How can a neutral atom attract electrons when it's supposed to have zero charge? A neutral atom is an atom with an equal number of protons and electrons

We know the force of attraction between the proton and electron is equal
and assuming that:

one proton's positive charge attracts one electron.
The "neutral" in a neutral atom means electrically neutral.

How can a neutral atom attract electrons when it's supposed to have zero charge ?

An atom is only neutral when viewed as a single object from large enough distance. But as an electron comes closer to the atom, it "notices" the electron cloud first. This cloud also "notices" the electron and deforms—the atom polarizes—so as to keep the atomic electrons farther on average from the extra electron, since like charges repel. But this deformation leads to less shielding of the nucleus than in isolated atom, so the extra electron begins to feel a non-negligible attracting electric field.


Can Gas diffuse according to its partial pressure, not according to its concentration in gas mixture? by Raoult's law,

Partial pressure  =  total pressure  X  concentration(mole fraction)

In gas exchanging situation of 2 gas mixtures which are different in total pressure from each other(like higher temperature?) and include gas A of same concentration, each gas A in 2 mixtures will be the same in its concentration but different in partial pressure. Then, gas A will diffuse from high partial pressure to low partial pressure even though this diffusion is from low concentration to high concentration. Is this right??

Sorry for any misconceptions.

Strictly speaking, diffusion occurs due to gradients in chemical potential.
One implication of this is that sometimes diffusion can occur from lower to higher concentration.
See, for example, this nice class summary

If the total pressure is different, the transport will also occur by fast flow convection instead of diffusion until total (mechanical) pressures are equal (unless thermal transpiration effects are significant).


Triple Bond Character in Acyl Chlorides I am told that because of the poor overlap between a chlorine atom and a carbon atom and chlorine's relatively high electronegativity, there is a strong partial positive character on the central carbon in an acyl chloride molecule.

This makes sense. I am then told that because of this high partial positive character created through the inductive effect of chlorine and oxygen and the poor overlap between chlorine and oxygen, the oxygen's lone pair(s) actually participate in a sort of resonance with the central carbon, thereby giving the C=O bond in an acyl chloride partial triple bond character.

Is this true? Also what exactly is this called? Backbonding?

The various resonance structures that one can draw for an acid chloride are shown below.

Structure IV, is the "triple bond" structure that you mentioned.  The acylium ion ($\ce{R-C#O^{+}}$) is very stable, so structure IV does contribute to the overall description of an acid chloride.  Note that the $\ce{C#O}$ bond in IV is not a real triple bond as the carbon remains $\ce{sp^2}$ hybridized, with the attendant 120 degree (non-linear) bond angles.  I've also drawn a fifth resonance structure involving hypeconjugation.  This is probably a reasonable structure as well since it places negative charge on oxygen and positive charge on the hydrogen.  It also helps explain why hydrogens on the carbon alpha to a carbonyl have increased acidity.

Usually the term "backbonding" is used when electron density is transferred from one atom to another through the use of antibonding orbitals.  See here for a better description.  In resonance structure IV, the lone pair from oxygen is being shared with an empty carbon $\ce{sp^2}$ orbital, the one that had been used to bond to the chlorine.  I've never heard that referred to as backbonding.


Kohlrausch Law and conductivity Im writing up a lab report at the minute and part of it concerns ions in solution and how they affect conductivity. I've been doing lots of reading and I think ive got the grasp of most of it, but its left me with a few questions.

  Symbols used:

  Resistance ($R$), Resistivity  ($\rho$), Conductance ($G$), Conductivity ($\kappa$), Area of cross-section of the electrode ($A$), distance between the electrodes ($L$), Molar conductivity ($\mathrm{Λ_{m}}$)\, The molar conductivity of the electrolyte at infinite dilution ($\mathrm{Λ_{M}}^0$), The molar conductivities of the component ions at infinite dilution
  ($\lambda{_{M+}}^0+\lambda{_{A+}}^0$)\, Concentration (c)

$$\;
[1]\;\; R = \rho \frac{L}{A}\\$$
$$[2]\;\;R = \frac1G\\
$$

$$\;[3]\;G = \frac{{\kappa}A}{L}$$

$$[4]\;\;\kappa = \frac{GL}{A} = G\kappa_{cell}$$

$$[5]\;\;\kappa_{cell} = \frac{L}{A}$$

$$[6]\;\;\mathrm{Λ_{m}} = \frac{k}{concentration}$$

$$[7]\;\;\mathrm{Λ_{m}} = \mathrm{Λ_{M}}^0 - k\sqrt{c}$$

$$[8]\;\;\mathrm{Λ_{M}}^0 = \lambda{_{M+}}^0+\lambda{_{A+}}^0$$

So as I understand it,

Conductance is the ability of a component to conduct electric current.

Conductivity is the ability of a material to conduct electric current, regardless of its dimensions.

Molar Conductivity is the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte, and so measures the efficiency with which a given electrolyte conducts electricity in solution.

Conductivity is related to the amount of ions in solution, the more ions, the higher the conductivity.

Molar conductivity varies with concentration, perhaps counter intuitively, as concentration increases the molar conductivity decreases. The molar conductivity also varies depending on the amount of dissociation (strong electrolyte/weak electrolyte).

The relationship between molar conductivity and concentration is caused by the asymmetric/relaxation effect and the electrophoretic effect.

Here are my questions:

1) For what condition do the equations [6] and [7] apply? I think that [7] is for strong electrolytes at low concentration though im not sure. So what about at high concentrations of strong electrolytes? What about weak electrolytes?

2) When would you use equation [6], when would you use equation [7]? Are they equivalent?

3) In what situation should you use equation [8]?

4) One of my questions is to calculate the concentration of impurities in tap water, using the conductivity measurement. How can I do that when I have more than one unknown in the equation?

Ive not been given a lecture on this topic so im struggling a bit here. Any help is greatly appreciated

For what condition do the equations 1 and 2 apply? I think that 2 is for strong electrolytes at low concentration though im not sure. So what about at high concentrations of strong electrolytes? What about weak electrolytes? 2) When would you use equation 1, when would you use equation 2? Are they equivalent?

Equation 1 defines the molar conductivity:
$$[6]\;\;\mathrm{Λ_{m}} = \frac{\kappa}{c}$$
It is generally applicable. You use it to compute the molar conductivity from experimental data (conductivity and molar concentration of electrolyte). Here $\kappa$ is the conductivity. For an ideal measurement cell and electrolyte it is defined as
$$\kappa = \frac{l}{RA}$$
where l and A describe the geometry of the cell. A more general definition is possible for an arbitrary geometry or sample composition.
Equation 2 is called Kohlrausch's law and sure enough was derived by Kohlrausch based on the behaviour of strong electrolytes, to which it commonly applies at low concentrations:
$$[7]\;\;\mathrm{Λ_{m}} = \mathrm{Λ_{M}^\circ} - k\sqrt{c}$$
You might fit this equation to concentration versus molar conductivity data to determine limiting molar conductivities $\mathrm{Λ_{m}^\circ}$, and from that obtain the limiting molar conductivities of the component ions. The parameter k (also obtained from the fit) may be modeled by applying the Debye-Huckel-Onsager theory.
For weak electrolytes one might invoke Ostwald's dilution law.

In what situation should you use equation [8]?

Equation [8] is the law of independent migration of ions:
$$[8]\;\;\mathrm{Λ_{M}}^0 = \lambda{_{M+}}^0+\lambda{_{A-}}^0$$
where the $\lambda$ are limiting ionic conductivities, properties of individual ionic species in a specific solvent at infinite dilution. As written the equation applies to a solution containing a single 1:1 electrolyte. More generally $\Lambda_{M}^0 = \sum_i \nu_i\lambda_i^0$, where the $\nu$ are stoichiometric coefficient of individual ions in the electrolyte solution. The equation allows prediction of the limiting conductivity from the electrolyte composition provided limiting conductivities have been compiled. You can also use the equation to parse limiting conductivities of individual components from data for different simple electrolytes.

One of my questions is to calculate the concentration of impurities in tap water, using the conductivity measurement. How can I do that when I have more than one unknown in the equation?

Conductivity measurements can be used to estimate the concentration of total dissolved solids in tap water. Typically standard conversions factors are used to convert from conductivity to TDS, the conversion factor depending on the water source, as explained in the wikipedia:

The relationship of TDS and specific conductance of groundwater can be approximated by the following equation:

TDS = keEC

where TDS is expressed in mg/L and EC is the electrical conductivity in microsiemens per centimeter at 25 °C. The correlation factor ke varies between 0.55 and 0.8.2.

There are also online calculators (such as from lenntech) that convert between TDS and conductivity. However if you have performed a water composition analysis you can combine this with tabulated conductivities for different ions, as the lenntech website explains:

The EC determination can be done in different ways. One possibility is the usage of an ion specific conductivity coefficient. This coefficient is listed in charts though for their usage it is necessary to have an exact water analysis because every single ion affects the conductivity. An example for such a chart can be found in the „Handbook of Chemistry and physics” 76th edition, S. 5-90“.

See also ISO document ISO 7888:1985 and links from the EPA.


Can Le Chatelier's principle be derived? Le Chatelier's principle says :

  If a constraint (such as a change in pressure, temperature, or concentration of a reactant) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.

This seems as an analogue of Newton's first law of motion where we talk of inertia resisting a change in the state of rest or motion of a body. Now in physics we have General Relativity whence we can derive Newton's laws. Is there a corresponding general theory in the domain of chemical kinetics too, based on sounder mathematical principles rather than just empirical observations, whence we can derive the Le Chatelier's principle as a special case?

Any suggestions are thankfully invited.

Yes. To prove Le Chatelier's principle doesn't require a general theory of chemical kinetics, just an understanding of (thermodynamic!) fluctuations and response. As usual, Callen's Thermodynamics and an Introduction to Thermostatistics is a good reference. Ch. 8.5 is relevant, and I shall simply recapitulate Callen's development below.

(Incidentally, I think this principle is closer to Newton's third law of action and reaction, but in general thermodynamic laws have a very different flavor from mechanical laws because of their underlying structure. Thermodynamics deals with nebulous relationships; mechanics with definite equations of motion.)

In fact, we will prove an extra statement, the Le Chatelier-Braun principle. This principle states that the secondary effects induced by a perturbation also serve to curtail it. This augments the Le Chatelier principle, that the primary effect induced by a perturbation serves to curtail it.

I will steal Callen's example of a system immersed within a pressure and temperature reservoir, with diathermal walls and a movable domain wall with which to control its volume. The wall is moved slightly outward, causing a positive volume change $\text{d}V$. The primary effect is the decrease in pressure of the system, which leads to a driving force to decrease the system's volume. A secondary effect is the change in temperature $\text{d}T$ resulting from this change in volume, $$\text{d}T = \left(\frac{\partial T}{\partial V}\right)_S\text{d}V = -\frac{T\alpha}{Nc_v\kappa_T}\text{d}V.$$ The prefactor is unimportant; we care only about the sign of the result. All variables are positive except for $\alpha$, which is of variable sign. For now, we can assume it positive. The reduction of temperature in the system then drives heat flow into it, which will itself affect the pressure of the system: $$\text{d}P = \left(\frac{\partial P}{\partial S}\right)_V\frac{\text{d}Q}{T} = \frac{\alpha}{NT^2c_v\kappa_T}\text{d}Q.$$ This change in pressure is positive and diminishes the effect of the original perturbation. The same result is obtained if one takes $\alpha$ negative.

Let us consider a thermodynamic system with first law $$\text{d}U = f_1\,\text{d}X_1 + f_2\,\text{d}X_2.$$ It is coupled to a reservoir, itself with first law $$\text{d}U' = f_1'\,\text{d}X_1' + f_2'\,\text{d}X_2' = -f_1'\,\text{d}X_1 - f_2'\,\text{d}X_2,$$ noting that $\text{d}X_i' = -\text{d}X_i$, because $X_i+X_i'$ is assumed fixed. A perturbation $\text{d}X_1^\text{pert}$ drives fluctuations $$\text{d}f_1^\text{fluc} = \frac{\partial f_1}{\partial X_1}\text{d}X_1^\text{pert} \quad \text{and} \quad \text{d}f_2^\text{fluc} = \frac{\partial f_2}{\partial X_1}\text{d}X_1^\text{pert}.$$ These fluctuations in intensive quantities themselves lead to responses $\text{d}X_1^\text{resp}$ and $\text{d}X_2^\text{resp}$. The signs of these responses can be determined by minimizing the total energy of the system and reservoir given the initial perturbation, $$\text{d}(U+U') = (f_1-f_1')\,\text{d}X_1^\text{resp} + (f_2-f_2')\,\text{d}X_2^\text{resp} = \text{d}f_1^\text{fluc}\text{d}X_1^\text{resp} + \text{d}f_2^\text{fluc}\text{d}X_2^\text{resp} \leq 0.$$ Because $X_1$ and $X_2$ are independent variables, each term in the sum must be negative, and we have $$\text{d}f_1^\text{fluc}\text{d}X_1^\text{resp} \leq 0 \quad \text{and} \quad \text{d}f_2^\text{fluc}\text{d}X_2^\text{resp} \leq 0.$$ The first equation yields $$0 \geq \text{d}f_1^\text{fluc}\text{d}X_1^\text{resp} \quad \Longleftrightarrow \quad 0 \geq \text{d}f_1^\text{fluc}\frac{\partial f_1}{\partial X_1}\text{d}X_1^\text{resp} = \text{d}f_1^\text{fluc}\text{d}f_1^{\text{resp}(1)},$$ where we have multiplied both sides of the inequality by $\partial f_1/\partial X_1$, which must be positive by stability. $\text{d}f_1^{\text{resp}(1)}$ is the response of $f_1$ to the fluctuation due to $X_1$ only, and so we have the Le Chatelier principle. The second equation yields $$0 \geq \text{d}f_2^\text{fluc}\text{d}X_2^\text{resp} = \frac{\partial f_2}{\partial X_1}\text{d}X_1^\text{pert}\text{d}X_2^\text{resp},$$ and hence that $$0 \geq \frac{\partial f_1}{\partial X_1}\text{d}X_1^\text{pert}\frac{\partial f_1}{\partial X_2}\text{d}X_2^\text{resp} = \text{d}f_1^\text{fluc}\text{d}f_1^{\text{resp}(2)},$$ where we have both used a Maxwell relation and multiplied both sides of the inequality by $\partial f_1/\partial X_1$. This is the Le Chatelier-Braun principle.


How to find the Cp, difference of Cp and heat of fusion from the dependence of vapor pressure on temperature? The vapor pressure of zinc varies with temperature as
$$\log P(\pu{mm Hg}) = -6850/T - 0.755\log T + 11.24 \label{eqn:1}\tag{1}$$
and that of liquid zinc as
$$\log P(\pu{mm Hg}) = -6620/T - 1.255\log T + 12.34 \label{eqn:2}\tag{2}$$
Calculate:
a) boiling point of $\ce{Zn};$
b) triple point;
c) heat of evaporation at boiling point;
d) heat of fusion;
e) the difference in $C_p$ of solid and liquid zinc.

I have found out the boiling point by substituting pressure as $\pu{760 mm Hg}$ on first equation \eqref{eqn:1}, from here boiling point $T$ can be found out as $\pu{1181 K}.$
I have also found out triple point by equating the two equations because at triple point both the pressure of the two phases must be equal.
I have also found out heat of vaporization by using the equation
$$\frac{\mathrm d\ln p}{\mathrm dT} = \frac{H_\mathrm{vap}}{RT^2}.$$
But to find the heat of fusion I must know the heat of sublimation, but the heat of sublimation is not provided here. Is there any other way to know the heat of fusion (melting)?
Also, how to find the $C_p$ of both the phases and the difference of $C_p?$

Use the equation for the vapor pressure of the solid to get the heat of sublimation.  The heat capacity of the vapor minus the heat capacity of the liquid is equal to the derivative of the heat of vaporization with respect to temperature. The heat capacity of the vapor minus the heat capacity of the solid is equal to the derivative of the heat of sublimation with respect to temperature.

ADDENDUM

$$\Delta H_{sublimation}=RT^2\frac{d\ln{P}}{dT}=(\ln{10})RT^2\frac{d\log{P}}{dT}$$where log P is given by Eqn. 1.


Do alkylhydroxylamines form N-nitrosamines? Usual secondary amines can form N-nitrosamines with nitrosating agents, e.g., $\ce{KNO2}$, and are therefore forbidden in metalworking fluids.

But can alkylhydroxylamines form N-nitrosamines with nitrosating agents in metalworking fluids?

I assume the basicity of the N-atom in alkylhydroxylamines is lowered by the inductive effect and the mesomeric effect. But is this enough to prevent formation of N-nitrosamines?

Yes they can in acidic solutions according to the abstract of this paper$^{[1]}$:

  O-Alkylhydroxylamines react with nitrous acid with a rate law $v = k_2\ce{[R^1NH2^+OR^2][HNO2]}$; the mechanism involves electrophilic nitrosation at the nitrogen atom of the free base. At acid concentrations below $\pu{2M}$-per-chloric acid, hydroxylamine and N-methylhydroxylamine react with nitrous acid by an acid-catalysed mechanism with a rate law $v = k_3\ce{[RNH2^+OH][HNO2]}h_\circ$; this involves initial nitrosation at the oxygen atom of the conjugate acid, followed by migration of the nitrosonium ion to the nitrogen. At higher acidities the O-nitrosation can be reversed, and the migration of the nitrosonium ion becomes the rate-determining stage.

Reference

T. D. B. Morgan, G. Stedman, M. N. Hughes, "Kinetics and mechanism of the reaction between nitrous acid and hydroxylamine," J. Chem. Soc. B 1968, 344-349 (DOI: https://doi.org/10.1039/J29680000344).


Miller indices: How does one translates the orientation to the origin of the unit cell? In a section discussing Miller indices, my textbook says the following:

  The rules for determining the Miller indices of a direction or an orientation in a crystal are as follows: translate the orientation to the origin of the unit cell, and take the normalized coordinates of its other vertex.

What I'm confused about is how one translates the orientation to the origin of the unit cell?

Since the textbook just mentions this without elaboration or illustration, it is not clear precisely what this means or how one would go about it. I would appreciate it if people could please take the time to explain and demonstrate how this is (mathematically) done. Sources would be appreciated, if you have them.

Let's say your structure contains a short piece of helical DNA, and it (the helix axis) starts at point P1 and goes to point P2. You want to know the Miller index associated with that direction. So you move the entire helix (without rotating) so that P1 is now at the origin. The new position of P2, lets call it P2(prime), will be at:

$$ P_2' = P_2 - P_1 $$

Just checking that is is a translation (I have to subtract the same vector from the entire structure, otherwise I'm rotating or warping it):

$$ P_1' = P_1 - P_1 = (0,0,0)$$

Again, the pair of two points still points in the same direction, but now $P_1$ is at the origin. Taking the fractional coordinates (i.e. along the unit cell axes) of $P_2'$, you can turn those into Miller indices $h$, $k$, and $l$ by multiplying with a large number and rounding to the next integer. Generally, the direction will be off because of the rounding. However, in special cases, e.g. if your DNA is on some crystallographic symmetry axis, the Miller indices will match your direction exactly.


Are there any acid-base indicators whose protonated and deprotonated forms have the same number of resonance structures? This question asks about the color change of 7-hydroxyphenoxazone, the active compound in litmus paper, and the relation of that color change to the conjugated system in the molecule.

In Klaus Warzecha's answer, he mentions 4-nitrophenol as another, simpler indicating system, analogous to 7-hydroxyphenoxazone.

In both of these species, as noted by Klaus, deprotonation of the hydroxyl group to an -olate moiety enables a resonance between the substituents, through the molecular cores, which is not possible in the protonated species:

I'm interested in knowing whether this property of an "opened-up resonance on deprotonation" and the concomitant influence on the excitation spectrum are necessary characteristics for indicating compounds.

So: Are there any known examples of acid-base indicators (e.g., compounds exhibiting a change in visible-spectrum absorbance upon deprotonation) where the removal of the proton does not "open up" additional resonance structures in this fashion?

Clicking through some of the links on Wikipedia's $\mathrm{pH}$ indicators page, I would guess that cresol red, bromocresol green, and bromophenol blue potentially fall into this category, but I'm not confident enough in my evaluation of resonance structures to be certain.

In the examples you drew, these resonance structures are not ‘opened up’; they are there in both the protonated and the deprotonated form. However, since the protonated form requires charge separation to access these resonance structures, they are less important.

It rarely makes sense to do resonance structure bookkeeping like you are trying to implement in your question.

Indicators need to display a colour change upon protonation/deprotonation. that can either happen by modifying the structure (e.g. bromocresol green) or by the electronic changes induced by the charge occurring. The latter case will mean some shifting in the orbital energies — most importantly in HOMO and LUMO — and can practically always be explained in a ‘resonance way’.


Pauli's Exclusion Principle While reading Pauli's Exclusion principal from here, they said that

  The Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers." Now again in we know that the The orbital angular momentum quantum number $l= 0,1,2... (n-1)$ – Source

So from here we can say that $l\neq n$. So here automatically no two electrons can have the same four quantum numbers.  So what's new in Pauli's principle?

I think you've misunderstood, Pauli's exclusion principle doesn't mean that the four numbers are different it means that no electron can have the same respective values for $n$, $l$, $m_s$, $m_l$ as another electron (in some case some of the numbers themselves can be the same). i.e. electrons in the same suborbital must have different spins.

$n$ indicates the orbital shell
$l$ indicates the type of suborbital, s, p etc
$m_l$ indicates the orientation of the suborbital, px, py etc
$m_s$ indicates the electron spin

Pauli's exclusion principle means that no electron around an atom can occupy identical orbitals, i.e two electron with up spin in a 2s1, as all their electron numbers would be the same as their respective equivalents ($n_1 =n_2 = 2$, $l_1 = l_2 = 0$, $m_{l,1} = m_{l,2} = 0$,  $m_{s,1} = m_{s,2} = -1/2$).


Fischer to Haworth projection: which carbon connects to the oxygen? For D-erythose, carbon 4, (the last carbon), connects to the O which connects to Carbon 1.

However, for D-glucose, carbon 5 (the 2nd last carbon) attaches to the O which attaches to carbon 1.

Why is it the last carbon for D-erythose but the second last carbon for D-glucose that is connected to the oxygen in the Haworth projection? How do I decide which one gets connected when trying to convert Fischer to  Haworth projections?

It just is. There are pyranose and furanose forms for all aldoses and hexoses (given they are long enough), including glucose:

$\hspace{3cm}$

You will just have to remember which connections correspond to which sugars. You can read more about anomers ($\alpha$ vs. $\beta$) here, and more about carbohydrate chirality (D vs. L) here.


How does a simple alkaline battery work? The basic + - electrolyte set up. Modern batteries and the like. How exactly do they work?

An alkaline battery (or more precise alkaline manganese cell) consists of a zinc anode, a $\gamma\!-\!\ce{MnO2}$ cathode and $\ce{KOH}$ (in paste form) electrolyte. Zinc is the negative electrode and $\ce{MnO2}$ the positive electrode.

During battery discharge, zinc at the anode is oxidized and dissolves in the electrolyte as hydroxozincate. When the electrolyte is saturated with zincate, zinc oxide starts to precipitate. With continued discharge and decreasing $\ce{OH-}$ concentration, zinc hydroxide is formed, which can slowly dehydrate to zinc oxide.

\begin{aligned}\ce{
Zn + 4OH- &->~ [Zn(OH)4]^{2-} + 2e- \\
[Zn(OH)4]^{2-} &->~ ZnO + 2OH- + H2O\\
Zn + 2OH- &->~ Zn(OH)2 + 2e- \\
Zn(OH)2 &->~ ZnO + H2O\\
}\end{aligned}

At the cathode, $\ce{MnO2}$ is in turn reduced, first to $\ce{MnO(OH)}$. When the battery is discharged slowly, this can be further reduced to manganese(II) hydroxide.

\begin{aligned}\ce{
MnO2 + H2O + e- &->~ MnO(OH) + OH- \\
MnO(OH) + H2O + e- &->~ Mn(OH)2 + OH- \\
}\end{aligned}

The overall reaction, when only considering the first reduction step at the cathode, is:

$$\ce{Zn + 2MnO2 + 2H2O + 2OH- ->~ [Zn(OH)4]^{2-} + 2MnO(OH)}$$

During discharge, zinc is irreversibly oxidized and water is consumed. As a consequence, the battery is "dead" when the electrolyte is dried out and/or all zinc has dissolved.

Sources: 1, 2


What is the oxidation state of nitrogen in N2O5? According to the formula it should be +5 but according to its structure shouldn't it be +4? How is it +5 then?

It is +5 .Even by the structure.

Notice that it is a resonance structure of:

Now one of the bonds are coordinate bonds where nitrogen donates BOTH its electrons to oxygen (otherwise, oxygen cannot have a single bond without a negative charge and no peroxy type linkage).

So, while determining oxidation state from the structure :

O.N=+2+2+1=+5

References:
https://en.wikipedia.org/wiki/Dinitrogen_pentoxide


The safety of homemade slime Disclaimer: I am not a chemist, but I figured this may be the place to ask.

So, recently, there has been a lot of news going around about the homemade slime burning a child's hands and I got curious as to whether the ingredients were harmful and if the parents just didn't know any better:

Borax (Sodium Borate)
Elmer's Glue-All
Warm Water

And realized that none of these are really harmful to handle on your own.

Would mixing these create a harmful compound?

Here's one of the news articles, they're mostly all about the same incident: 11-year-old girl's hands covered in third-degree burns after making "slime"

Short Answer(as it applies to humans):

Not in short exposures.

Longer Answer:

Polyvinyl acetate (PVA or PVAc) is the substance that makes up the active portion of Elmer's / school / wood glue -- the rest being water. Sodium borate (borax) can be used to cross-link the polymers present in the emulsion to form a slimy substance.

Some polymer cross-linking processes can be exothermic, but I'm going to set that aside for now because... extended exposure to borax alone can cause the symptoms described. Some of the articles relating to the incident linked to in the question state that the child had been making and handling this material for weeks.

There are a few studies in which volunteers were given lotion/shampoo containing >1% sodium borate. After about a week, some experienced the described symptoms.

DOI:10.3109/10915818309142004

  A cumulative irritancy test was used to study the effect of repeated exposures
  of a hair preparation containing 3.2% Sodium Borate and a cleansing cream containing
  1.7% Sodium Borate on the skin of 12 and 14 subjects, respectively (Table
  2). The test material was applied under an occlusive patch to the backs of subjects
  daily for 21 consecutive days. Sites were scored one hour after patch
  removal. Applications 4 to 21 of the hair preparation produced erythema and
  papules in most subjects; the total cumulative irritancy score was 571 (maximum
  score = 630). The cleansing cream caused slight erythema in two subjects only,
  resulting in a total irritancy score of 6.4. The investigators concluded that, under
  the conditions of the study, the hair preparation was a "mild to moderate"
  cumulative irritant, whereas the cleansing cream was practically non irritating.

  In a similar study, the contact-sensitizing potential of a cleansing cream containing
  1.7% Sodium Borate was tested in 22 subjects (Table 2). Preliminary irritancy
  testing revealed no irritation to a 48 h patch containing this product;
  therefore, skin sites were pretreated with 5% SLS for 24 h prior to application of
  the initial induction patch. The product was applied under occlusion for 48 h,
  every other day for 10 days (five applications). Following a 10- to 14-day rest,
  48-hour occlusive patches containing the test material were applied to fresh sites
  (with and without SLS pretreatment). SLS controls were also applied. Sites were
  scored at 48 and 72 h. More than half of the subjects reacted to SLS treatment
  (with or without application of test material). No significant irritation was observed
  at sites tested with the cleansing cream alone. The product containing 1.7%
  Sodium Borate was nonirritating and nonsensitizing.

In summary, this material should be pretty safe for short term handling with unbroken skin. Continuous use increases the risk of mishandling (broken/compromised skin contact) leading to boron absorption which may result in skin irritation (redness, bumps).

Note: Abstract does not mention the specific study referenced.


How does aqueous ammonia react with hypochlorous acid in water? Here is what I have so far:
$$\ce{NH3 + HOCl -> NH2Cl + H2O}$$
Since ammonia reacts with hypochlorous acid to form the chloramine $\ce{NH2Cl}$.
But I don't know if this is correct.

Yes, aqueous ammonia will react with hypochlorous acid to form chloramine gas as suggested by @Waylander. But the reaction will also form an intermediate product i.e  ammonium hypochlorite($\ce{NH4ClO}$).

$$\ce{HClO + NH3•H2O → NH4ClO + H2O}$$

  Hypochlorous acid react with ammonium hydroxide to produce ammonium
  hypochlorite and water. Ammonium hydroxide - diluted solution.(chemiday)

The mechanism of this reaction is explained in this Reddit post. You will find the insight of the post useful. I am just pointing out one information.

  Ammonia solution and hypochlorite react (in solution) to produce toxic
  chloramine gas. Ammonium hypochlorite is also formed during the course of
  the reaction, but it decomposes somewhat rapidly to chloramine.

I am assuming this is an exercise question. Do not try to attempt this reaction if you are not an experienced chemist as it also produce deadly byproducts like chlorine gas, hydrazine, dichloroamine and probably nitrogen trichloride. Chloramine is itself poisonous.


Enthalpies of formation of alkali metal halides Shriver and Atkins's Inorganic Chemistry has this graph:

How can I explain the different direction of the trends for fluorides and chlorides? I know that the decreasing lattice energy (as M goes from Li to Cs) is one factor that needs to be taken into account, but the trend in lattice energy should be similar for all the metals.

The equation for the formation of $\ce{MX(s)}$ from its constituent elements is:

$$\ce{M(s) + 1/2 X2(g) -> MX (s)}$$

In general, $\ce{X2}$ could be liquid ($\ce{X} = \ce{Br}$) or solid ($\ce{X} = \ce{I}$). However, since your question is asking about the fluorides and chlorides, we can just limit it to the gaseous state. The principles that will be described can be applied just as easily to the bromides and iodides without any problem.

For any question that asks about trends in thermodynamics, the best way to go about solving it is always to draw a cycle. This will allow you to identify which terms contribute to the change in the trend, and from there rationalise the observed trend. So, without further ado:

$$\large\require{AMScd}\begin{CD}
\ce{M (s) + 1/2 X2 (g)} @>\color{blue}{\Delta_\mathrm{f}H(\ce{MX})}>> \ce{MX (s)} \\
@V\color{red}{\Delta_\mathrm{atm}H(\ce{M})}VV  @AA\color{red}{-\Delta_\mathrm{L}H(\ce{MX})}A \\
\color{red}{\ce{M (g) + 1/2 X2 (g)}} @. \color{red}{\ce{M+ (g) + X- (g)}} \\
@V\color{red}{I_1(\ce{M})}VV @AA\color{red}{E_\mathrm{a}(\ce{X})}A \\
\color{red}{\ce{M+ (g) + e- + 1/2 X2 (g)}} @>\color{red}{\frac{1}{2}B(\ce{X-X})}>> \color{red}{\ce{M+ (g) + e- + X (g)}}
\end{CD}$$

By Hess's law, the enthalpy change for the blue path must be equal to the enthalpy change for the red path. This follows from the fact that enthalpy, $H$, is a state function. So:

$$\Delta_\mathrm{f}H(\ce{MX}) = \Delta_\mathrm{atm}H(\ce{M}) + I_1(\ce{M}) + \frac{1}{2}B(\ce{X-X}) + E_\mathrm{a}(\ce{X}) - \Delta_\mathrm{L}H(\ce{MX})$$

Note the sign convention I am using for the electron affinity and lattice enthalpy (these seem to be defined differently in different places). The lattice enthalpy used here is always positive (endothermic).

Clearly, the terms $\Delta_\mathrm{atm}H(\ce{M})$ and $I_1(\ce{M})$ only depend on the identity of the metal $\ce{M}$. Therefore, it is helpful to lump them together and consider them as one term, called $\Delta H_\ce{M}$. Similarly, the next two terms only depend on the halide $\ce{X}$ and we can collect them together and call it another term, $\Delta H_\ce{X}$. It is by no means necessary to do this; however, it saves some clutter in the expressions:

$$\Delta H_\mathrm{tot} = \Delta H_\ce{M} + \Delta H_\ce{X} + \Delta H_\ce{MX}$$

with

$$\begin{align}
\Delta H_\mathrm{tot} &= \Delta_\mathrm{f}H(\ce{MX}) \\
\Delta H_\ce{M} &= \Delta_\mathrm{atm}H(\ce{M}) + I_1(\ce{M}) \\
\Delta H_\ce{X} &= \frac{1}{2}B(\ce{X-X}) + E_\mathrm{a}(\ce{X}) \\
\Delta H_\ce{MX} &= -\Delta_\mathrm{L}H(\ce{MX})
\end{align}$$

The fluorides

As shown in your graph, the trend for the metal fluorides is that going from $\ce{LiF}$ to $\ce{CsF}$, $-\Delta_\mathrm{f}H(\ce{MF})$ decreases, i.e. $\Delta_\mathrm{f}H(\ce{MF}) = \Delta H_\mathrm{tot}$ increases (it becomes less negative). Here, the halide ion $\ce{X-}$ is not changing, so $\Delta H_\ce{X}$ is a constant and we only need to consider the other two terms:

Going from $\ce{Li}$ to $\ce{Cs}$, the atomisation enthalpy decreases and the first ionisation energy decreases. That means that $\Delta H_\ce{M}$ decreases.
Going from $\ce{Li}$ to $\ce{Cs}$, the radius of the metal ion increases, and so the lattice enthalpy decreases. This means that $\Delta H_\ce{MX}$ (which is the negative of the lattice enthalpy) increases.

So, there are two opposing factors. The first factor says that $\Delta H_\mathrm{tot}$ should decrease, and the other says that it should increase. Since the observed trend is that $\Delta H_\mathrm{tot}$ increases, this means that the second factor is greater than the first. That is to say, the increase in $\Delta H_\ce{MX}$ outweighs the decrease in $\Delta H_\ce{M}$, leading to an overall increase in $\Delta H_\mathrm{tot}$.

The chlorides

The analysis here - in terms of the individual trends in $\Delta H_\ce{M}$ and $\Delta H_\ce{MX}$ - is exactly the same. However, this time, the overall trend is reversed: $\Delta H_\mathrm{tot}$ decreases. That can only mean that, when $\ce{X} = \ce{Cl}$, the decrease in $\Delta H_\ce{M}$ now outweighs the increase in $\Delta H_\ce{MX}$. Why is that so?

First, we note that $\Delta H_\ce{M}$ is entirely dependent on the identity of the metal ion. It is independent of the identity of the halide. That means that the decrease in $\Delta H_\ce{M}$ when $\ce{X} = \ce{F}$ must be exactly the same when $\ce{X} = \ce{Cl}$ - there is absolutely no reason for it to change.

Logically, since the decrease in $\Delta H_\ce{M}$ is not changing in magnitude, it must mean that the increase in $\Delta H_\ce{MX}$ must have become smaller in the chlorides. Otherwise, there is no way for the decrease in $\Delta H_\ce{M}$ to suddenly outweigh the increase in $\Delta H_\ce{MX}$. And we can explain this very easily using what we know about lattice energies, because that's exactly what $\Delta H_\ce{MX}$ is: a lattice energy (with a minus sign).

The lattice energy of an ionic salt can be estimated using the Kapustinskii equation. For a metal halide, this reduces to the equation;

$$\Delta_\mathrm{L}H \propto \frac{1}{r(\ce{M+}) + r(\ce{X-})}$$

Clearly, going from $\ce{LiX}$ to $\ce{CsX}$, $r(\ce{M+})$ increases and $\Delta_\mathrm{L}H$ decreases. The difference is that when $\ce{X} = \ce{F}$, $r(\ce{X-})$ is smaller than when $\ce{X} = \ce{Cl}$. Mathematically, his means that in the fluorides, the lattice enthalpy is more sensitive to changes in $r(\ce{M+})$. This makes sense - if we went the other way and took the limit of $r(\ce{X-}) \to \infty$, the lattice enthalpy would not depend on $r(\ce{M+})$ at all.

That means that the decrease in $\Delta_\mathrm{L}H$ - which is equivalent to the increase in $\Delta H_\ce{MX}$ - must be larger in the fluorides, and smaller in the chlorides. That is why in the chlorides, the trend in $\Delta H_\ce{M}$ overrules the trend in $\Delta H_\ce{MX}$, and the reversal of the overall trend is observed.

Ending words

How would you know all of this if you were not given the graph? Well, the truth is, without looking up values of thermochemical radii, ionisation enthalpies, etc., there is no real way to know. The fact that the overall trend is reversed in the chlorides means that the variations in $\Delta H_\ce{M}$ and $\Delta H_\ce{MX}$ must be pretty similar in magnitude. However, if you are given data and asked to rationalise it, you can definitely do so by constructing a diagram and relating the variation in the various energetic terms to variation in physical properties that you are familiar with (ionisation energy, ionic radii, and so on).


Salicylic acid - polarity and solubility Really confused about this, since there is carbon and two other types of atoms: should be polar, right? However, some people online say it is non-polar. Also, is it soluble in water?

Well it has polar groups - carboxylic and phenolic but benzene ring reduces its polarity. One could say carbon atoms win 7:3 with oxygens, but it's still not bad :) PubChem say it's solubility is about 2g/L, but this site gives more info - it dissolves much better in hot water and its solubility in methanol, which is quite polar, is good, and in benzene (non-polar) bad. Generally it seems to be somewhere in the middle between strongly polar and strongly non-polar compounds.


Deduce the possible structure given spectroscopic data for the aromatic compound For $\ce{C11H13NO}$, I have the following $\ce{^1H}$ NMR data:

shift (ppm)

3.00 (6H, s)
6.55 (1H, dd, $J=15.5, \pu{9 Hz}$)
6.70 (2H, d, $J=\pu{8Hz}$)
7.35 (1H, d, $J=\pu{15.5Hz}$)
7.45 (2H, d, $J=\pu{8Hz}$)
9.60 (1H, d, $J=\pu{9Hz}$)

I also know there are two absorptions from IR at $1600$ and $\pu{1660cm^{-1}}$.

From the IR it seems reasonable to suggest an aromatic amide ($\pu{1660cm^{-1}}$). The shift of $\pu{9.60ppm}$ is huge and I think this could really only come from a proton connected to the carbon of a carbonyl. The 6H peak could be two equivalent $\ce{CH3}$ groups, and as it is singlet, I thought they could hang off the nitrogen of the amide, then it wouldn't split. The two sets of 2H atoms I thought could be from the aromatic ring. The problem is piecing all of it together!

The closest structure I have is this:

but I know this is wrong as it lacks a $\pu{9.60ppm}$ shift. How can I arrive at the correct structure?

It always helps to start with the ‘blooming obvious’: the signal at $\pu{9.60ppm}$. This is obviously an aldehyde. It also obviously couples ($^3J=\pu{9Hz}$) with the signal at $\pu{6.55 ppm}$. This proton is either in the range of a dihalogenated carbon, an electron-rich aromatic system or an alkene. We can rule out all possibilities except for the alkene: it can’t be aromatic because there would be no $^3J$ coupling to a non-aromatic proton (there has to be at least one quarternary carbon in-between).

Moving on, we have a very high coupling constant of $J = \pu{15.5 Hz}$. While it would, in theory, be possible that this arises from a geminal $^2J$ coupling of two diastereotopic protons (although those are typically on the order of $\pu{12Hz}$, so slightly less), the more reasonable explanation is an (E)-configured alkene. If it were a $\ce{CH2}$ group, we not only would get into trouble explaining the extremely high chemical shift but also why the aldehyde only couples to one of those protons. Thankfully, an alkene is what we already guessed from the chemical shift. So we now have three protons out of the way and a $\ce{?-CH=CH-CH=O}$ substructure.

Now we might choose to take double bond equivalents into play. The sum formula $\ce{C11H13NO}$ evaluates to 6 double bond equivalents. We have used two for the alkene and the aldehyde. The four remaning double bond equivalents together with the region the chemical shifts are in scream substituted benzene. We may now take a look at the actual integrals and couplings of the protons we deem can still be aromatic (all except for the 6 singlet protons at $\pu{3.00ppm}$). We have two sets of two and each are an apparant doublet. Therefore, we must have a disubstituted with a para-substitution pattern.

The two fragments we have so far, a $\ce{C6H4}$ and a $\ce{C3H3O}$ one, add up to $\ce{C9H7O}$, meaning that $\ce{C2H6N}$ is still missing. We should also note that all these six hydrogens are chemically equivalent. You can play around with this, but the only way to have six chemically equivalent hydrogens and a nitrogen in a group that connects only once is in a dimethylamino group. Thus, the final structure is:

Finally, allow me a few words on why your structure is incorrect apart from the missing proton corresponding to $\pu{9.60ppm}$:

The phenyl protons would show up as a set of three (so far so good). However, they would be a doublet and two triplets (one of the triplets corresponding to the single proton). Your spectrum does not contain any apparant triplets. Also, monosubstituted phenyl groups tend to have much more similar shifts to the point where they often only give a $\ce{5H}$ multiplet.
The six protons on the $\ce{NMe}$ groups must be equivalent. Since they are obviously on different carbons, there must be some rotation that maps one set onto the other. In my structure, you can easily rotate the $\ce{Ph-N}$ bond to transform one group onto the other. In your structure, rotation around the $\ce{C(O)-N}$ bond is restricted due to the partial double bond character.  Therefore, the two methyl groups would show up as two separate signals of three hydrogens each. Look up the spectrum of DMF (dimethylformamide) to confirm this.

Final note: I know nothing about IR and I do not consider it a helpful tool in structure determination. Thus, I cannot comment on the IR frequences and what they can, would or cannot correspond to.


Converting an Alcohol to Alkyl Chloride (Ibuprofen) In the synthesis of Ibuprofen you can shake 1-(4-Isobutylphenyl)ethanol to produce 1-Chloro-1-(4-isobutylphenyl)ethane

$\hspace{7 mm}$

In a separatory funnel at room temperature.

Usually its only tertiary alcohols that can proceed this way so fast, under mild but in this reaction its a secondary alcohol. Is it to do with the alkyl group with the hydroxyl activating the aromatic ring or maybe the para isopropyl group?

Any advice on this would be greatly appreciated :)

The presence of the aromatic ring next to the position where the reaction is happening it what would allow the reaction to take place under the mild conditions described.

The HCl first protonates the alcohol, causing it to leave (i.e. the first step of an SN1 reaction). This is made 'favourable' by the fact that a cationic charge can be stabilised at the benzylic position (you can push the arrows so that the benzenes pi-electrons move, delocalising the positive charge around the ring). The chlorine anion (produced when the alcohol is protonated) may then attack the cation at the benzylic position, giving the product.


Gibbs free energy and electrical potential The relation between the Gibbs free energy and the electrical potential is:
$$\Delta G=-zFE$$
$z$ is the valence of the electrochemical reaction and $F$ is the Faraday constant.

If at equilibrium the $\Delta G$ is zero this means that also $E$ is zero but this is not true because $E$ at equilibrium is the equilibrium potential of the electrochemical reaction considered.

This makes some confusion in my mind and for sure I'm not properly consider something.
Can someone help to clarify my mind?

Your $\Delta G$ is really $\Delta G_{\mathrm{reaction}}$, and the reaction does not happen alone: it is accompanied by charging of electrodes. The Gibbs free energy of charging is just the associated change in the electroststic energy. That is, it equals $zF$ (the amount of charge moving against the electric potential for 1 mole of reaction) times $E$. Therefore, at equilibrium, the total change of the Gibbs free energy satisfies
\begin{equation}
\begin{split}
\Delta G_{\mathrm{total}} &= \Delta G_{\mathrm{reaction}} + \Delta G_{\mathrm{charging}}\\
&= 0
\end{split}
\end{equation}
just as it should.

In fact, when deriving the expression for the cell potential $E$, the logic goes backwards. We know that the electrodes are charged until equilibrium is reached, in which case the sum of $\Delta G_{\mathrm{reaction}}$ and $\Delta G_{\mathrm{charging}} = zFE$ should vanish. Thus, the cell potential is given by
\begin{equation}
E = -\frac{\Delta G_{\mathrm{reaction}}}{zF}.
\end{equation}


Simple practical, non-toxic, non-flammable reaction for slow release of CO2 Someone I know (not me) has recently had problems with bed bugs. I have advised him his best bet is to (1) Buy a kick-ass steam cleaner and nuke the brutes from orbit, (2) get a professional pest controller in, and (3) Stop buying second hand furniture in future!

But being an inventive sort, he prefers a more subtle approach and is now intent on designing the perfect indoor bed bug trap.

It appears these critters are attracted by a localized $\ce{CO2}$ excess and by body heat, and are repelled by light and (presumably) excess heat. So assuming this trap won't involve a tethered goat or suchlike, he will need a compact, non-flammable, non-toxic, readily and cheaply available substance that can be persuaded to release a steady stream of $\ce{CO2}$ for several hours (without releasing CO or any other noxious or smelly biproduct).

I was thinking perhaps water dripping slowly onto calcium carbide, if there was some way of converting the resulting acetylene to $\ce{CO2}$ without burning it at a readily discernable rate. But a "slow burn", if it could be attained, might provide the heat source (although I reckon a more controllable electric heater would be better for that).

Any ideas? I suppose one obvious answer would be to simply buy a $\ce{CO2}$ cylinder, but could one then buy a sufficiently slow release valve to go with it?

I would recommend not touching or coming near calcium carbide at any rate. The acetylene it creates is explosive in certain concentrations (with a flame to ignite it) and notorious for extremely loud bangs. And it also is rather a diversion for what you want to do.

Instead, consider carbonates. Baking soda is one of the prime household sources of carbonates, and you just need to add an acid to release carbon dioxide; and nicely, the only by-product is water. You should use rather dilute acids so that you can slow down the reaction rate. Vinegar might not be the acid of choice (it has a distinct smell) but either essence of vinegar diluted or citric acid (which you should be able to buy from baking suppliers, too) should work. Or lemon juice.

To make the reaction sufficiently slow, you would need to (as I said) dilute the acid let it be poured into the baking soda solution slowly. Experiment a little! For a heat source, you could likely just put your average heating cushion underneath.

(Personally, though, I would use the steam cleaner because it sounds most kick-arse!)


What is the influence of glass type on silver mirror making? I'm considering the reaction wherein ionic silver (from the dissolved nitrate) is reduced by glucose in the presence of NaOH and forms a metalic silver layer on glass.
Reading through several topics here on stackexchange it seems that it still hasn't been clearly elucidated what exactly is happening on the glass surface for reduced silver to deposit onto / bind to it (instead of just precipitating out as a powder in solution).
I'd like to know if/how glass type might affect the process:
Can I get as perfect of a mirror on a borosilicate surface as on pure SiO2 glass?
Is the silver layer in either case uniform or will is be porous at the molecular level?
I've read that a slow reduction is key. Is it so that a silver mirror is qualitatively better produced at lower temperatures (or with highly diluted reagents) over a long time than by gently heating the solution?
Could a large temperature gradient between the solution and the glass surface be advantageous for the reaction to proceed much faster on the surface than in solution (e.g. mixing a solution at 10C into a glass container whose surface is brought to 80C)?
How can the process be optimized to produce a perfectly insulating silver layer at the molecular level?

It is not so much the glass, it is the contact between the substrate and the reflecting layer.
Some of the problems with the silver/glucose approach, like their optical imperfections is related to the bad adhesion of elemental silver to glass.  The «normal glass mirror» in the bathroom thus uses a tin layer between the cleansed glass, and the subsequently added silver (mostly) responsible for the reflection.  Discovery channel once  aired brief documentary about this; the more interesting part may be still seen on youtube here.
Modern mirrors for telescopes and optical instruments in general do not use silver, but aluminum; the reflective surface is now not behind the glass, but on top (first surface mirrors).  One of the reasons to use aluminum instead of silver is that silver darkens over time (think about the formation silver sulfide, already on cutlery), while aluminum oxide forming on top of aluminum still is transparent enough (for visible light) that these mirrors are aluminized about once every two years only.  Willi Koorts shows the procedure in a about 20 min video here on small scale: washing the mirror, remove of Al by KOH, washing, deposit of Al in high (real high) vacuum.  (This shows the operations at a larger, $\pu{8 m}$ scale.)


Mass of monomer needed to obtain a polymer of a given molecular weight In a condensation polymerization between two different monomers, the average molecular weight of the product depends on the ratio of the monomers used. This question illustrates the effect of using different ratios of ethylenediamine ($\ce{A}$) and 1,4-butanoic acid chloride ($\ce{B}$):

The reaction of $\ce{A}$ ($\pu{2 mol}$) with $\ce{B}$ ($\pu{1 mol}$) gives a compound with molecular weight $\pu{202.3 g mol-1}$ as the major product.
The reaction of $\ce{A}$ ($\pu{1 mol}$) with $\ce{B}$ ($\pu{2 mol}$) gives a compound with molecular weight $\pu{297.1 g mol-1}$ as the major product.
What mass of $\ce{B}$ (in grams) should be combined with $\pu{400 g}$ of pure $\ce{A}$ in order to obtain a high molecular weight polymer?

This question is part of a sample test for my course. I calculated the answer as follows.
For the higher molecular weight polymer, we'll need a $1 : 2$ molar ratio between $\ce{A}$ and $\ce{B}$.
That means we'll need $\pu{2067 g}$ of $\ce{B}$. However, this doesn't seem to be the right answer, which is given as $\pu{1032 g}$.
I'm really not sure what I'm doing wrong at this point.

You have a $2:1$ ratio. You need a $1:1$.

The $\bar{X}_m$ resulting (the number average degree of polymerisation) is derived from the Carothers equation:

$$\bar{X}_m = \frac{1+p}{1-p}$$

where $p$ is the mole ratio. For a $2:1$ this is $0.5$, and the effective $\bar{X}_m = 3$. If you mix at $2:1$, you'll have the $\ce{BAB}$ trimer as the major product with a $M_m$ of $\pu{297.1 g/mol}$.

If you have one component in $1\%$ excess ($1.01:1$) the $\bar{X}_m$ will be $199$, at $.1\%$ excess the $\bar{X}_m$ will be $1,999$ etc.


Calculating equilibrium partial pressures of a gas given Kp My work:

${K_p = 28.4 = \frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{[NOBr]^2}{[107]^2[160]}} \rightarrow$$\ce{[NOBr]=7212 torr}$.

This doesn't match up with the given answer. I don't see why 262 torr is the correct answer. I correctly set up the equilibrium constant expression and everything in the reaction equation is in the gas phase so they are all included and I used only equilibrium concentrations.

Another problem with the same issue:

In general $k_{p}$ is not a dimensionless quantity. In the case of the reaction $$\ce{2NO + Br2 <=> 2NOBr}$$ the unit of measurement for $k_{p}$ is in fact $$\frac{pressure^2}{pressure^2 * pressure} = pressure^{-1}.$$ Thus the value 28.4 of the equilibrium constant given in the assignment is incomplete.

Given the correct answer, it turns out that $k_p$ is actually given in $\mathrm{atm^{-1}}$. Since $$\mathrm{28.4~atm^{-1}} = 28.4*(\mathrm{760~Torr})^{-1} = 0.0374~\mathrm{Torr}^{-1}$$ you should use this value for $k_p$ if you prefer to do your calculation with $\mathrm{Torr}$.


How to extract thymol from oregano leaves? This is my second graders in highschool project of science report.

I try to a make natural hand sanitizer using thymol from oregano leaves as  the antiseptic ingredient instead of alcohol.

I want to extract using a maceration method.

What solvent should I use?
How to separate thymol from the solvent?

This sounds like a nice project and I'm tempted to answer in too much detail, but since this is homework […]. Here are some hints:

Find information about thymol. This is crucial!

What is the structure?
In which solvents is it soluble?
Is thymol soluble in water?

Is it soluble in acidic or alkaline solution?

What about the boiling point and the melting point?

Plant leaves have a cell structure in which your thymol is somehow bound and hidden. In order to break of the structure, you may want to

dry and powderize the leaves or
cut the fresh leaves into pieces and grind them in a mortar with sea sand

Oregano leaves are green. You know that they contain chlorophyll. There's probably a lot of other compounds too, not just thymol.
Supposed that you treat the leaves as described in 2. and start to extract using ethanol in a Soxhlet extractor, you'll end up with a nice cocktail. Do you need to separate thymol from the other extracted compounds? If so, how will you proceed? Knowing the properties of thymol will help you here!

Remember that the easiest work is the one that you can avoid ;-)

Think in analogies. Thymol is a monoterpene. A lot of other members from this family have a typical smell and are used as fragrances. There must be an established, mild procedure to obtain these compounds from plant material.

   Yes there is! Search for steam distillation :)

If there is an established prodecure that avoids column chromatography or other purification techniques, maybe somebody has already used it for thymol and wrote about it.

Back in the days, our professors used to say with a smile:

  A week in the lab saves you an hour in the library

Replace library with internet and have another search for a suitable and easy procedure already published.

Update

In a follow-up comment, you asked:

  Then, how much ethanol should I use?

My suggestion is: Avoid subsequent laborious workup and do not use any ethanol at all!

Read Introducing Organic Chemistry Students to Natural Product Isolation Using Steam Distillation and Liquid Phase Extraction of Thymol, Camphor, and Citral, Monoterpenes Sharing a Unified Biosynthetic Precursor by Katherine A. McLain, Kenneth A. Miller, and William R. Collins, J. Chem. Educ., 2015, 92, 1226-1228. Follow the given procedure, which was deliberately designed to be performed in a students' lab:

  One, 3 h laboratory period is required to complete the experiment. Dried plant material (10 g) or fresh plant material (20 g) is placed in a high-speed blender with distilled water (200 mL) and blended to a fine
  slurry. The slurry is transferred to a round-bottom flask (500 mL) with a distillation apparatus attached and a stir bar, and it
  is placed directly on a hot plate turned to high heat. A beaker
  (200 mL) is used as the receiving vessel to collect
  approximately 100 mL of distillate. The bioactive molecule is
  extracted from the distillate with an appropriate organic solvent
  (30 mL). The organic phase is dried with magnesium sulfate,
  and the organic solvent is removed on a hot plate at medium
  heat.

For further details, have a look at the Supporting Info.


Why does the standard enthalpy of formation diverge so far from the standard Gibbs free energy of formation for some substances? If you look at a Table of Thermodynamic Values for chemical substances, most substances have very close values for their standard enthalpy of formation ($\Delta H_{\mathrm{f}}^\circ$) and their standard Gibbs free energy of formation ($\Delta G_{\mathrm{f}}^\circ$). This makes intuitive sense because differences in enthalpy, which in chemical reactions is mostly just a signifier of internal energy because so little work is done by reactions, under normal circumstances correlates with the change in free energy of that system. When heat is lost through a reaction, that system has both less internal energy but also less free energy because the production of heat and then exporting of heat to the surroundings requires the 'use' of free energy.

My question: why do some substances have widely diverging values for $\Delta H_{\mathrm{f}}^\circ$ and ($\Delta G_{\mathrm{f}}^\circ$)? It appears that some substances containing nitrogen have the widest divergence for standard conditions of 25 °C and 1 bar, whereas most substances have a $\Delta H_{\mathrm{f}}^\circ$ and ($\Delta G_{\mathrm{f}}^\circ$) differing by no more than 30%. For example, $H_\mathrm{f}(\ce{AgNO3}) = −124.39$ kJ/mol and $G_\mathrm{f}(\ce{AgNO3}) = -33.41$ kJ/mol, and $H_\mathrm{f}(\ce{N2O4}) = 9.16$ kJ/mol and $G_\mathrm{f}(\ce{N2O4}) = 97.89$ kJ/mol.

A mathematical explanation would be helpful, but I'm most interested understanding this conceptually and why these nitrogen-containing compounds exhibit these properties. Are these properties more common among other substances at extremely high or low temperatures and pressures?

Continuing from the comments above.

The difference is due to the entropy appearing in $\Delta G$ as the energy $-T\Delta S$. Classical thermodynamics cannot tell us what the entropy is in molecular terms. This is discovered using statistical mechanics via Boltzmann's entropy equation $S=k\ln(\Omega)$ where $\Omega$ is the number of arrangements of particles into their energy levels, sometimes called 'complexions'.

Different types of molecule have different natures, because they contain different number of atoms, which means that there are different numbers of molecular vibrations, or they contain atoms of different types, which means that vibrations are of different frequencies to one another. Hence molecules of different types have a unique set of energy levels (ignoring for simplicity degenerate vibrations).

As energy is added, the number of ways of filling up the energy levels is clearly different between different types of molecules because the stack of energy levels differs one to another. Single bonds between heavy atoms, for example, have lower frequencies, than those between lighter atoms, and so more levels are populated at the same temperature than in the heavier atom molecule. In this case the entropy is greater than that for a molecule with widely spaced levels.

The same type of argument can be made for rotational and translational energy levels, and can also be made for phase changes, such as solid-solid, or for melting or vaporisation. Which of these are to be counted depends upon whether the substance is gas, liquid, or solid at the required temperature.

Notes:

Technically the entropy is calculated as
$$S=R\ln(Z)+RT\left ( \frac{\partial \ln(Z)}{\partial T}  \right)_V$$

where Z is the partition function and $Z=\Sigma _i g_i \exp(-\epsilon _i/(kT))$ where $\epsilon_i$ is the energy of level i with degeneracy $g_i$ and k is the Boltzmann constant.


Which is the specific heat of human body? I'm studying about heat transfer and I wanted to know if there exists a specific heat of the human body or how one studies the human body's heat transfer.

A specific heat for humans typically quoted in the literature is 3.5 kJ/(kg K) (for example, pg 16 of Herman's "Physics of the human body", 2nd ed).   This is below the value of water (4.2 kJ/(kg K)) primarily due to body fat, which has a lower specific heat than water.

Despite its title, the article:

Faber and Garby, "Fat content affects heat capacity: a study in mice", Acta. Physiol. Scand., 1995, 153, 185-187, https://doi.org/10.1111/j.1748-1716.1995.tb09850.x

contains a brief summary of the literature concerning the average specific heat of humans.  The authors find that the specific heats of lean and fat mice agree reasonably well with a relationship originally proposed for humans:

$$ c_P = (3.72 - 1.84 \: m_f ) \text{ kJ/(kg K)}$$

where $m_f$ is the fraction of body fat by weight.  Wikipedia  ( https://en.wikipedia.org/wiki/Body_fat_percentage ) indicates that body fat percentage (by mass) is 17-21% and 25-30% for "average" men and women respectively (both neither "fit" or "overweight").  Substituting $m_f = 0.22 $ into the formula above gives 3.3 kJ/(kg K), in rough agreement with typical value of 3.5 kJ/(kg K) quoted in the literature (the literature value corresponding to $m_f = 0.1$, a value typical for a male athlete).

There is some uncertainty in my mind about the accuracy of the formula above, as it is based on a heat capacity of fat significantly smaller than recently reported (see database reference below).  However, I have trouble imagining a situation where the average specific heat capacity of a human would need to be known to great accuracy (it is needed for the artificial, but instructive Problem 1.54 in Schroeder's Introduction to Thermal Physics).  On the other hand, the heat capacity of various components of the human body could be of significant utility (the influence of localized heating etc...).  These specific heats (blood, bone, etc...) are tabulated (with citations) in the "Tissue Properties Database"  http://dx.doi.org/10.13099/VIP21000-03-0 (V3.0).


Why do α-hydroxy ketones give Tollens' test? I know that tautomerism causes the $\alpha$-hydroxy ketones to give the all the tests of an aldehyde. But how does it exactly tautomerise and what's the intermediate?

Does benzoin gives Tollens' test? If not, why? I'm really confused with this. I read somewhere that benzoin doesn't give Tollens' test as it does not have an alpha carbon to form enediol intermediate. Please help.

The Tollens' test is done in base, which mediates tautomerization through the enolate:

The misconception that you have is that all alpha-hydroxy ketones can give a positive Tollens' test. Aldehydes give a positive test, and so only alpha-hydroxy ketones that can tautomerize to aldehydes will give a positive Tollen's test. My example above is 1-hydroxy-2-propanone, which isomerizes to 2-hydroxypropanol. Most alpha-hydroxy ketones with names that match the pattern 1-hydroxy-2-alkanone will give a positive Tollens' test.

Benzoin will not give a positive test, since it cannot isomerize to an aldehyde. What does benzoin form upon tautomerization?


Reaction of cyclohexene with hydrochloric acid What happens when cyclohexene reacts with concentrated hydrochloric acid?
I formed the carbocation intermediate by protonating cyclohexene.
But now I am stuck. I know the end product will be a dimer.
I tried to form a bond between the intermediate and cyclohexene like this.

Then through rearrangement the plus charge will shift to 3 degree carbon.
In the answer there is a double bond between the two rings.
That's where I am stuck . I can not think anything further.

Have a look at the mechanism for the dimerization of cyclohexene in strongly acidic medium:

After the attack of a nucleophile (pi bond) on the electrophilic carbocation, we have:

A: joining of the two rings with a bond, and the consequent carbocation (why did it get formed on that position? why not on the adjacent ones?)
B: rearrangement of the carbocation (why did it happen?)
C: loss of a proton to form the thermodynamically most favorable product (most substituted alkene)

The double bond was formed because of the loss of the $\ce{H+}$ ion in step C, formation of a negative charge at that position, and then the delocalisation of that negative charge into the adjacent empty p-orbital (of the $\ce{C+}$ position).


The chemical reaction between lemon juice and polystyrene foam cups I have noticed that when I have lemon juice on my fingers and I touch a polystyrene foam cup (I will refrain from using the term 'styrofoam'), the cup begins to break down, i.e., melt. I can leave fingerprints on the cup in this manner. What is it about lemon juice and foam cups? What is happening, other than what I see; the cup melting.

I have never tried replicating this using other citric acids, such as an orange. So, I don't know yet if it's just lemon juice that will do this.

It is not any acid in the juice, but limonene, the natural, fragrant, oil found in citrus fruit, and similar terpenes that are good organic solvents.

In fact, these oils are sold for the purpose of cleaning. Note, though limonene itself is considered comparatively nontoxic, after it has dissolved the polystyrene, it might be a slightly more dangerous, since any unreacted styrene monomer in solution is somewhat more toxic.


LiOH base ionization reaction What is the base ionization reaction for LiOH? Usually, the reaction is like this: $$Base + \ce{H2O->H}[Base] + \ce{OH-} $$

But $\ce{LiOH2+}$ makes no sense. So what could the reaction be?

What you call bases is in fact a subset thereof. Besides that, there is a whole class of inorganic bases, also known as alkali. These are the hydroxides (sometimes called basic salts) of alkaline metals. Their ionization is somewhat different:
$$\ce{NaOH->Na+ + OH-}$$


How to determine the vapor pressure of a glycerine and propylene glycol mixture? I am looking to calculate the minimum heat needed to create vapor from a specific mixture. I have a mixture of 60 % $\ce{C3H8O3}$ (glycerine) and 40 % $\ce{C3H8O2} $ (propylene glycol, propane-1,2-diol). Total volume of the mixture is 3 ml. Here are some facts that I know

 C3H8O2
 Density = 1.04 g/cm³
 Boiling point = 370.8 °F (188.2 °C)
 Molar mass = 76.09 g/mol
 Specific Heat = 2.5 kJ/(kg·K) or 0.60 Btu/(lb·°F)

 C3H8O3
 Density = 1.26 g/cm³
 Boiling point = 554 °F (290 °C)
 Molar mass = 92.09382 g/mol
 Viscosity = 1.412 Pa·s
 Specific Heat = 2.43 kJ/(kg·K) or 0.576 Btu/(lb·°F)

At what heat will the mixture begin to create vapor? I am looking for an answer, although I am not looking for someone to come in and just leave an answer. I would rather understand how you came about the answer, i.e. equations, additional info, so that I can adjust the percentages and volume on my own. The percentages of the mixture will vary. The volume of the mixture will vary depending on the amount of heat needed to create vapor, i.e. decrease volume if not enough heat is present.

Your best bet for this will be to use the Clausius-Clapeyron relation. This is an exponential function that describes the vapor pressure of a substance as a function of temperature, and uses the enthalpy of vaporization (latent heat) and a constant as parameters.

The constant is dependent on the particular substance, but there is a way to eliminate it from the equation if you can get one data point: the vapor pressure at a specific temperature. If you can do that, then you can use the two-point form to predict vapor pressure at another temperature:

$\ln\frac{p_2}{p_1} = \frac{-\Delta H_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

This version only has one parameter - the enthalpy of vaporization ($\Delta H_{vap}$). $p_1$ and $T_1$ are the vapor pressure and temperature at point 1 (an arbitrary point), and $p_2$ and $T_2$ are the vapor pressure and temperature at a different arbitrary point.

For a pure substance, if you know the enthalpy of vaporization and the boiling point, you an calculate the vapor pressure at any other temperature. That is because the boiling point is the temperature at which the vapor pressure equals one atmosphere. In other words, you would set:

$p_1 = 1 \space \mathrm{atm}$

$T_1 = T_b$,

Then solve for $p_2$. The equation would be:
$p_2 = p_1\exp([\frac{-\Delta H_{vap}}{R}](\frac{1}{T_2}-\frac{1}{T_1})) \tag{1}$

When you use this equation, make sure you first convert the temperatures to Kelvin. This only works with an absolute temperature scale. Pressure can be in whatever units you want, as long as both pressures have the same units. In other words, convert 1 atm to whatever units you want to use for vapor pressure.

The data you listed didn't have the enthalpy of vaporization for either component, but they are available from NIST:

$\Delta H_{vap} = 91.7 \space \mathrm{\frac{kJ}{mol}}$ - glycerol

$\Delta H_{vap} = 71.2 \space \mathrm{\frac{kJ}{mol}}$ - ppg

Incidentally, NIST lists different values for the boiling point than you have. I would double check those before plugging numbers in.

Now, you don't have a pure substance; you have a mixture. Depending on how non-ideal it is (which basically means how much the molecules of one component will affect the other components) you might have to account for that if you want to get very accurate results. On the other hand, if you just want an estimate and are planning to experimentally adjust the power output from there, you can get a good starting point by assuming that the mixture and the vapor are both ideal. Under those conditions, the total vapor pressure of the mixture will be equal to the sum of the partial pressures...

$p_t = p_{gly} + p_{ppg}$

... and you can use the Clausius Clapeyron equation from above for each component individually to find the respective vapor pressures. Then you can add them together if you want the total vapor pressure. However, you may be more interested in the vapor pressure of each component individually, since (from what I have heard) glycerol is more responsible for the "cloud" that forms, while propylene glycol is the better organic solvent. It would be helpful to know how much vapor of each component a given mixture would produce at a given temperature.

That's how you can predict vapor pressure from temperature. To relate this to the amount of heat you need, you will need to do another step. Temperature is a function of heat supplied and the specific heat of the substance:

$q = mC\Delta T \tag{2}$

where $q$ is the amount of heat transferred, $m$ is mass, and $C$ is the specific heat. You know the specific heat of the mixture, and the mass is an independent variable since you will control the composition and amount of the mixture. If you assume that volumes and heat capacities are additive, that mixing is fast, and that heat transfer is fast (at least locally), you can estimate the temperature (or rather, the maximum temperature) by connecting this to the power supplied by your device. Power in watts is the same as Joules/second. Therefore, you can multiply the power output by the time to get a total (maximum) amount of heat transfer (in Joules):

$q = P*t \tag{3}$

We can combine equations $(3)$ and $(2)$ and solve for the operating temperature:

$T=\frac{Pt}{mC}+T_i \tag{4}$

where $T_i$ is the starting temperature of the mixture (before you apply power).

Now we can substitute this back into $(1)$ to come up with an estimate of the vapor pressure of each component as a function of the mass of the liquid, the power applied, and the length of time that it is applied for:

$p_{vap} = (1 \space \mathrm{atm}) \exp([\frac{-\Delta H_{vap}}{R}](\frac{1}{\frac{Pt}{mC}+T_i}-\frac{1}{T_b})) \tag{5}$

The real vapor pressure will likely be lower, but I think this would get you to a good starting point.

A couple of things to remember:

$m$ refers to the mass of each component that is in close proximity to the heating element.
$Delta H_{vap}$ is in kJ/mol - make sure that the $R$ you use has the right units.
$T$ must be in Kelvin (both of them)
The units of $P$, $m$, and $C$ have to cancel out to give K.

If you want to get a (rough) upper estimate of the mass that will be vaporized, you could assume an ideal gas and use

$m=\frac{pVM}{RT}$

where $V$ is the volume of the vaporizer chamber, and $M$ is the molar mass of the component.


Can an improvised torch be made to last for a prolonged period of time? In the movie First Blood, there is a scene where Rambo is trapped in a cave after escaping a bazooka rocket that caused the mine he was hiding in to collapse. After realizing the entrance has been blocked, he decides to head down into the cave, and using the fire he built, a branch and some cloth he has, he prepares a torch (I can't remember now if he dips the cloth in some flammable liquid or not). Then he wanders through the cave for some time until he finds his way out, and the torch is holding up rather nicely... (he is seen feeding it with more strips of cloth from time to time)

My question is: is it realistic? Can a torch made from a branch and a burning cloth (whether or not dipped in gasoline or something else) be used for a prolonged period of time while remaining useful (i.e. not being burned-down, too hot to handle, etc.)?

The key is whether or not he added some kind of fuel to the cloth. Without that, a flammable cloth would burn itself out very quickly. If he had kerosene or some other fuel to soak the cloth, then the cloth would act somewhat like a wick in an oil lamp (except that it would not be "wicking" fuel up from any fuel source) in that mostly the fuel would burn rather than the cloth.

If he had a liquid fuel source, even a non-flammable cloth like his clothing would have been made of would have sufficed. In fact, he probably wouldn't have had to keep adding cloth in that case, but he would have had to add more fuel periodically.

So, the bottom line answer is: Without a liquid fuel source to soak the cloth, no chance. With a liquid fuel source to soak the cloth, no problem.


Inductive vs conjugation effects in ester hydrolysis I just came across the following problem about alkaline ester hydrolysis:

Since all of the substituents are in the para position, I immediately thought about the ability of those substituents to contribute electron density (through conjugation) to the ring and, therefore, the carbonyl group. Thus, I ruled out the p-methoxy and the p-toluene compounds since their conjugation effects lowers the electrophilicty of the carbonyl group.

Since I knew that chloro groups are ortho-para directors in electrophilic aromatic substitution, I also concluded that the chloro group also donates electron density (as shown below) and should also be less reactive at the carbonyl group than the non-substituted benzene ester, which I concluded to be the most reactive compound.

However, the answer key says that the p-chloro compound is the most reactive due to its electron-withdrawing abilities (which should be through induction only).

However, wouldn't those effects be minimal in comparision to the conjugation effects, or does the large nature of chlorine's 3p orbital vs carbon's 2p orbital make conjugation mininmal?

Clearly, you have not read benzene halogenation before. In the electrophilic substitution of halogens in a benzene ring, the halogen, taking Cl for example, deactivates the ring precisely because its inductive effect of withdrawing electron density from the benzene ring outweighs the resonance effect of Cl donating its lone pair into the ring. It is experimentally proven that chlorobenzenes are less reactive, and you can understand the reason why inductive effect outweighs the resonance effect as the Cl atom, being highly electronegative, is "not willing" to readily donate its lone pair of electrons into the ring; High nuclear charge causes Cl atom to hold tightly to its valence electrons, thus they are not available for donation. Also explains why Cl- ion is a weak conjugate base and weak nucleophile.


Can an uncharged particle pass through a molecular bond? Do particles ever pass through or interact with the molecular bond? Is there a physical size limit? What happens to the atoms and the particle and does this ever happen in real life?

The molecular bond is not a solid thing. It is merely an energetically suitable arrangement of atoms and their electrons. So another particle that happens to come close to that bond can only interact with those atomic nuclei and electrons. Whether it does interact depends.

Which particles can interact?

Considering electron-magnetic forces only:

Charged particles can interact.
Uncharged small particles can not interact.
Larger uncharged particles (e.g. other small molecules) could get polarized. This means they develop partial charges which then can interact.

How can charged particles interact with the bond?

Slow particles (with low kinetic energy) can get attracted or be repelled by the molecule, depending on the (partial) charges of the particle and molecule. If it gets attracted a chemical reaction can happen.
Fast particles (with high kinetic energy) might just pass through the molecular bond. The bond has actually much less electron density then any atom. Depending on how close it gets to an atomic nuclei, it might get deflect. Or it may even deflect the atom and break the bond. The particle could also grab or deflect an electron of the bond and thus ionize the molecule.
As a third option: there is always a slight chance of quantum tunneling.

Does this ever happen in real life?

Actually a lot: Neutrinos have no electric charge and almost no mass. As a result they can pass through matter without any consequences. They are emitted by stars and pass through earth by millions per second.

An example for charged particles would be using an electron gun to ionize molecules for mass spectrometry. The electron gun creates a beam of fast electrons, which can remove an electron from the targeted molecule upon collision (1 goes in, 2 go out).

Is there a physical size limit?

If the particle becomes larger than the molecular bond is long ... well I would not call it "passing through the bond" anymore. But the above concepts would still be the same.


How to write half equations based on oxidizing/reducing agents pairs? In most videos, people use oxidation numbers to find the oxidizing agent and the reducing agent, It's a great way to use, but our teacher never used it before he would give us a reaction and ask us directly to write the half reactions and what are the oxidizing and reducing agent on each. Using oxidization numbers I can easily find the oxidizing/reducing agents and write half reaction for each.

However, sometimes we don't have the final reaction, we only have oxidizing/reducing agents and are asked to write half equations. For example:  Given the pair ($\ce{CO2/H2C2O4}$) as an example, should I write the half equation for the $\ce{CO2}$ to turn into $\ce{H2C2O4}$, or the inverse?

Another question is how can I find oxidation numbers on compound elements such as: $\ce{C6H12O6}$?

Firstly, you need to understand that there is a maximum and minimum oxidation number.

For metals, the minimum is usually $0$ while the maximum is usually the number of their outermost electrons. For example, $\ce{Na}$ which has $1$ outermost shell electron would have a minimum of $0$ and a maximum of $1$.

For non-metals, the minimum is usually the number of their outermost shell electrons minus $8$, while the maximum is usually the number of their outermost shell electrons. For example, $\ce{S}$ which has 6 outermost shell electrons would have a minimum of $-2$ and a maximum of $6$.

For the pair $\ce{CO2}$ and $\ce{H2C2O4}$:

Firstly, note that the most electronegative element in each compound is $\ce{O}$. You can learn more about electronegativity here.

Therefore, it takes the minimum, which is $-2$.

Secondly, usually hydrogen is $1$ (in most covalent compounds) or $-1$ (in most ionic compounds). In this case, hydrogen is $1$.

It follows from calculation (that the sum of the electronegativity of each element in a compound must be zero) that the $\ce{C}$ in the first compound has oxidation number $4$ while that in the second compound has oxidation number $3$.

Since the oxidation number has dropped, this is a reduction reaction.

We write the unbalanced equation as a starter:

$$\ce{2CO2 -> H2C2O4}$$

Balancing the number of $\ce{C}$.

Now, we are in an acidic environment, so we add $\ce{H2O}$ to balance the oxygen and $\ce{H+} to balance the hydrogen. The oxygen is already balanced, so:

$$\ce{2H+ + 2CO2 -> H2C2O4}$$

Note that the charges are still not balanced while the elements are balanced, so we add electrons:

$$\ce{2H+ + 2CO2 + 2e- -> H2C2O4}$$

Thus our half equation is complete.

The method to find the oxidation number of each element in $\ce{C6H12O6}$ is detailed above, and I leave this for you as an exercise.

PS: in an alkaline environment, we add $\ce{OH-}$ to both sides of the equation to eliminate the $\ce{H+}$, i.e. (the following is wrong because we are in an acidic environment thanks to the acid $\ce{H2C2O4}$ (acetic acid):

$$\ce{2H+ + 2OH- + 2CO2 + 2e- -> H2C2O4 + 2OH-}$$
$$\ce{2H2O + 2CO2 + 2e- -> H2C2O4 + 2OH-}$$


What is the Lewis Structure for CO? In many sites I see that they take $\ce{C}$ with valence 4, bur our teacher takes it with valence 2. With what valence should I take $\ce{C}$?

There is resonance between various Lewis structures as:

$$\ce{:\!\overset{\ominus}{C}#\overset{\oplus}{O}\!: ~<->~ :\!C=O\!:: ~<->~:\!\overset{\oplus}{C}-\overset{\ominus}{O}\!:::}$$

The term "valence" is not used much nowadays, and "oxidation state" is used but has a different meaning.

The IUPAC definition of oxidation state requires the oxygen have a -2 oxidation state except in peroxides.  All the oxidation states must add up to 0 for a neutral molecule.  Therefore the oxidation state of carbon in CO must be +2.

There is an IUPAC definition of valence:

  The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.

This imposes a unique valence upon each element, as explained in this article, in which case valence is always 4 for carbon.

There are other definitions of "valence" such as:

  The number of hydrogen atoms that can combine with an element in a binary hydride or twice the number of oxygen atoms combining with an element in its oxide or oxides

(Chemistry of the Elements by Greenwood et al.)

According to this definition, carbon would have a valence of 2 in CO.


finding cation of unknown salt in solution from precipitate I should probably add that I was originally given an unknown solid mixture of 2 salts. I separated the insoluble salt out by centrifuging and then decanting. I found that the insoluble salt is $\ce{MgCO_3}$ by adding $\ce{HCl}$ (bubbles) and $\ce{NaOH}$ into the solution (jelly like white ppt). The decanted liquid was then what I observed as having 1 soluble salt. The soluble salt I know by testing for $\ce{SO4^{2-}}$ with $\ce{BaCl_2}$ and observing a violet flame in a flame test. Then I had a second bottle with 3 unknown, 2 of which I discovered above. I know that the 3rd one has 2 soluble and then my original question follows.

I have 2 soluble salts in aqueous distilled water solution. I know that one of the salts is potassium sulfate ($\ce{K_2SO_4}$) and I know that the anion of the second salt is chloride ($\ce{Cl^{-}}$). I know that the second salt is either magnesium, sodium, calcium or potassium chloride ($\ce{MgCl_2}$, $\ce{NaCl}$, $\ce{CaCl_2}$, or $\ce{KCl}$).

I tested for the anion for the second salt by adding silver nitrate ($\ce{AgNO_3}$) into the solution and observing a white precipitate. When I added sodium hydroxide ($\ce{NaOH}$) to it, I observed that the cloudy white precipitate disappeared and turned into small black precipitate which then settled to the bottom of the tube. Is there any way to get the cation of the second salt with this information, and are there any other test(s) I can do to find what is the cation for the second salt?

Among the options for the second salt in solution, one is less likely to be found in the presence of $\ce{K_2SO_4}$.

These solutions for analysis are prepared by adding decent amounts of two salts with good solubility to water. Under these conditions, a double displacement reaction may occur and in one case, a precipitate will be formed:

$\ce{K_2SO_4 + CaCl_2 -> 2KCl + CaSO_4}$

I doubt that the person preparing the solutions for analysis used calcium chloride, since

this would have required to remove the precipitate by filtration or decanting off the supernatant solution
the concentration of $\ce{Ca^{2+}}$ in the solution would be rather low. Remember that these qualitative tests are not necessary suitable for trace analysis.

The second salt therefore is most likely sodium chloride ($\ce{NaCl}$), potassium chloride ($\ce{KCl}$) or magnesium chloride ($\ce{MgCl_2}$).

I would do a flame test first. If you see a yellow coloured flame, the second cation is the sodium cation ($\ce{Na+}$). Similarly if you see a lilac coloured flame, the second cation is the potassium cation ($\ce{K+})$ and if you initially see a colourless flame and finally see an intense white coloured flame,the second cation is the magnesium cation ($\ce{Mg^{2+}}$).
If you want to test for the magnesium cation ($\ce{Mg^{2+}}$), place a drop of the solution on a spot plate, make sure it's alkaline by adding aqueous $\ce{NaOH}$ if necessary, and add a bit of a titan yellow solution. This azo dye with its misleading name (there is no titanium in it) will form a red precipitate with the magnesium cation ($\ce{Mg^{2+}}$).


Regioselectivity for electrophilic aromatic substitution with multiple strongly electron withdrawing groups At which position will electrophilic substitution on p-cyanonitrobenzene occur?

As both are -M groups, there are two possible sites of reaction, meta to each of the groups.
The answer is meta to $\ce{CN-}$. Why is this so? According to me it should be meta to $\ce{NO2}$, because $\ce{NO2}$ is the stronger -M group?

As Mithoron suggests, the correct answer is probably "no reaction." There are no examples of electrophilic aromatic substitution of p-cyanonitrobenzene in SciFinder, by the introductory organic chemistry mechanism at least.

There is one report of halogenation (with N-halosuccinimide) of this compound with palladium catalysis. The cyano group serves as a directing group to place a halogen ortho to the cyano. Chlorination and bromination are both reported.

Reference: Palladium-Catalyzed Highly Selective ortho-Halogenation (I, Br, Cl) of Arylnitriles via sp2 C–H Bond Activation Using Cyano as Directing Group


Is the product of Lindlar hydrogenation of (2Z,4S)-4-methylhept-2-en-5-yne optically active? According to me the poisoned catalyst will only reduce the triple bond so the compound produced must be optically active.. But the answer given is that the final product is optically inactive.. is it correct and if so how?

this is the final product I made:

The triple bond gets reduced to double bond as Lindlar's catalyst has been used so the reduction stops at this stage.. is this correct?

The two cis-configured pro-1-en-1-yl groups present on the central carbon remove the asymmetry. You can put a plane of symmetry through the $\ce{Me-C-H}$ plane of the central carbon to transform the molecule onto itself. You can confirm that by rotating the $\ce{(C=)C-C(HMeR)}$ bond.

Thus, this hydrogenation removes chirality.


What is the logical way to understand Iso, Sec, Tert and Neo prefixes? I'm only familiar with Alkanes at a basic level currently.
What seems logical to me is that an Alkyl molecule with the:

"iso" prefix means its an isomer of that alkyl, the usual way i'd see it would be the alkyl group with a substituent (the way an hexane could be isopentane if  I got it right)

as for the "sec" prefix, the alkyl group is connected to the main molecule through a secondary carbon (meaning the carbon on the alkyl is bonded with two other carbons in the alkyl, no consideration given to the main chain here)

I find the "tert" prefix similar to the latter, it is connected to the maine chain through a tertiary carbon (bonded with three carbons on the alkyl)

I couldn't find a basic logical way to address the definition of the "neo" prefix.

Are the ways I described the rest of the prefixes correct?

The prefixes, iso-, sec-, tert-, neo- are all traditional names when structural organic chemistry was still developing. You have already rationalized all except neo-. Neo- is Greek term for new. I looked up the OED etymological dictionary for older meanings, and the earliest use is traced to 1876. Now you can see in no 2, that neo- was a good option.

Used in the names of chemical compounds, minerals, etc., which are closely related to, or new forms of, known substances.
neocatechu  n. (Obsolete) a kind of catechu.
Used in the names of compounds and radicals in which one carbon atom is linked to four others.
Not part of standard chemical nomenclature although well established in neohexane and neopentane (and related terms).
(a)
Categories »

† neo-acid  n. Obsolete
1876   W. Odling in London, Edinb., & Dublin Philos. Mag. 5th Ser. 1 216   The names which call to mind that the normal, iso-, and neo-acids are mono-, di-, and trimethyl & etc. derivatives would seem to be most advantageous.

Neo- even it is used today represents a terminal tert-butyl group, just like the prefix iso- represents a iso-propyl group at the end.

IUPAC names have superseded all.


Why is sulfuric acid used in crystallization I read that:

  Take a cupful of water in a beaker and add a few drops of dilute
  sulfuric acid. Heat the water. When it starts boiling add copper
  sulfate powder slowly while stirring continuously.
  Continue adding copper sulfate powder till no more powder can be
  dissolved. Filter the solution. Allow it to cool. Do not disturb the
  solution when it is cooling. Look at the solution after some time. Can
  you see the crystals of copper sulfate? If not, wait for some more
  time.

So why is sulfuric acid required in this experiment?

The sulfuric acid will dissolve any residual copper metal or any copper oxides that exist as impurities in the copper sulfate, converting those impurities into copper sulfate.

$$\ce{H2SO4(aq) + Cu(s) -> CuSO4(aq) + H2(g)}$$
$$\ce{H2SO4(aq) + CuO(s) -> CuSO4(aq) + H2O(l)}$$
$$\ce{2H2SO4(aq) + Cu2O(s) -> 2CuSO4(aq) + H2O(l) +H2(g)}$$


How could chlorine dioxide be stabilized in aqueous solution? There is a water-based mouthwash product which is advertised as having the ingredient "stabilized chlorine dioxide." The only other listed ingredients are trisodium phosphate and citric acid. I assume this is a buffering solution or something.

In the company's documentation, it says it contains

  a stabilized molecular form of chlorine dioxide.

But in another location, the company claims that

  the solution is naturally activated by the amino acids in saliva,
  creating a small and very safe amount of chlorine dioxide.

So that sounds to me like there's actually something else in the solution, and the that the $\ce{ClO_2}$ is being produced ("activated") during use.

I've done some reading, and I have learned that $\ce{ClO_2}$ can be produced in several ways, but all of these methods seem to involve unpleasant reagents like chlorine gas, HCl, or oxalic acid.

I'd like to know what actual compound might be referred to as "stabilized chlorine dioxide" and what reactions might be involved.

I'm not asking about the effectiveness of the product, or what value $\ce{ClO_2}$ might have as a disinfectant, I'm just curious as to whether the claims made about the ingredients are plausible, and if so, what reactions might account for this?

Your assumption is correct. The term “stabilized chlorine dioxide” is misleading since the solution does not contain relevant amounts of chlorine dioxide $\ce{ClO2}$.

The solution actually contains sodium chlorite $\ce{NaClO2}$, which is the salt of the unstable chlorous acid $\ce{HClO2}$. At low pH, chlorite solutions contain chlorous acid
$$\ce{ClO2- + H+ <=> HClO2}$$
which decomposes to form chlorine dioxide approximately according to
$$\ce{5HClO2 -> 4ClO2 + HCl + 2H2O}$$
or
$$\ce{4HClO2 -> 2ClO2 + HClO3 + HCl + H2O}$$

At higher pH, decomposition of sodium chlorite slows down. Therefore, sodium chlorite is mainly stabilized by using a buffered solution. In your product, the buffer consists of trisodium phosphate and citric acid. Thus, a better description would be “stabilized sodium chlorite” instead of “stabilized chlorine dioxide”.


Will sodium thiosulfate react with nitrate and/or nitrite ions I am doing nitrate and nitrite quantification in raw water and treated water using a colorimetric method. The manufacturer of the instrument states that water with chloride will interfere with the test result. I was thinking to inactivate the chloride in the water with sodium thiosulfate. However, I am not sure how is sodium thiosulfate react with the nitrate and nitrite in my sample. Thanks in advanced for any guidance provided! Cheers!

First, typically only high concentrations of chloride will interfere with the colorimetric test...I assume you are using NitraVer or something similar? If you aren't using seawater (i.e., $\ce{Cl-} \gt \pu{100 mg/L}$) then you will be okay. But, let's say you are using seawater. There is an easy solution, simply make your standards in an aqueous matrix containing the same amount of chloride. (I hope you are making standards or have a check standard and are not just blindly trusting the UV-Vis result...)

Second, thiosulfate is a reductant and thus could potentially reduce nitrate and nitrite. However, your most powerful answer will be your own tests - you need to do control experiments. Make some nitrate and nitrite solutions of known concentrations and then mix them with thiosulfate at concentrations you would use. Check the nitrate/nitrite concentrations after mixing and determine if thiosulfate reduced them.


Interpretation of Coloumb operator in Hartree Fock equation I have read in a textbook (Modern Quantum Chemistry Szabo and Ostlund) that the Coulomb operator of the form
\begin{equation}
\mathcal{J}_{j}\left(\mathbf{x}_{1}\right)=\int d \mathbf{x}_{2}\left|\chi_{j}\left(\mathbf{x}_{2}\right)\right|^{2}r_{12}^{-1}
\end{equation}
describes the average local potential at $\mathbf{x}_{1}$ arising from the charge distribution from an electron in $\chi_{j}$. And therefore the term $\sum_{j \neq i}\left[\int d \mathbf{x}_{2}\left|\chi_{j}\left(\mathbf{x}_{2}\right)\right|^{2} r_{12}^{-1}\right] \chi_{i}\left(\mathbf{x}_{1}\right)$ for an N-electron system gives the total averaged potential of $N-1$ electrons in other spin orbitals on the electron in $\chi_{i}$.
However, if one expands the sum
\begin{equation}
\sum_{j \neq i}\left[\int d \mathbf{x}_{2}\left|\chi_{j}\left(\mathbf{x}_{2}\right)\right|^{2} r_{12}^{-1}\right] \chi_{i}\left(\mathbf{x}_{1}\right) = \left(\int d \mathbf{x}_{2}\left|\chi_{1}\left(\mathbf{x}_{2}\right)\right|^{2}r_{12}^{-1} + \int d \mathbf{x}_{2}\left|\chi_{2}\left(\mathbf{x}_{2}\right)\right|^{2}r_{12}^{-1} +...\right) \chi_{i}\left(\mathbf{x}_{1}\right)
\end{equation}
it is indeed a sum over the spin orbital but not of the electrons in the system. My interpretation for the sum is as follows: the total averaged potential acting on electron 1 due to the charge distribution of electron 2 in N-1 spin orbitals . I think it has something to do with the fact that electrons are indistinguishable so one cannot assign an electron to an orbital. However, I think the Hartree Fock integro-differential equation
\begin{equation}
h\left(\mathbf{x}_{1}\right) \chi_{i}\left(\mathbf{x}_{1}\right)+\sum_{j \neq i}\left[\int d \mathbf{x}_{2}\left|\chi_{j}\left(\mathbf{x}_{2}\right)\right|^{2} r_{12}^{-1}\right] \chi_{i}\left(\mathbf{x}_{1}\right)-\sum_{j \neq i}\left[\int d \mathbf{x}_{2} \chi_{j}^{*}\left(\mathbf{x}_{2}\right) \chi_{i}\left(\mathbf{x}_{2}\right) r_{12}^{-1}\right] \chi_{j}\left(\mathbf{x}_{1}\right)=\epsilon_{i} \chi_{i}\left(\mathbf{x}_{1}\right)
\end{equation}
was derived for hydrogen molecule. Is it the same for an N-electron system?

From the labels ($i,j$), I am guessing that $\chi$ is a molecular orbital (MO). When formulating the equations for molecular orbitals, one assumes that there are as many (spin-)orbitals as there are electrons. In any event, when writing down the Coulomb operator in the MO representation for the ground state, only occupied orbitals appear.
Therefore, I believe that the equations you are showing here are for the general problem, not just $\ce{H2}$. In some sense, the equations are the same for either, it is just a question of how many summands there are. Finally, note that the last equation you showed is valid for any system, but that it is actually part of a set of such equations, one for each MO.