayashi, Chem. Eur. J. 2010, 16, 12616. DOI: 10.1002/chem.201001108.
High-Yielding Synthesis of the Anti-Influenza Neuraminidase Inhibitor (−)-Oseltamivir by Two “One-Pot” Sequences

Dave Evans taught me that you should generally analyze these types of problems by drawing out allyl conformations.

I've drawn the nitronate to minimize 1,3-allylic strain. Now, the top face of the nitronate is largely blocked by the t-butyl carboxylate. I strongly suspect that having a bulking alkyl group there is important to getting good selectivity.
This means that the electrophile will react on the bottom face, and that will lead to the desired stereochemistry depicted in 7 with the nitro group down on the ring.


text:
Steady-state approximation for the destruction of ozone
As a personal problem, I am trying to find a rate law for the reaction $\ce{2 O3 -> 3 O2}$ using the steady-state approximation.
The elementary reactions are, \eqref{eq:equilib} (forward rate constant $k_1$, reverse rate constant $k_{-1}$), and \eqref{eq:destroy} (forward rate constant $k_2$):
\begin{align}
\ce{O3 &<=> O2 + O} \tag1\label{eq:equilib}\\
\ce{O + O3 &-> 2 O2} \tag2\label{eq:destroy}
\end{align}
I go about by setting the total rate of disappearance of $\ce{O}$ equal to that of its formation. Thus,
$$k_1[\ce{O3}] = k_{-1}[\ce{O2}][\ce{O}] + k_2[\ce{O}][\ce{O3}]$$
and rearranging gives
$$[\ce{O}] = \frac{k_1[\ce{O3}]}{k_{-1}[\ce{O2}] + k_2[\ce{O3}]}$$
Next, I express the overall rate in terms of the disappearance of $\ce{O3}$. This means that
$$\text{Rate} = -\frac{\mathrm{d}[\ce{O3}]}{\mathrm{d}t} = k_2[\ce{O}][\ce{O3}].$$
Then I can substitute the expression for $\ce{[O]}$ to get an expression for the overall rate, and I get
$$\text{Rate} = \frac{k_1 k_2[\ce{O3}]^2}{k_{-1}[\ce{O2}] + k_2[\ce{O3}]}$$
With the step above lies my question. Is $-\frac{\mathrm{d}[\ce{O3}]}{\mathrm{d}t}$ equal to what I have above, or does it also include the rate of disappearance of $\ce{[O3]}$ from the first elementary reaction?
$$\text{Rate} = -\frac{\mathrm{d}[\ce{O3}]}{\mathrm{d}t} = k_1[\ce{O3}] + k_2[\ce{O}][\ce{O3}],$$
as opposed to just
$$\text{Rate} = k_2[\ce{O}][\ce{O3}]?$$
I suppose the former is correct, because when $k_2$ is small, the expression reduces to one obtained by simpler methods ($\text{Rate} = \frac{k_1k_2[\ce{O3}]^2}{k_{-1}[\ce{O2}]}$). However, it seems that I do need to include the first elementary step, since that is involved in consuming $\ce{[O3]}$.

The rate of change of ozone has three terms,  two from the equilibrium reaction and one from the reaction with O atoms, $\frac{d\ce{[O_3]}}{dt} = -\ce{k_1[O_3]  +k_{-1}[O][O2] -  k_2[O][O_3] } $  (To get steady state the general way would be to  put $d[O]/dt = 0$ which gives the terms you quote)

Yet if I defined the rate of reaction based on the rate of appearance of $\ce{O_2}$, as $\ce{\frac{1}{2}\frac{d[O_2]}{dt}}$, then I get a different expression: Rate = $\ce{k_2[O][O_3]}$, which differs from that given by using -$\ce{\frac{d[O_3]}{dt}}$

The rate of change of dioxygen  also has three terms.

Applying the steady-state and rate-determinign-step aproximations, we have:
$$\ce{2 O3 -> 3 O2}$$
$$\ce{O3 + M <=> O2 + O + M}$$
$$\ce{O + O3 -> 2 O2}$$
$$R=\frac{1}{2}\frac{\mathrm{d}[\ce{O2}]}{\mathrm{d}t} = k_2[\ce{O}][\ce{O3}]$$
$$\frac{\mathrm{d}[\ce{O}]}{\mathrm{d}t} =0= k_1[\ce{O3}][\ce{M}] -k_{-1}[\ce{O2}][\ce{O}][\ce{M}] - k_2[\ce{O}][\ce{O3}]$$
$$[\ce{O}] = \frac{k_1[\ce{O3}][\ce{M}]}{k_{-1}[\ce{O2}][\ce{M}] + k_2[\ce{O3}]}$$
$$R=\frac{1}{2}\frac{\mathrm{d}[\ce{O2}]}{\mathrm{d}t} =\frac{k_1k_2[\ce{O3}]^2[\ce{M}]}{k_{-1}[\ce{O2}][\ce{M}] + k_2[\ce{O3}]}$$
$$k_{-1}[\ce{O2}][\ce{M}] >> k_2[\ce{O3}]$$
$${Rate} = \frac{k_1k_2[\ce{O3}]^2}{k_{-1}[\ce{O2}]}=k_{eff}\frac{[\ce{O3}]^2}{[\ce{O2}]}$$

why are you assuming that k−1[O2][M]>>k2[O3]? That is not a given here. If you make that assumption, you don't even need to do the steady-state approximation, as you can just use the equilibrium expression for the first step.

Due to the fact that k-1>>k2, because the second elementary reaction is the rate determining step. This means that its rate is much slower, so the k2 is also tiny

Nowhere in the question does it say that the second elementary reaction is the rate determining step

Well, it's just how it is in real life this specific reaction.

And that's the way reaction rates are deduced.

The error in your logic is when you write:

Next, I express the overall rate in terms of the disappearance of $\ce{O3}$. This means that
Rate$=−\frac{d[\ce{O3}]}{dt}=k_2[\ce{O}][\ce{O3}]$.

For any reaction $\ce{aA + bB -> cC}$, the overall rate is given by
Rate $=-\frac{1}{a}\frac{d[\ce{A}]}{dt}=-\frac{1}{b}\frac{d[\ce{B}]}{dt}=\frac{1}{c}\frac{d[\ce{C}]}{dt}$.
So in your case, the overall stoichiometry is $\ce{3O2 -> 2 O3}$, so your overall rate is equal to $-\frac{1}{2}\frac{d[\ce{O3}]}{dt}$.
Going back to your quoted expression, we instead have
Rate$=−\frac{1}{2}\frac{d[\ce{O3}]}{dt}=k_2[\ce{O}][\ce{O3}]$, so $=\frac{d[\ce{O3}]}{dt}=-2k_2[\ce{O}][\ce{O3}]$.
From there, substitute just as you did to get the final answer.
UPDATE: Apparently this was problem 7.3 of the International Chemistry Olympiad exam in 2005. Their derivation of the answer is different from mine (though giving the same result), so I'm adding it here for completeness, with some explanatory text for the steps.
Start with the expression for the net rate of decomposition of ozone:
$$Rate=−\frac{d[\ce{O3}]}{dt}=k_1[\ce{O3}]-k_{-1}[\ce{O2}][\ce{O}]+k_2[\ce{O}][\ce{O3}]$$
Make a steady-state assumption for the intermediate O (as shown in the question above) to get the equation
$$\frac{d[\ce{O}]}{dt}=0=k_1[\ce{O3}]−k_{−1}[\ce{O2}][\ce{O}]−k_2[\ce{O}][\ce{O3}]$$
Rearranging terms, we get
$$k_1[\ce{O3}]−k_{−1}[\ce{O2}][\ce{O}]=k_2[\ce{O}][\ce{O3}]$$
The left side represents two of the terms in our original equation, so we can substitute the right side term for them:
$$Rate=−\frac{d[\ce{O3}]}{dt}=k_1[\ce{O3}]-k_{-1}[\ce{O2}][\ce{O}]+k_2[\ce{O}][\ce{O3}]$$$$=k_2[\ce{O}][\ce{O3}]+k_2[\ce{O}][\ce{O3}]$$$$=2k_2[\ce{O}][\ce{O3}]$$
To finish, we substitute in the expression for [O] that can also be derived from the steady-state equation (as shown in the question) and get a final result of
$$Rate=\frac{2k_1k_2[\ce{O3}]^2}{k_{−1}[\ce{O2}]+k_2[\ce{O3}]}$$

I'm sorry, but I think you're wrong.

First, if you consider the global reaction rate as the decreasing of the ozone, you must also consider the 3 terms involved with it.

Second, is a mistake to say that the stoichiometric coefficient of the ozone is 2, because the definition is only valid for elementary steps (you are considering the global reaction), so it would be 1.

The reason you do not need to consider all three terms is the steady-state assumption, which requires that step 1 happens only once (net) per reaction 2. If forward step 1 is faster than step 2, it is offset by the reverse step 1, so that O that is produced in step 1 is consumed in step 2 at the same rate. That means that the overall rate of O3 consumption is intuitively just 2 times the rate of step 2.

Well, that's another way to solve it, but less practical, due to its complexity.

@Andrew solution is correct and is the common solution scheme for these mechanism steady-state problems in chemistry competitions. Sometimes, the question would then ask if k2 >> k1, what is the simplified rate expression which is what you got.


text:
How to determine what compounds form hydrogen bonds?
Which of the following compounds can form hydrogen bonds?
a. $\ce{H2S}$
b. $\ce{CH3OCH3}$
c. $\ce{HCl}$
d. $\ce{BH3}$
e. $\ce{NH3}$
Why isn't $\ce{CH3OCH3}$ since $\ce{H}$ is bonded with $\ce{O}$ whereas $\ce{H}$ in $\ce{NH3}$ is not bonded to any oxygens?

In dimethyl ether, $\ce {CH_3OCH_3}$, all the hydrogen atoms are bonded to carbon in methyl groups.  Only the carbon atoms are bonded to oxygen.
Ammonia has no oxygen atoms.  But nitrogen is electronegative enough and compact enough, so that hydrogen bonded to the nitrogen can hydrogen-bond to an adjacent ammonia molecule.
Generally the strongest hydrogen bonds form between molecules in which hydrogen is bonded to nitrogen, oxygen, or fluorine.

Then what about BH3. It has 3 H like NH3.

Read that last sentence.  You need a compact electronegative atom to polarize the hydrogen.

I did not see the rest of the answer, thanks.

In general: There is a proton and another atom with lone pairs, there is hydrogen bonding. The determining quality is here how strong it'll be.

Whoever downvoted please check the last paragraph I edited, is that OK now?


text:
How to measure/know the moles of carbon dioxide (as dry ice) when in KOH solution
I am doing an experiment where I am dropping dry ice in a $\ce{KOH}$ solution. I want to know how I can figure out how many moles of $\ce{CO2}$ reacted with the $\ce{KOH}$. I am pouring $25 \,\mathrm{g}$ of $\ce{KOH}$ in $100 \,\mathrm{mL}$ of water and I am using $50 \,\mathrm{g}$ of dry ice. The experiment is conducted at room temperature.
I know one way to do this is by isolating the product of $\ce{K2CO3}$ but because $\ce{K2CO3}$ is soluble, it is going to be very hard to isolate this product and weight it and use simple stoichiometry to figure out the moles of $\ce{CO2}$ reacted.
I was thinking maybe the pH meter will help but again, I do not know how to use the pH to calculate how many moles of carbon dioxide reacted.
Observations
During experimentation, I saw the pH decreased when I added the $\ce{CO2}$.Will this fact be useful to calculate moles of $\ce{K2CO3}$ formed? The pH decreased from 13 to 10.
Can someone please help?

weigh-difference of the solution from before and after adding $CO_2$ and letting it react?

The pH meter will only help if you know there is some direct correlation between the addition of $CO_2$ and the pH of the solution during the reaction.  Do you know definitively if this is the case?

No I do not I was only guessing that CO2 in solution behaves as acid so maybe there might be an increase in pH

@Fl.pf. wouldn't there be conversation of mass

@user510 Yes? But that has nothing to do with anyhting here.
You have 100 ml of water with 25 ml KOH = 100g + how much 25 g KOH weighs.
After it has reacted with the CO2 you have 100 ml Water + rest KOH + K2CO3 - whatever leaves the reaction. Then you can easily calculate how much CO2 reacted..

In this particular case weighing would work with limited accuracy (I am going to explain the limitations later in the answer).
Since the only compound leaving the reaction is the unreacted $\ce{CO2}$, it is valid to use weighing. Everything else is either sticking around or is already included in the initial weighing. So by weighing everything prior to reaction and after "all unreacted $\ce{CO2}$ is vaporized", you can calculate the amount of "reacted $\ce{CO2}$".
(weight of reacted $\ce{CO2}$) = (Weight of initial $\ce{CO2}$) - (weight of unreacted $\ce{CO2}$)
(Weight of initial $\ce{CO2}$) = (50 g) - (condensed water from atmosphere [estimate])
(Weight of unreacted $\ce{CO2}$) = (deficit in pre vs post reaction weighing)
pre reaction weight: 50 g $\ce{CO2}$ + 25 g $\ce{KOH}$ +100 g water = 175 g
post reaction weight: weight of the whole solution after all unreacted $\ce{CO2}$ is escaped from the solution (make sure the whole solution is at room temperature and make sure there is no bubbling occurring).
I understand there are some practical limitations, mainly :
1- There might be unreacted/frozen $\ce{CO2}$ still hanging around, Try to shake well. Also when weighing post reaction, make sure your solution is slightly warm to assist unreacted $\ce{CO2}$ to evaporate fully from the reaction.
2-Since you have dry ice, some water from atmosphere will condense into your reaction, this will lead to underestimation of the "leaving compound". The way to overcome, is to keep your reaction vessel as small as possible and control its inlet by a cold trap to make sure no water from atmosphere gets in. Also account for amount of water usually present in 50 g dry ice, by examining the water content on a similar sample beforehand.
pH:
Your initial pH is clearly not right. concentration of your $\ce{OH-}$ is way higher than just 0.1 M. Let's assume that it is due to high pH value being out of the calibration range of the pH-meter and let's assume that your pH=10 after addition of $\ce{CO2}$ is correct.
One will be able to calculate the amount of unreacted $\ce{CO2}$ from reaction balance. You are starting with 25 g (0.446 moles) of $\ce{KOH}$ and 50 g (1.14 moles) of $\ce{CO2}$. For every $\ce{CO2}$ reacting, 2 $\ce{KOH}$ are used up. So IF all(or most) of $\ce{KOH}$ is used up, you should end up with 0.446/2~ 0.22 moles of $\ce{K2CO3}$, which means the reacted CO2 is about 0.22 moles(~9.7 g), and unreacted is 0.92 moles(~40.3 g).
Why do I make the assumption that all the $\ce{KOH}$ is reacted, notice your final pH is 10 , meaning that your [OH-] concentration is about 0.0001 M, your starting concentration was 4.4 M.
What stops the reaction from using all $\ce{KOH}$ to be used up and dropping the pH to neutral zone? You are using excess $\ce{CO2}$, so we expect all the $\ce{KOH}$ to be used up. This has to do with part of $\ce{CO2}$ escaping the reaction as gas, also since the temperature is freezing the water (by dry ice) you may not have complete agitation which hinders the reaction procession to some extent.
I still think that you should take the weighing method seriously. Using weighing methods will provide confirmation for your calculations.

so you want me to evaporate the solution. I do not get how I will weight "evaporation of all unreacted CO2"

No no, don't evaporate the whole solution. Just letting the sit and reach room temperature will make most of the CO2 to evaporate.

i edited my question after i saw some observation when i did the experiment. Can you let me know if the observations help to go the pH way?

The pH should decrease after the CO2 addition, right? so please edit the question again. I will put up something for calculation.

i did: it went from 13 pH to 10

In the observation part please change "pH increased" to "pH decreased", first sentence. I added the calculation in the answer based on CO2 being in excess.

i thought more of the CO2 would react under a lower temperature?

The theoretical maximum for CO2 is about 0.223 moles, there would not be any KOH beyond that to react with CO2. Your starting KOH is 0.446 moles, and each CO2 molecule reacts with two molecules of KOH to make one K2CO3 molecule. No matter what the temperature is, it would bot be possible for CO2 to react beyond 0.223 moles.

is there a way to back calculate from moles of Co2 in solution (which you said was 0.22 moles) to what the pH should be of the KOH + CO2 reaction?

I meant the theoretically maximum amount of CO2 that would react is 0.223 moles. That is because the amount of KOH is 0.446, and CO2 can not react more than half of that amount. And theoretically there should be a way to back calculate remaining CO2 in the solution from pH. But since your pH is very basic (10), I don't see any such calculation to be reliable. Since you are starting with very high concentration of KOH, it makes things difficult. If you start with low KOH and a pH of about 8-9 and then drop the pH by adding CO2 to lower than 7 , then it would be possible to do that.

Right now, pH is still so basic that makes any CO2 react and make K2CO3, effectively eliminating the possibility of any existence of unreacted CO2 inside the solution.

okay, so you said that the initial weight accounts for all the reactants and products, I am not sure if that is true. My initial weight is just KOH and water. My final weight I would measure would be the KOH+CO2 reaction weight. Could you please explain why the initial weight includes the weight of K2CO3 and produced water. Also, wouldn't the post minus pre reaction weighting give me the weight of CO2 reacted?

I revised the answer to address your points. By pre reaction weight, I meant the weight of everything before the mixing. The reason I broke it down like above was to account for condensed water on dry ice. If you exclude CO2 from initial weighing and wait long enough for all unreacted CO2 to evaporate and then weight post reaction, then yes: post minus pre reaction (without 50 g CO2 weight) would be weight of reacted CO2.


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Why are some elements more abundant than others in the universe?
Hydrogen is the most abundant element in the universe. But what makes it so? At the time of big bang, what made certain elements more abundant than the others?

I don't find this order of abundance too predictable. What makes neon more abundant than nitrogen?

Most of those elements were not created by the big bang, but later by nuclear reaction within stars.

because those elements are easier to make

This can be considered borderline off-topic and is probably already answered in much better way on Physics and Astronomy.SE

WHAT MAKES HYDROGEN ABUNDANT IN UNIVERSE:
After few minutes of creation of the universe, protons and neutrons began to react with each other to form deuterium, an isotope of hydrogen. Deuterium, soon collected another neutron to form tritium. Rapidly following this reaction was the addition of another proton which produced a helium nucleus.
Sources say that there was one helium nucleus for every ten protons within the first three minutes of the universe. After further cooling, these excess protons would be able to capture an electron to create common hydrogen. Consequently, the universe today is observed to contain one helium atom for every ten or eleven atoms of hydrogen.
WHAT MAKES NEON STAND 4TH ON THE ABUNDANCE LIST:
The star in its first stage is powered by a series of nuclear fusion reactions that convert hydrogen to helium. The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two γ rays, and a great amount of energy. It then takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium.
Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. The after reaching high temperature the helium-4 nuclei begin to fuse, producing beryllium-8.
The neon production:

Beryllium-8 has low neutron-to-proton ratio makes it very unstable. So
  it reacts with a third helium-4 nucleus to form neon, magnesium and
  carbon. Neon production doesnt stop here. Again from the reaction of
  carbon with helium-4 to initiate a series of reactions, neon, oxygen
  and some more elements are produced. So in the stages of stars, neon
  is produced multiple times in huge amounts making it one of the
  abundant element in the universe.

Could you please see the edits?

I think this is a very good answer to the original question, that was then drastically altered to include 10 specific elements rather than 1, of course including elements that were not produced only by the big bang, as was specified in the original question. +1 for the answer to the original question.

I have added few points on neon being abundant in universe.

I think the abundance chart of https://en.wikipedia.org/wiki/Nucleosynthesis might make a nice addition here, seeing how strong the number of protons being odd or even is influencing the nucleosynthesis.

OT: when I read "Sources say that there was ..." I thought for a moment that there were some written records of this :-D


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Why is it that the number of nucleon in an element is the mass of the element
Why is it that the number of nucleon in an element is the mass of the element for example the mass of oxygen is 32g and the nucleon number is 32.

Roughly speaking, because proton weighs 1, neutron weighs 1, and electron weighs nothing. Also, it is 16, not 32.

As stated by Ivan Neretin, the reason that the mass of an atom (in atomic mass units or amu) corresponds to the number of nucleons (hence the name mass number for this quantity) is that the mass of the proton and neutron are roughly 1 amu and the mass of the electron is negligible. The fact that the mass of the atom in amu corresponds to the molar mass in g/mol is a consequence of the definition of the number of particles in one mole. The conversion factor for amu to g is $1.660539\times10^{-24}$ g/amu this is equal to the reciprocal of Avogadro's number (this is how much mole one single particle represent) $1/N_\text{A}=1/6.0221409\times 10^{23}=1.660539\times10^{-24}$ mol. Because these factors are equal the atomic mass corresponds to the molar mass.
Note that the deviations of the real mass to the mass number are larger for the heavier elements because the number of electrons increases and also the binding energy per nucleon increases (the so-called mass defect). Note also that the masses in the periodic table often represent the mass averaged over the natural isotopes.


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Bond Lengths: Equatorial vs Axial in Trigonal Bipyramidal
Why is it that the axial bond length is less than equatorial bond length in $\ce{PF2Cl3}$ and $\ce{SF2Cl2}$ even though both have trigonal bi-pyramidal geometry?
I know that in general axial bond length is greater than equatorial bond length in trigonal bi-pyramidal structures (as in the case of $\ce{PCl5}$). (See: PBP vs TBP geometry?)

Possible duplicate of PBP vs TBP geometry?

That question (PBP vs TBP geometry?) answered only about PCl5 type molecules (pentagonal bi-pyramidal having all attached atoms same). But I've seen something different about PF2Cl2 and SF2Cl2 which cannot be explained by the same reasoning.

Why does F replace the axial bond in PCl5? may be helpful.

For trigonal bipyramidal structure, we know that lone pairs are preferred first to be positioned in the equatorial position. So in $\ce{SF2Cl2}$ lone pairs will be positioned in the equatorial position.
After placing lone pairs in the equatorial position, double bonds are then preferred in the equatorial position but as there are no double bonds in the structures we are considering, so we jump on to the next rule i.e. atoms having higher electronegativity is positioned at axial position and atoms having low electronegativity is positioned at equatorial position. You can also interpret this in another way. You can also say that atoms having higher atomic size is positioned in the equatorial plane and atoms having lower atomic size is positioned in the axial plane.
Now coming back to your structures, $\ce{PF2Cl3}$ and $\ce{SF2CL2}$ we will see structures in this manner:

$\ce{F}$ is more electronegative than $\ce{Cl}$, hence the $\ce{P-F}$ bonds are going to be shorter than $\ce{P-Cl}$ bonds and $\ce{S-F}$ bonds are going to be shorter than $\ce{S-Cl}$ bonds. As $\ce{F}$s (fluorines) are located in the axial position, hence these $\ce{P-F}$ or $\ce{S-F}$ bonds which are actually axial bonds are shorter than that of equatorial bonds (i.e. $\ce{P-Cl}$ or $\ce{S-Cl}$ bonds).
In $\ce{PCl5}$ all the atoms around $\ce{P}$ are basically the same i.e. $\ce{Cl}$. So the electronegativity factor of $\ce{Cl}$ doesn't play any role in reasoning out why axial bonds are longer than equatorial bonds in case $\ce{PCl5}$. The reason is - in $\ce{PCl5}$ the axial bonds suffer repulsion from the equatorial bonds so they become long. The equatorial bonds providing repulsion to axial bonds also happens in case of $\ce{PF3Cl3}$ and $\ce{SF2Cl2}$, however this factor is overlooked/ not considered in cases of $\ce{PF2Cl3}$ and $\ce{SF2Cl2}$ because there this factor/ effect is not dominating. In those cases as the type of atoms attached (i.e. electronegative atoms) are different, hence the effect of bond length shortening due to the electronegative atom attached is the dominating factor.

What about the molecule XeF3Cl4. I have a reference that it has 2 Cl in axial positions and rest atoms in equatorial positions. Our teacher gave us a rule that more electronegative atom is present at a greater distance i.e. the bond which is longer in these molecules and thus explained the structure of XeF3Cl4 by saying the axial bond length is shorter due to less repulsion (angle is 90 compared to 72 in equatorial) and so Cl being less electronegative will occupy axial position and F will take equatorial position (including the remaining Cl atoms). How far is this reasoning true?

If you are concerned with octahedral structure, then there is no such axial and equatorial planes because all the planes in octahedral are identical. In case of octahedral structures, you can have the most electronegative atom or lone pair or double bond any where you want. Axial and equatorial terms are only used in the case of trigonal bi pyramidal structures. So if your given molecule (Xef3Cl4) exist, then you can have your fluorine any where you want because most probably i feel your molecule will be octahedral and hence you can have fluorine at any position you want.

Can you say me were did you found XeF3Cl4 molecule? If that structure exist then i think it would be in octahedral form and hence you can have fluorine at any position you want because all the positions in octahedral geometry is identical.

Well our teacher explained using that molecule as a reference (I still have it in my notebook). But he gave a pentagonal bi-pyramidal structure for it and hence the axial and equatorial bonds.

Though now I think it must have been a hypothetical example just for explanation as I cannot find any such molecule over the internet.

By the way, what do think about the rule of the more electronegative atom being present at a greater distance. Is this true or it's just a trick which only works for a few molecules (though I've just seen the contradiction in PF2Cl3 and SF2Cl2).

Well more electronegative atom at greater distance - this statement doesn't make sense its actually incorrect to say that. It is better if you remember:  more electronegative atom is always placed at axial position. As I said earlier: axial position is not always the greater distance from the central atom.