text:
Potassium permanganate to analyze iron compounds
In my lab, I used potassium permanganate in order to analyze the amount of iron in a compound with iron in the +2 oxidation state using titration. My question is how would this analysis be different if we were analyzing iron in a different oxidation state like +3?

There would simply be no reaction observable. In my introductory lab, we were given an iron(III) sample to titrate and the first step was to reduce it to iron(II).

Assuming you have enough sample, first simply titrate to find how much iron(II) you have. Then convert all iron(III) to iron(II). Titrate again, and the difference you find is iron(III). (Not sure how reliable this is though.)

The Standard Methods for the Examination of Water and Wastewater 3500-Fe is a method for determining total $\ce{Fe}$, $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ in water using a colorimetric detection technique.  As in your titrimetric method, this method can only directly measure $\ce{Fe^{2+}}$.  In order to measure $\ce{Fe^{3+}}$ (the only other reasonably stable oxidation state of $\ce{Fe}$) you must first reduce the $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$.  I believe you could use the same reduction technique to prepare a solution of $\ce{Fe^{3+}}$ for your titration method.
The procedure sited above states that:

"[the $\ce{Fe}$ in the sample is] reduced to the ferrous state by boiling
  with acid and hydroxylamine."

The hydroxylamine solution is prepared by dissolving 10 g $\ce{NH2OH⋅HCl}$ in 100 mL water.  50 mL of the water sample is acidified with 2.0 mL of concentrated $\ce{HCl}$, and 1 mL of the $\ce{NH2OH}$ solution is added to the acidified water sample.
As $\ce{Fe^{3+}}$ species are frequently insoluble, the procedure described above calls for boiling the solution down to 15-20 mL to get all of the iron both dissolved and reduced before bringing the solution back up to 50 mL for analysis.  If the solution were prepared from a soluble $\ce{Fe^{3+}}$ salt like $\ce{FeCl3}$ for example, simply boiling the solution for 5 minute or so would likely be sufficient.
So, I suspect you could measure the $\ce{Fe^{3+}}$ in your solution using your titration method if you first reduced the $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$ using the hydroxylamine reduction described above.


text:
Is it possible to use the school supplied algorithm to build the Lewis diagram of the iodate ion?
I am upgrading the grade 12 chemistry credit and have to use the school supplied algorithm to build the Lewis diagram of $\ce{IO3-}$ ion. It seems to me that iodine oxidation state in this ion should be 5, therefore it would have one single and two double bonds with oxygen. When I follow the steps below, I am arriving at a different formula.
I do not need to answer whether the ion is polar or not and what is its shape.

Step 1: Count all of the valence electrons in the molecule or ion. In the case of an ion, add or subtract electrons to account for the ionic charge.

$$\ce{I = 7e^-, O = 18e^-}; \text{ionic charge} = \ce{1e^-}; \text{total} = \ce{26e^-}$$

Step 2: Arrange the peripheral atoms symmetrically around the central atom. Use a pair oil electrons to form a bond that links these atoms to the central atom.

6 electrons placed on $\ce{I}$, $\ce{26e^- - 6e^- = 20e^-}$ remain

Step 3: Add pairs of electrons to complete the octet of the peripheral atoms.

$3\cdot 6$ electrons placed on $\ce{O}$, $\ce{20e^- - 18e^- = 2e^-}$ remain

Step 4: Place any unassigned electrons on the central atom.

2 electrons placed on $\ce{I}$, $\ce{2e^- - 2e^- = 0e^-}$ remain

Step 5: If the octet of the central atom is incomplete, move a lone pair of electrons from a peripheral atom to a new position between the central and peripheral atom.

After completing step 4 of the rules above I am arriving at $\ce{I}$ having 8 electron dots. At this point all electrons have been used up, the octet of $\ce{I}$ is complete, and I can only create a structural diagram with two single and one double bonds.
In the course I have taken previously we were taught that $\ce{I}$ was an exception to the octet rule, but when I asked this teacher, I was told that I was not allowed to use the exceptions and had to apply the octets only.
Am I missing anything? Is it possible to comply with the algorithm above and come up with the correct answer?

Possible duplicate of Iodate ion (IO3-). Is it polar or non-polar?

@Mithoron How's that relevant?

Read precisely question and answers (at least Jan's answer) and see.

@Mithoron My question is quite different. I do not care about the shape and polarity, only of how I must follow the school's algorithm. Did you actually read my question? I am going through the course where sulphite ion has 3 single bonds with O, so I really have to put up with weird stuff.

The algorithm is naturally as flawed as the Lewis-theory for these kinds of molecules and one structure alone will never sufficiently describe the bonding situation. However, in this particular case, the description of all single bonds is the closest to the actual bonding. The so-called octet expansion through the involvement of d-orbitals has been disproven many times. Most of the times the octet rule prevails. Similarly 'sulphite ion has 3 single bonds with O' is correct and not at all weird. Double bonds in those compounds are weird.

@Martin-マーチン Please prived a recognized source for "double bonds in these compounds are weird"

When I follow your school's algorithm for iodate I get a central iodine atom with three single bonds to oxygen.  I've shown my work below.  This is problematic, as you well know, but I can potentially see where your teacher is coming from.

When I teach formal charge and resonance structures and the like, I often have students use their intro chemistry rules to determine the structure of a polyatomic ion such as this.  If they follow the rules they get a pretty structure that looks proper - I then introduce the idea of formal charge by pointing out the wild imbalance of charges in the ion and how energetically unfavorable that would be, then we work out a way to rearrange the electrons to minimize that problem.
I also have a lot of students who immediately go to google to find Lewis structures rather than drawing them themselves, so when a student tells me that they think the structure should be something different, it takes an actual argument from them before I believe that they know why that should be.  I don't know that that is the case here; it also seems entirely possible that your instructor is just wrong.

The structure is in no way problematic; instead it is the most correct.

@Jan The idea of resonance structures in this sort of thing is entirely incorrect then?

Resonance structures would imply octet expansion which in turn would imply iodine d-orbital participation. We are talking about the 5d orbitals for an element of the fifth period, i.e. the other relevant orbitals are 5s and 5p. (4d are fully populated and function as core orbitals.) But these are energetically far away; even 6s has less energy than 5d (yet nobody ever suggests any $\mathrm{s^2p^3}$ hybridisation). Look around a few answers by Martin to see the d orbital participation theory disproven.

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text:
Difference between Coupled Cluster and Full CI
I'm looking at some high precision quantum chemistry methods like Coupled Cluster (CC) and FULL CI (FCI).
It seems both CC and FCI start from Hartree-Fock and then excite the electron to higher states.
If I understand correctly, CC uses an exponential operator to raise the states of the electron while FCI explicitly calculates the coefficients of each determinants.
What makes the difference between CC and FCI? Can someone give an intuitive explanation? Thanks.

Related but solidly different, this answer discusses FCI as it relates to CASSCF/RASSCF.

Well I am sure one of the more experienced users will weigh in on this, but given that I have recently started reading into this stuff myself , I would say configuration interaction is not size extensive (say for 2 non-interacting $\ce{H2}$s), while CCSD is exact for N non-interacting $\ce{H2}$. My 2 cents.

@getafix FCI is also horrifically expensive for all but the tiniest of systems. To be clear, AFAIK full CI (where every single orbital is correlated) is actually size extensive. It's the truncated versions that are not.

@hBy2Py I see. Would you happen to know how the two methods scale ? $\mathcal{O}(N^6)$ or $\mathcal{O}(N^7)$ or higher?

@getafix Pretty sure FCI is O(N!) or higher. Not sure about CC. Would have to pull Szabo and Ostlund out.

Full Coupled Cluster (FCC) vs. Full Configuration Interaction (FCI)
The main theoretical difference is the way excitations are used. "Excitation" refers to putting one or more electrons in higher orbitals than the "reference" calculation, which is often a Hartree-Fock calculation, would. CI uses a "simple", linear excitation operator, whereas CC uses a more complicated, exponential one. One can think of CI as a sum of excitations while CC results in a product of excitations.
Symbolic equations for the wave function ($\Psi$) ansatz of either method read:
$$
\Psi^{\text{CI}} = (1 +C_1+C_2+C_3+...) \Psi^{\text{HF}} \\
\Psi^{\text{CC}} = \exp(T_1+T_2+T_3+...) \Psi^{\text{HF}} \\
= (1 +T_1+T_2+T_3+\frac{1}{2}T_1^2+\frac{1}{2}T_2^2+T_2T_1+\frac{1}{2}T_3^2+...+\frac{1}{6}T_1^3+...) \Psi^{\text{HF}}
$$
In CI, the excitation operator is often denoted $C$ and the prefactor of determinants are called CI coefficients, whereas CC uses $T$ and "amplitudes". In either, the subscript refers to the number of excited electrons.
For "full" calculations, that is, calculations that take into account every possible excitation, the results of CC and CI will be identical.
Full calculations are too expensive for all but the smallest systems, because the cost rises as $O(N!)$, where $N$ is some measure of the system size involving electron number and basis set size. For this reason, one typically decides to "truncate" the excitations to lower the cost.
Single-reference Truncated CC vs. Single-reference Truncated CI (this is what CC and CI commonly refer to)
Truncation first involves collecting the excitations into "classes" of excitations. Within a class, one gets similar cost and hopes for similar contributions. Thus, one may sort of gauge the cost and accuracy of the calculation. Standard ways of doing this have evolved, and this is what CCSD, CCSD(T), CISD etc. refer to. CASSCF/RASSCF can be categorized like this, as well.
With truncation, another difference between CI and CC arises: CC is no longer variational and CI is no longer size-extensize. Variationality is the property that the resulting energy has the true energy (of the given basis set) as a lower bound - one approaches the energy monotonically from above as more excitations become involved. Size-extensivity is the property that two identical systems at infinite separation from each other yield twice the energy of a single system. The latter property is thought of as the more important one, mainly because of its implications for closer, interacting systems. Variationality is also less of a concern because finite basis sets must be used, thus, one is typically far above the true energy anyway.
Another technical difference is that truncated CC and CI lead to different equations, for which different methods of solving have been developed. Overall, CC implementation is, or at least used to be, more difficult. Properties other than the total energy can also be more difficult to implement.
Scaling of CC and CI
Basically, when truncation is performed at similar levels, the two methods scale the same, but the prefactor may be different - this also depends on the quality of the implementation. For reasons stated above, CC has been worked on more intensively since, say, the mid 1990s. So, CISD and CCSD scale similarly, CISDT and CCSDT would as well. Bottom line: CCSD(T) will still be affordable for molecules where full CI would require specialized implementation on a dedicated supercomputer.

The reference wave function does not have to be a Slater determinant from a Hartree-Fock solution; that's just convenient to use. The former gives rise to multiconfigurational approaches. An early attempt of correcting CISD for size-extensivity was the Davidson Correction. Another approach is QCISD. It actually is hard to believe that CISD scales similar to CCSD, as the latter has exponentially more terms. You should review the edit, your Taylor expansion is incomplete.

To @Martin-マーチン 's comment: Yes, other references are possible, (including DFT-based determinants). The answer will deal with single-reference. Frank Jensen (Introduction to Computational Chemistry, 2nd ed., Wiley) [FJ] notes that the Davidson correction is not correct for two-electron systems. QCISD was shown to be an approximation to CCSD [FJ]. Can you be more specific concerning the Taylor expansion?

My comment was in no way critique, just an extension, maybe asking for some more explanation for the scaling part. I remember from my courses on post-HF methods that even though CCSD(T) is more expensive than CISD(T), we use it since it recovers more of the correlation energy. As for the expansion, I quite liked it how you grouped the clusters according to their excitation $= (1 +T_1+(T_2+T_1^2)+(T_3+T_2T_1+T_1^3)+...)$, now it looks quite odd since it does not have much resemblance with the Taylor expansion $\exp{ \mathbf{T} } = \sum_n\frac1{n!} \mathbf{T}^n$, i.e. the terms appear random.

I apologize for the curt comment I gave before, it was a lot to get to. Note that scaling is only one part of the cost, the other is the prefactor, which may easily be higher for CC and is also very much implementation-dependent. An edit of the Taylor expansion was suggested by other users - the expansion you gave can be found in most books, anyway. Thank you for your suggestions!

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If the ferric ion is more stable than the ferrous ion, then why is it readily reduced to the latter?
The standard reduction potential value for the ferric ion:
\begin{align}
\ce{Fe^3+ + e- &<=> Fe^2+} & E^\circ &= \pu{0.77 V}
\end{align}
The value is positive, hence it means that ferric ion is readily reduced to ferrous ion, i.e. ferrous ion must be more stable than ferric ion.
But in a book I have also seen that ferric ion is more stable than ferrous ion as it has stable $\mathrm d^5$ configuration.
I am little bit confused that if ferric is more stable, then why ferric is readily reduced to ferrous. Please explain it in detail.

related: Stability of ferrous ions and iron(II) hydroxide in acidic conditions

The problem with nearly every question about "stability" is that it is unclear what exactly you mean by "stable".  This is no exception.  For example, do you think the stability of ferric ion aqueous solution would be different if it were under an atomsphere of pure hydrogen, or if it were under an atmosphere of pure oxygen?

As already pointed out, the notion of stability needs to be defined much more accurately.
The fact that the standard reduction potential is positive has no direct bearing on the gas-phase stabilities of the two ions. The only thing that the standard reduction potential actually reflects, is that the standard Gibbs free energy of the following reaction
$$\ce{Fe^3+ (aq) + 1/2H2 (g) -> Fe^2+ (aq) + H+ (aq)}$$
is negative.
So, for example, Curt already hinted in the comments that under an atmosphere of oxygen, $\ce{Fe^3+}$ would be more stable. This is simply because the reaction
$$\ce{4Fe^2+(aq) + O2 (g) + 4H+ (aq) -> 4Fe^3+ (aq) + H2O (l)}$$
(or some variation of it) has a negative Gibbs free energy. That is very easy to prove, since we know that $E^\circ(\ce{O2}/\ce{H2O}) = +1.23~\mathrm{V}$ and $E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) = +0.77~\mathrm{V}$.
In addition if you were to carry out the reaction in base, the stability of $\ce{Fe(III)}$ relative to $\ce{Fe(II)}$ will be increased greatly, since any $\ce{Fe^3+}$ will be precipitated as $\ce{Fe(OH)3}$.
Remember that a standard reduction potential is measured against an arbitrarily determined zero. If, instead of the standard hydrogen electrode, we chose to use the $\ce{O2/H2O}$ reaction as the standard, then $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ would have a negative value.

Let's also get this myth out of the way. $\ce{Fe^2+}$ does not, under any circumstances, have a $\mathrm{(3d)^5(4s)^1}$ configuration. Period.

Ok, with all of that settled, let's answer the question you were actually trying to ask. Yes, $\ce{Fe^3+}$ is a decently good oxidant. There are plenty of half-reactions that have $E^\circ$ below the $\pu{0.77 V}$ of $\ce{Fe^3+/Fe^2+}$. If you want to use $\ce{H+/H2O}$ as a reference point (which is perfectly fine, but you have to make it very clear that this is a relative stability, not an absolute stability), then yes, $\ce{Fe^3+}$ is a better oxidant than $\ce{H+}$.
The reason behind that has nothing to do with $\mathrm{d}^5$ stability. It is simply because $\ce{Fe^3+}$ has a pretty high charge and would quite like having another electron so that it doesn't have quite so high a charge.
More formally, you could construct a cycle for the half-reaction $\ce{Fe^3+ + e- -> Fe^2+}$, neglecting entropy for the time being.
$$\require{AMScd}
\begin{CD}
\ce{Fe^3+ (aq) + e-} @>\large\Delta H^\circ>> \ce{Fe^2+ (aq)} \\
@V\large-\Delta_\mathrm{hyd}H(\ce{Fe^3+})VV & @AA\large\Delta_\mathrm{hyd}H(\ce{Fe^2+})A \\
\ce{Fe^3+ (g) + e-} @>>\large -I_3(\ce{Fe})> \ce{Fe^2+ (g)} \\
\end{CD}
$$
The quantity $\Delta_\mathrm{hyd}H(\ce{Fe^2+}) - \Delta_\mathrm{hyd}H(\ce{Fe^3+})$ is positive since $\Delta_\mathrm{hyd}H(\ce{Fe^3+})$ is more negative than $\Delta_\mathrm{hyd}H(\ce{Fe^2+})$. So, the hydration enthalpy terms disfavour the reduction of $\ce{Fe^3+}$ to $\ce{Fe^2+}$.
However, the third ionisation energy of Fe is pretty large, and it turns out that in the 3d block, the third ionisation energy is the major factor that controls the reduction potential of $\ce{M^3+}$ to $\ce{M^2+}$. In particular, it is quite large because Fe is the 6th element in the 3d-block; you should know that $Z_\mathrm{eff}$, as well as ionisation energies, steadily increase across the 3d block due to the poor shielding provided by 3d electrons.
Put simply, the large value of $I_3$ makes it difficult to oxidise $\ce{Fe^2+}$ to $\ce{Fe^3+}$.
The $\mathrm{d^5}$ configuration of $\ce{Fe^3+}$ does have an impact. It can explain why $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ is relatively small compared to its neighbours: $E^\circ(\ce{Mn^3+}/\ce{Mn^2+})$ and $E^\circ(\ce{Co^3+}/\ce{Co^2+})$ are both much larger than $\pu{0.77 V}$. However, this "half-filled shell stabilisation" (more properly phrased in terms of exchange energy) is not quite a large enough factor to make $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ negative.
The question that I linked above has slightly more information.

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Simple way to understand why chromophores often have double bonds?
Chromophores — atoms or groups of atoms within a molecule that absorb some visible wavelengths better than others — result in compounds that have color. For example, see this answer about azo dyes, and this mention of bleaching stains by oxidation of double bonds.
Without going too deep into quantum mechanics and/or molecular orbital theory (MO), is there a way to understand in a qualitative way why chromophores are often related to the presence of double carbon or double nitrogen bonds?

Related: http://chemistry.stackexchange.com/questions/8910

@KlausWarzecha I've been admiring those answers. Here I'm asking for something a little simpler to start with. Rather than a long discussion of the HOMO-LUMO gap or wavelength shift, I'm asking what it is about a double bond that gives it a wavelength dependent photon cross-section (i.e. color) to begin with, as compared to a single bond.

No. Without going into quantum mechanics and molecular orbital theory, there is no way to understand why chromophores typically have double bonds.
Everything related to colour is related to the excitation of electrons from one orbital to another and therefore any explanation that does not rely on orbitals in at least some way is doomed to fail.

If you decide to accept a little MO theory then you can be helped. Double bonds are formed by an underlying σ bond (or single bond) and a π bond. Each bond is formed by two atomic orbitals (one from each atom) that combine to give a bonding and an antibonding orbital. The energy difference between the bonding and antibonding orbital corresponds to the excitation energy: a photon that carries exactly that energy will be absorbed and the electron excited to the higher level. The greater the overlap between the atomic orbitals, the higher the energy difference between bonding and antibonding.
Now in π orbitals, the atomic p orbitals are oriented perpendicular to the bond axis, so any overlap is doomed to be much smaller than the corresponding single bond’s σ overlap. Therefore, the energy difference from the bonding to the antibonding π orbital is not far into the ultraviolet region; typical values would correspond to a wavelength of $150$ to $200~\mathrm{nm}$. Extending the π network will move those orbitals closer together energetically (again, the complete answer requires much more MO theory) which is why benzene has an energy difference corresponding to a wavelength of $250~\mathrm{nm}$. Thenceforth, it is not a long way until you reach visual wavelengths ($400{-} 700~\mathrm{nm}$).

Thanks! This answer is really helpful to me. It gives a starting point from which I can try to read further, and hopefully ask further, more quantum-mechanically-focused questions. Do note I've asked "without going too deep into quantum mechanics..." not "without quantum mechanics", which suggests I had already decided to accept a little quantum mechanics. :)


text:
How does a photochemical reaction of lignin with oxygen (yellowing of paper) produce chromophores?
Edit:  This question is about chromophore production in lignin as found in low cost or short-term use paper such as newsprint. A different, excellent answer describes oxidation yellowing of the cellulose found in books.

This answer contains a quote from "Stuff Matters" by Mark Miodownik:

Lignin reacts with oxygen in the presence of light to create chromophores ( meaning, literally, 'color carriers'), which turn the paper yellow as they increase in concentration. This type of paper is used for cheap and disposable paper products, and is why newspapers yellow quickly in light.

I've read here that bleach and peroxide can oxidize chromophores by reacting with double bonds, reducing or removing the appearance of color. In this particular case of what sounds like a photochemical reaction of lignin with oxygen, how is a chromophore produced?
Or, is it really even true that a chromophore is responsible for the yellow color of yellowed paper? Or is it the oxidation and loss of a fluorophore that was originally making the paper look white? I'm not sure of that, because the phenomenon seems to be pretty old, possibly before the use of fluorophores in paper manufacture.

Pure cellulose will not turn yellow in the presence of oxygen air light, while paper containing lignin does. The effect results from the oxydation of the phenols to quinones.

@KlausWarzecha That certainly sounds helpful, why not leave a proper answer? (I've asked here as well.)

@KlausWarzecha is there anything I can do to convince you to expand on that just a little in the form of an answer? Sometimes people don't leave answers because they don't have time to write a thorough one, but in this case, a short answer will be greatly appreciated!


text:
Why is ΔS the same regardless of whether a process is irreversible or reversible?
Why do we take the entropy change of the system to be the same in both irreversible and reversible processes? What is the idea behind this?
It would be nice if anyone could give me an example with some data.

But change in entropy is not same for reversible & irreversible process. A irreversible process is always accompanied by increase in entropy of universe while in case of reversible process, entropy remains constant.

@Avi what you said is true, but that applies to the entropy of the universe. The question (admittedly rather unclear) is concerned about the entropy change of the system, which is not necessarily equal to zero.

In an irreversible process, we can't use the integral of dQ/T to calculate the change in entropy between the initial and final states of the system, because there is entropy being generated within the system (as a result of mechanical energy dissipation to internal energy) which the integral does not account for.  However, since entropy is a function of state, we can focus on the exact same initial and final states of the system, and devise a path between these two states that does not involve dissipation of mechanical energy, and calculate the integral of dQ/T for this alternate path.  Such a path is a reversible path, and the integral along this (or any other) reversible path gives us the change in entropy.


text:
Comparing nucleophilicity
We have to compare the nucleophilicity of $\ce{OH-,NH2-,CH3CH2-}$ and $\ce{CH#C^{-}}$. I managed to find out that $\ce{CH3CH2-}>\ce{NH2-}>\ce{OH-}$ and that $\ce{CH3CH2-}>\ce{CH#C-}$. But I don't know how to compare $\ce{CH#C-}$ with either $\ce{OH-}$ or $\ce{NH2-}$.
My attempt:
I know that the acidity order of the 2nd period is $\ce{CH4}<\ce{NH3}<\ce{H2O,HF}$. Now the stronger an acid is, the weaker its conjugate base will be. Therefore, the conjugate base order will be:
$$\ce{CH3-}>\ce{NH2-}>\ce{HO-}>\ce{F-}$$
We know that nucleophilicity parallels basicity along the period, therefore the nucelophilicty order will be as the basicity order:
$$\ce{CH3-}>\ce{NH2-}>\ce{HO-}$$
Now we can compare $\ce{CH3CH2-}$ with $\ce{CH3-}$.
Since methyl is an electron donating group, it will increase the negative charge on the carbanion and therefore $\ce{CH3CH2-}$ will be more nuceleophilic than $\ce{CH3-}$.
Since $\ce{CH3-}$ is more nucleophilic than $\ce{NH2-}$ and $\ce{OH-}$. Also $\ce{CH3CH2-}$ is more nuceleophilic than $\ce{CH3-}$. As a result $\ce{CH3CH2-}$ will be more nuceleophilic than $\ce{NH2-}$ and $\ce{OH-}$ and $\ce{CH3CH2-}$.
We can also compare $\ce{CH3CH2-}$ and $\ce{CH#C-}$. Since $\ce{CH#C-}$ has sp hybridization, it will be less nuceleophilic than $\ce{CH3CH2-}$.
But how can I compare $\ce{CH#C-}$ to either $\ce{OH-}$ or $\ce{NH2-}$?

I don't know the reason but I have acidity order as H2O >$\ce{CH#CH}$ >NH3. Hence nucleophilicity will be OH- < $\ce{CH#C^{-}}$ < NH3.

Do you have a order of nucleophilicity list? I could really use one of those.

Sorry to say but I don't have any such list.

But nucleophilicity and basicity are not the same.

But nucleophilicity runs in parallel with basicity in a period

@Abhishek Maybe, but I won't take that rule of thumb too seriously. There can be subtle but significant differences between nucleophilicity and basicity. For one thing, nucleophilicity is defined in terms of kinetics and basicity in terms of equilibrium. This makes the two properties possibly respond differently to reaction conditions such as changing the temperature. My point is don't take their correlation as a stricter rule than it is.

You also may want to consider Hard-Soft Acid-Base Theory. The carbanion is generally said to be a better nucleophile than hydroxide, presumably as measured in a highly controlled setting using a standard electrophile like methyl iodide. I'm under the impression nucleophilicity is very nearly defined as how fast an electron donor reacts with $\ce{CH3I}$. Note that there are no hydrogens available for elimination on this electrophile, which precludes the possibility of the nucleophile acting as a base instead.
In practical reaction conditions (protic solvents, beta-hydrogens on electrophile, etc), we may need to worry about our nucleophile acting as a base. This is where HSAB Theorycomes in. Hydroxide is considered a "hard base", which means it may be more inclined to react with "hard acids" like $\ce{H+}$ rather than "soft acids" like methyl iodide. As a "soft base", a carbanion will probably prefer to bond with methyl iodide (as a classic nucleophile) than with any available protons (as a Brønsted-Lowry base). So in practice, the difference between a carbanion's nucleophilicity and hydroxide's might be larger than expected due to competing "hard base" reactions (Brønsted-Lowry rather than Lewis). Just food for thought.
Qualitative nucleophilicty ratiomale

By removing H ions, negative charge on more electronegative more stable so acidic strength order that those EN is less easily lose its electron so order of nucleophilicity EN of NH3 is 3 o is 3.5 $\ce{C#C}$ is 3.25


text:
How do I turn room temperature water into boiling water for drinking using an chemical/additive?
I am trying to create instant boiling water using a chemical of sorts, so that it can be consumed by a human. For example; someone could make tea without a kettle or any available water heating systems.
It must be completely safe and if possible, leave no bad tasting remains. Preferably, the chemicals used would have to be easily accessible and stable in its bought state.
I am in high school and just want to do a cool science experiment. I guess this could have endless potential applications. I am also open to other ideas, like making a disposable bottle, that when you twisted the lid, the water would heat up as the chemicals dropped down or the bottle itself heated up. Just any way of doing it would be awesome.
I saw this page but is wasn't consumable ideas:
Can you heat water with additives?
Thanks,
Charles

You mean you want an idea that would surely make you an instant billionaire? Think again. ;-)

https://www.google.com/search?q=self+heating+coffee

It would work,if the impurities have a lower boiling point than water,which I can tell you almost certainly are not.

Adding something to water that heats it to 100°C and leaves only water is of course nonsense.
A self-heating cup is easy: You take a double-walled beaker, fill the space between with two compounds that react with great heat but no gas formation, separated by an extra wall with a valve.
Google the self-cooling beer keg. It works very similar, only it gets hot outside, not inside.
Or just google "self-heating coffe cup".

As the comment above suggests, if there was an obvious way to do this it would already be making someone wealthy.
Self heating cans (and gloves) are seen where an exothermic chemical reaction or heat of solution is used to thermally warm food or gloves. The most common application of this method are 'flameless ration heaters' used to heat MRE (military rations).
Typically the ration heater employs an oxidation-reduction reaction e.g. a small amount of water is added to finely powdered magnesium metal, alloyed with a small amount of iron, and table salt. This forms an electrolyte and because the magnesium and iron are in contact it is analogous to many short circuited batteries. The boiling point of water is quickly reached as the reaction proceeds.
Note that the exothermic chemicals are physically isolated from the food being heated. A flameless ration heater might raise the temperature of a 230g meal, 38°C in 10 minutes.

Long shot but if you can engineer a strong enough metal container and isolate the expanding gas' cooling effect from the system, you can maybe just maybe mix HN3 and nitrous acid. Byproducts are a lot of gas that should escape the system vessel (thus cooling it so big fail unless you can shield your system from the cooling effect) and FDA approved N2O and water. You will need explosives license to ship the product but hey you said could it be done and I am pretty sure it can't be done but maybe it can be done. I dunno man water also takes too much energy to heat up and boil too so idk about all of it

Don't let the others here discourage you from thinking big.  Having said that, the problem with your scheme is that generally we want to avoid adding stuff to drinking water, in order to keep it pure (except in very small amounts, such as chlorination to disinfect it).  And you'd need to add a lot of reacting substance to water to get it to boil. That's why all the existing schemes to heat up food and drinks using chemicals keep the chemicals isolated from the water.
You'd need substances that are in solid or liquid form, and that reacted exothermically without leaving anything in the water.  I don't know of any.
The only thing I can think of is hydrogen and oxygen gas, which combine exothermically to produce water.  So you'd need some sort of a very heavy-walled container that could store oxygen and hydrogen under very high pressures, and could slowly flow them into a third container to gradually produce water. This would be very expensive, and wildly impractical for consumer use.  The same effect could be much more simply accomplished with a heating element and a battery.


text:
Washing glasswares
What do most labs use to wash their glasswares? I usually use plain clothes detergent. But, I'm starting to suspect that there are residual detergent even though I've brushed the glassware; rinsed it until there is no slipperi-ness and then after that, rinsing it in cold water again.
Is there anything wrong with this?

The usual brand of detergent is T!de Original.

Why cold instead of hot water rinse?

Coz hot water is more likely to have more copper oxides with it (or whatever hot water pipes in your part of the world is made out of).

Also for final rinsing, might as well use the less energy intensive cold one.

Good point.  I kind of assumed that you were using DI water for the rinse.  As far as removing detergent though, you could probably remove more detergent with less water if it were piping hot.

No de-ionized water here, sir. I'm just a man of meager resources. I could get a copper pot and tubes to do distillation, and that's as far as I'll go for a source reasonably pure water.

Impossible to answer. What are you using your glassware for? How perfectly clean do you need them? Are detergents a problem? Not using DI water for the final rinse leaves carbonate stains on the glass, which collect more dirt. Unless you wipe the glassware dry, but that could spread more contamination. Etc.

Pool money from your colleagues to get a DI water source. Give out the DI in ten liter cannisters, keep a record on it, and distribute the running costs accordingly.

If you're measuring fluorescence or luminescence, for example, most laundry detergents have fluorescent "clothes brighteners" added, which might linger in glassware.

Soap and water. Then rinse with DI. Then rinse with acetone.

Hygiene procedures are application specific.  For example,  a metals lab will acid wash glassware,  while an organics lab will solvent wash glassware.  Perform the most economical hygiene that suits your purpose.  If residual detergent could affect your experiment/analysis, it's time to change your wash procedure, regardless if there is a residual after washing.  However,  if there is no affect on your application,  you're doing fine.

Not "hygiene procedures" but "contamination procedures." You should never eat/cook in glassware used for chemistry experiments.

wow. that is a great distinction. i had always errantly lumped glassware and handling under chemical hygiene, but hygiene does specifically referred to protection of person. safety and nomenclature are two of my favorite things. thanks!


text:
Why Li can't make it's Peroxide when heated with more O2?
Every first group metal reacts with $\ce{O2}$ and makes there monoxide at room temperature.
But it says that $\ce{Li}$ does not make peroxide when heated with more $\ce{O2}$. But other metals in that group do.
Why $\ce{Li}$ can't make it's Peroxide when heated with more $\ce{O2}$?
And if we give the heat continuously with more $\ce{O2, K, Rb, Cs}$ give there Super-oxides. In that time Both $\ce{Li}$ and $\ce{Na}$ do not gives there Super-oxides. What is the reason for it?

related: Why do the alkali metals form different products upon combustion in air?

It has to do with the small size and high charge density of the lithium ion, which effectively polarizes any peroxide ion that might form and thus pulls it apart.  See http://www.chemguide.co.uk/inorganic/group1/reacto2.html.


text:
Is it possible to have same mass and volume for a solvent and its solution
Is there any solute-solvent interaction possible, in such a way that a solvent with mass X grams and volume Y litres, and a solute with mass A grams and volume B litres, to form a solution that has exactly the same mass and volume as that of the solvent (X grams and Y litres)?

Not exactly exactly, but if $X \gg A$ ... ;-)

I think you wrote this down wrongly, because the mass does of course not change . Do you mean you want to add something to 1 litre of solvent, and still have exactly one liter of solution? Or do you want that the volume of the solution is exaclty the sum of the volumes of solute and solvent?

As Klaus suggests above, you can't add mass to another mass without it increasing.  That would defy the first law of thermodynamics that energy cannot be destroyed since mass and energy are directly linked.
However, much of this depends on what degree of precision you're measuring to.  If you're working with a cubic metre of water and only measuring to the first two significant figures, mass would be 1000kg and volume 1000l, and you add 40kg of sodium chloride to make a 4% saline solution, you would still only have 1000kg and 1000l because you're not counting the third significant figure.
It's all about context ;-)


text:
Why is the dipole moment of chloromethane larger than the dipole moment of fluoromethane?
On the Pauling electronegativity scale, fluorine and chlorine are 3.98 and 3.16, respectively.
Since the dipole moment is dependent on electronegativity, why is the dipole moment of chloromethane larger than the dipole moment of fluoromethane?
I guess that the smaller size of the s orbital in the fluorine atom has something to do with this.

Dipole moment is not just about charges, it also has $L$ term. Bond length of $\ce{C-Cl}$ is greater than $\ce{C-F}$ and in this case, that is more dominating factor.
The dipole moment is in order
$$\ce{CH3Cl} \gt \ce{CH3F} \gt \ce{CH3Br} \gt \ce{CH3I}$$
You can see that electronegativity plays a more dominating role in $\ce{CH3X}$ when $\ce{X}$ is $\ce{Br}$/$\ce{I}$.


text:
Modeling the dissolution of an oil droplet with detergent

How to model the dissolution of an oil droplet on the surface of water when a detergent is (gradually) added to the water?

I believe this includes an interplay of coalescence keeping the droplet together and dissolution tearing it apart. Adding the detergent then changes the system so that the dissolution becomes stronger.
So far I tried a partial differential equation approach with the Cahn–Hilliard equation. However this model seems not to account for the coalescence, so that the droplet is always dissolved regardless of the parameters.


text:
How is the nucleophilicity of alkoxy groups related to acidity and +/- inductive effects due to substituents?
I found a nucleophilicity order as
$\ce{HO-} > \ce{CH3-O-} > \ce{(CH3)2CH-O-} > \ce{(CH3)3C-O-}$
The reason was given that bulky groups reduce nucleophilicity. But why?
The order of the $+I$ effect is $\text{tertiary} > \text{secondary} > \text{primary}$. Due to the greater $+I$ effect, the electron on oxygen must experience greater repulsion and thus the order of neucleophilicity must be reversed. Also, according to the given order of nucleophilicity a tertiary alcohol is more acidic than a primary alcohol. But I have read that presence of more $-I$ effect increase acidity of alcohols like $\ce{Cl-CH2-OH}$ is more acidic than $\ce{CH3-OH}$. Similarly due to more $+I$ effect in $\ce{(CH3)3C-OH}$, it should be less acidic than $\ce{(CH3)2CH-OH}$ which should be less acidic than $\ce{CH3-CH2-OH}$.

Nucleophilicity is a kinetic property. It is about how good a species is at attacking a carbon centre. If the species in question is very large it just can't get close enough to the electrophilic centre to react.

Then what about acidity of alcohols? Is tertiary alcohols more acidic than primary alcohol.

No. Acidity is a thermodynamic property relating to the relative stabilities of the acid and the conjugate base. Basicity and nucleophilicity are not the same thing.

Does this mean that in this case Basicity order is opposite of the given order?

Yes it does. It is not unusual for the order of basicity to be different to the order of nucleophilicity.

Thanks for answering so patiently but I want to ask , Is this difference in neocleophilicity and basicity is only for alcohols or there are other compounds too. If yes then please give some examples.

$$\Large\color{red}{\text{nucleophilicity} \ne \text{basicity}}$$
Please memorise firmly that nucleophilicity and basicity, albeit related in many ways, are different physico-chemical properties. Most importantly, basicity is a thermodynamic property while nucleophilicity is a kinetic property. Therefore, it is not uncommon for compound $\ce{A}$ to be more basic but less nucleophilic than compound $\ce{B}$.
This is the case here, too. It is unquestioned and supported by the corresponding $\mathrm{p}K_\mathrm{a}$ that tert-butanolate is the strongest base, followed by isopropanolate followed by ethanolate. This is due to the $+I$ effects you correctly outlined.
Nucleophilicity does not parallel this series since the it does not abide by the same rules. For a compound to be nucleophilic, the nucleophilic atom’s lone pairs must be able to actually attack, i.e. reach, the electrophilic site it wishes to attack. This means two molecules interacting and not just one molecule interacting with a proton which is so small it can even tunnel. In tert-butanolate, you have three α-methyl groups that take up space and make said interaction harder. Due to this steric interaction, tert-butanolate is an absolutely lousy nucleophile (while it is a great base).
It is similar for isopropanolate which still has too much steric hindrance to be an effective nucleophile although the conditions are better than for tert-butanolate.


text:
Why is temperature considered as an important factor in deciding chemical reaction kinetics?
According to the Arrhenius equation, the rate of reaction is proportional to $e^{-1/RT}$.
Here is a plot of the equation and its derivative with respect to temperature, both with respect to temperature. All constants are set to unity.

It seems to suggest that there is almost no change in the rate of reaction with increasing temperature except at very-low-temperature range.
Then, why is temperature considered an important factor in deciding the kinetics of a chemical reaction?

The Arrhenius eqn. describes how the rate constant changes with temperature as $k=A\exp(-E_a/(RT))$ where $E_a$ is the activation energy (an energy barrier) and A is a constant. At higher temperatures more molecules (by the Boltzmann distribution) have enough energy to surmount the activation barrier and so reaction becomes faster. Negative temperature as in your figure do not make sense. At very high temperatures the exponential tends to 1 as all  molecules have enough energy to exceed the barrier energy and so the rate constant tends to A.

It is possible to produce negative temperatures, (inverted population) such as in a laser, in certain NMR experiments, and in a chemical laser. But the first two are not chemical reactions and the latter the population inversion is in the products of an exothermic reaction. To answer your last question the rate constant does change by a large amount from low to high temperatures, basically from zero to from finite limiting value at high temperature.

If the activation energy is lower (<<50kJ/mol),  then temperature has a much lower effect than when activation energy is higher (>>100kJ/mol). Your comment would be true for the former case


text:
Why isn't ammonium strong acid, since it's charged and ammonia is neutral?
Why is ammonium such a weak acid $(\mathrm pK_\mathrm a = 9.24)$, if $\ce{NH4+}$ has a positive charge on nitrogen, while $\ce{NH3}$ is neutral. It means that conjugate base (ammonia) is really stable, which makes conjugate acid a strong acid.
Another argument why ammonium should be strong acid is comparison with hydronium ion $\ce{H3O+}$. Both nitrogen (EN = 3.04) and oxygen (EN = 3.44) are significantly electronegative atoms and "don't like" to be positively charged. That explains pretty good hydronium ion $\mathrm pK_\mathrm a = -1.74$, but why is then ammonium $\mathrm pK_\mathrm a$ is just $9.24$?

Ammonia is rather apolar (compared to water!). So apolar its Bp. is -33°C. It's really only soluble in water because it can take up a proton.

Check your acid-base chemistry: A very weak base is conjugate to a strong acid.

As chemists we are given experimental observations, such as the difference in measured pKa between ammonium and hydronium ions, and we invent concepts such as "electronegativity" to explain this result in conjunction with many other similar results. If this concept is insufficient in your mind to explain the experimental data, you may be motivated to seek a more satisfying and self-consistent explanation. For me electronegativity is sufficient in this case. The fact that ammonia is "stable" and can be put in a bottle does not mean that it is somehow unstable when protonated to give ammonium ion. Many ammonium salts can be put in a bottle, even salts of weak acids like carbonic. Having a positive charge on an atom is not in itself an unstable state-note how stable salts like sodium chloride (with a full positive charge on sodium) are, as well as species like tetramethylammonium ion-- you can make the hydroxide and put it in a bottle.


text:
Why does the nature and oxidation state of a central metal affect field split?
I know the following factors affect the value of field splitting in a coordination compound:

No. of ligands surrounding the central atom.
Strength of the ligand (as per spectrochemical series)
Nature of the metal ion
Oxidation State/charge on central atom

I have understood why the first two are contributing factors, but the 3rd and 4th points do not ring a bell.
For the 4th point, i.e., ox. state of central atom, my guess was that if there is a lower oxidation state present, then less number of electrons will be present in the orbitals of the central atom, and hence due to lesser repulsion between ligands, value of field splitting will decrease.
Am I correct in this?
As for the 3rd point, i.e., the nature of the metal ion, I am completely at a loss. What does nature specify here, and how does it affect the value of field splitting?

This Chemistry LibreTexts article gives a brief explanation on the different factors which affect $\Delta_o$.

Let me start this answer — as usually for an answer I give about coordination complexes — by showing the molecular orbital scheme of a typical octahedral complex ignoring $\ce{L\bond{->}M}$ π interactions.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.
It is easy to see where field splitting comes from: two of the ligand group orbitals transform as $\mathrm{e_g}$ as do two of the metal’s $\mathrm{d}$ orbitals ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$). These interact in a bonding/antibonding manner to give $\mathrm{e_g}$ (considered a ligand-centred orbital due to the electronegativity difference) and $\mathrm{e_g^*}$ (a metal-centred orbital).
It is important to note that this scheme depends strongly on the relative positions of the orbitals. The closer the energies of the ligand and metal group orbitals, the larget the split will be.
If you consider different aquacomplexes, you can always be sure that the ligand orbitals start off from the same energy (they are the same water, after all). However, the metal orbitals will be differently positioned. The higher the first ionisation enthalpy (typically corresponding to the removal of an electron from $\mathrm{4s}$ — here $\mathrm{a_{1g}}$) the lower these orbitals will be in this scheme. Loosely, the ionisation enthalpy depends on the number of protons in the nucleus, so we might expect nickel to have lower-lying orbitals than titanium. Whatever the actual difference is (there are many more effects), no two different metals will ever have the same orbital energies and therefore any two different metals will always have different field splits corresponding to a different colour. You know this: hexaaquachromium(III) is green while hexaaquairon(III) is yellow-orange.
The oxidation state influences the orbital scheme in a similar way. Each oxidation corresponds to the removal of an electron meaning less electronic repulsion meaning lower-lying orbitals (in slightly larger contexts, this corresponds to the contraction of cations). Thus, if you change the oxidation state of a metal, e.g. from $\mathrm{+II}$ to $\mathrm{+III}$, the metal orbitals will be lowered in energy and the field split will most likely increase since there is better overlap with the ligand orbitals. Therefore, iron(II) and iron(III) have different colours in solution.


text:
How to determine the potential of a cadmium zinc cell under standard conditions?
I write the half-equations:
\begin{align}
\ce{Zn &<=> Zn^2+ + 2e^-}& E^\circ &= +0.763~\pu{V}\\
\ce{Cd^2+  + 2e- &<=> Cd}& E^\circ &= +0.403~\pu{V}
\end{align}
In order to find the $E_\mathrm{cell}$ is it right to say:
$$E_\mathrm{cell} = 0.763~\pu{V} + 0.403~\pu{V} = 1.166~\pu{V}?$$
Or should I say $0.403~\pu{V} - 0.763~\pu{V}$?

0.763+0.403 = 1.166V is correct. To use -0.763 V you'd need to flip the reaction with Zn. Both cations can't be reduced obviously...

Woops! You have a mistake in the Cd half cell reaction. See Avi's answer.

The standard reduction potential of cd is -0.403V not +0.403V.

I will like to correct you that there is one problem with your question. The correct standard reduction potential is:
\begin{align}
\ce{Cd^2+  + 2e- &<=> Cd}& E^\circ &= -0.403~\pu{V}.
\end{align}
 And another thing is, you have used symbol $E^\circ$ for oxidation potential of $\ce{Zn}$ which should not be done. You can use there just $E = 0.7628~\pu{V}$ or in a better way, you can just write it with it's reduction potential value as:
\begin{align}
\ce{Zn^2+ + 2e- &<=> Zn}& E^\circ &= -0.7628~\pu{V}.
\end{align}
Now coming to your question, for these two reactions,
\begin{align}
\ce{Zn^2+ + 2e- &<=> Zn}& E^\circ &= -0.7628~\pu{V}\tag1\label{Zn}\\
\ce{Cd^2+ + 2e- &<=> Cd}& E^\circ &= -0.403~\pu{V}\tag2\label{Cd}
\end{align}
$E_\mathrm{cell}$ of any cell reaction is given as
$$E^\circ(\text{cell}) = E^\circ(\text{reduced species}) - E^\circ(\text{oxidized species})$$
or
$$E^\circ(\text{cell}) = E^\circ(\text{cathode}) - E^\circ(\text{annode}).$$
Here as $E^\circ$ value of reaction $\eqref{Cd}$ is greater (i.e. towards more positive). Hence $\ce{Cd}$  has greater tendency to accept electrons and get reduced. So the $\ce{Zn}$ is going to give electrons and gets oxidized. Now the half cell reaction will be,
\begin{align}
\ce{Zn &<=> Zn^2+ + 2e^-}\tag{at anode}\label{anode}\\
\ce{Cd^2+ + 2e- &<=> Cd}.\tag{at cathode}\label{cathode}
\end{align}
So according to formula
\begin{align}
E^\circ(\text{cell}) &= E^\circ(\text{cathode}) - E^\circ(\text{annode})\\
E^\circ(\text{cell}) &= -0.403~\pu{V} - (-0.7628~\pu{V}) = 0.3598~\pu{V}\\
\end{align}
The correct answer is $0.3598~\pu{V}$.

To answer this, lets get some basic things clear.

Reduction potential = - Oxidation potential
Oxidation means losing of electrons and reduction means gain of
electrons (OIL RIG)
And finally, Oxidation occurs at anode and reduction occurs at
cathode (CR AO)

Standard oxidation potential of Zn is +0.763V  and the standard oxidation of potential of Cd is 0.403V.
Since, the standard oxidation potential of Zn is more than that of Cd, Zn electrode will act as anode (where oxidation takes place) and Cd will act as cathode (where reduction takes place).
Ecell = (Eoxidation)anode - (Eoxidation)cathode
      = +0.763 - (0.403)
      = 0.36V

Alternatively,
Standard reduction potential of Cd electrode (-0.403V) is greater than that of Zn electrode (-0.763V) so Cd electrode will act as cathode and Zn electrode will act as anode.
Ecell = (Ereduction)cathode - (Ereduction)anode
      = -0.403 - (-0.763)
      = 0.36V

And one more important thing, If only electrode potential is mentioned then that electrode potential is to be assumed the reduction potential by default (not the oxidation potential).

The mnemonic I was taught for #3 was "AN OX, RED CAT". Also, just worth mentioning that electrode potentials are much more commonly quoted as reduction potentials - not that it invalidates anything you've said, of course. I'd just be hesitant to use oxidation potentials to calculate anything. Not because it's wrong, but just because it's possible to lead to confusion.


text:
During the decomposition of sodium bicarbonate lab, the mass of the final solid I received was less than expected. Errors?
The lab procedure that my class used can be found in the photo attachment.
The purpose of the lab was to determine the equation for the decomposition of sodium bicarbonate. The two possible equations were:

$\ce{2NaHCO -> Na2CO3 + CO2 + H2O}$
$\ce{2NaHCO -> Na2O + 2CO2 + H2O}$

I understand that the correct equation is supposed to be the first one where sodium carbonate was produced. However, through calculations using the ratio between reactant and solid product, my experiment showed that equation 2 where sodium oxide was produced is supposed to be correct.
I know that this is caused by the mass of product I produced which was 0.716 grams rather than the expected expected 1.187 grams. However, what I do not understand is what errors could have potentially lead to this less than expected mass?

You calculated your expected weight wrong. Edit your question to show that calculation.

Your equations contain an incorrect formula for sodium bicarbonate. If you used that in your calculations, it would give the wrong expected weight if product.

I understand that the correct equation is supposed to be the first one where sodium carbonate was produced. However, through calculations using the ratio between reactant and solid product, my experiment showed that equation 2 where sodium oxide was produced is supposed to be correct.

So now you're faced with a decision. You can either:

Massage the result of your experiment until it comes up with the answer you think is "supposed" to be correct.
Report the result of your experiment as it actually happened.

If you take the first route, I'm sure we can come up with some convincing explanations. Maybe your starting sodium bicarbonate contained water (so it wasn't 100% pure) and that water was driven off in the heating. Maybe you spilled some as you put it in the crucible. Maybe some of the mixture in the crucible splashed out during the reaction. Maybe you read the scale wrong. Maybe there is another reaction that also happens, perhaps involving the air, or the crucible, or the gas from the bunsen burner.
But I think it would be much more honest to report your result as it actually happened. Your data supports the hypothesis that the sodium bicarbonate decomposed to sodium oxide. In fact, the agreement with that hypothesis is quite strong, and the agreement with the supposedly correct hypothesis is terrible. It could be that this is due to an error you made... or maybe the stupid sodium bicarbonate didn't get the memo that, according to the teacher (or the text or Wikipedia or whatever), it was supposed to decompose to sodium carbonate, so it just went ahead and decomposed all the way to sodium oxide.
I realize you're probably worried about being marked down on your lab report because you came to the "wrong" conclusion. Any decent teacher, though, would give you much more credit for correctly interpreting your data and coming to a conclusion based on the data, even if it's wrong, than for massaging your data to support a preconceived hypothesis.

maybe the OP could also study the calculations to find the error.

We should use this answer as a go-to in case we encounter others asking how to explain unexpected results by fabrication.

The reactions would be better written as:

$\ce{2NaHCO3 ->[\Delta] Na2CO3 + CO2 ^ + H2O ^}$
$\ce{2NaHCO3 ->[\Delta] Na2O + 2CO2 ^ + H2O ^}$

$\ce{NaHCO3}$, MW = 84.01
$\ce{Na2CO3}$, MW = 105.99
$\ce{Na2O}$, MW = 61.98

$\ce{2NaHCO3 ->[\Delta] Na2CO3 + CO2 ^ + H2O ^}$

Starting with 2.0 grams $\ce{NaHCO3}$, the expected amount of $\ce{Na2CO3}$ is:
$g_{\ce{Na2CO3}} = \dfrac{1}{2}\cdot\dfrac{2.0}{84.01}\cdot105.99 = 1.26$

$\ce{2NaHCO3 ->[\Delta] Na2O + 2CO2 ^ + H2O ^}$

Starting with 2.0 grams $\ce{NaHCO3}$, the expected amount of $\ce{Na2O}$ is:
$g_{\ce{Na2CO3}} = \dfrac{1}{2}\cdot\dfrac{2.0}{84.01}\cdot61.98 = 0.74$

What you may have done was subtracted your final resulting weight of your crucible and Sodium Hydrogen Carbonate from the original weight of your Sodium Hydrogen Carbonate and Crucible before heating. This would have yielded an answer similar to the one you got. However, the correct procedure would be to subtract the weight of the original crucible from your final resulting weight of your Sodium Hydrogen Carbonate and Crucible. This would give you just the weight of how much Sodium Hydrogen Carbonate you have left, instead of your original mistake which only gives you the difference between your starting amount of Sodium Hydrogen Carbonate and your final amount.


text:
Will the reaction shift left or right?

The picture shows the correct answer.
At Eq., I have:

3 Mol of A
1 Mol of B
12 Mol of C

I was looking at each answer in terms of ratio of reactants to products, but its clearly not correct since I got 0.4/2.0. How should I think when approaching this problem?

Add your equation for the equilibrium to your question.


text:
Correction formula for alcoholmeter
I need to find the formula, if there is one, which would calculate the correction of alcoholmeter.
Basically convert observed %ABV (Alcohol by volume) to actual %ABV using the sample's temperature.
I know the correction tables, but I am looking for the formula or do I have to use tables?.

No doubt there is a formula. Not knowing what you are measuring or what table you are using, I can't find the formula for you.

So this is a table I am talking about
http://homedistiller.org/correctiontable.pdf
They even have online calculator for it
http://homedistiller.org/calcs/husker_temp_convert.php

and I am measuring  'fruit spirit' if it is called like that in english. I get distillate measure percante content  by volume of ethanol at some temperature, but then I need to correct that measurment to 20°C.

@MaxW Forget to mention you :]

You are probably better off using the tables.  Are you trying to put this formula into a spreadsheet or something?  The best you can probably do is plot a bunch of data from the table and come up with some high-order polynomial that fits it. Just be sure you understand the consequences of doing it that way (i.e. massive errors).  Take a look at the equations used to come up with that table.  Usually if there is a simple equation available, nobody would go through the trouble of making a table ;)

Well, I just noticed that the online calculator has the option of looking values up from the table or using an equation.  This is where I would start to dig.

I'm too lazy to dig for this. The technique uses a hydrometer which is probably calibrated to read %ABV at 20 C. So you need to find an equation for the density of an ethanol-water mixture as a function of ethanol concentration and temperature.

@airhuff - they make a table because calculating a polynomial expansion blows most folks mind. The computer program at the site no doubt uses an equation.  See: http://zd2.chem.uni.wroc.pl/pliki/alcohm.pdf

starting at http://web2.airmail.net/sgross/fermcalc/fermcalc_alcohol.html you should be able to find what you need...

@MaxW, probably shouldn't have even brought up the polynomial thing. That said, the problem I've seen is that with modern spreadsheets it's too easy for anyone to calculate up to a 9th order polynomial and say ~"look, I've fitted it with r^2 =0.9999" !

@MaxW, airhuff thank you both. I am trying to put this formula into computer program. I will probably try that formula at page 12, if it won't work I will map the data table somehow.

You'd have better luck using equation 7 on this webpage http://web2.airmail.net/sgross/fermcalc/fermcalc_alcohol.html

@MaxW I know it is more simple. But let's say I can not obtain brix value.
But I have solved the density equation. Is there a way how to get percentage content of ethanol by volume from that density value? Sorry I am really noob in this.

To convert between ABW (% by weight) and ABV (% by volume) either use Table 3b from OIML Alcoholometric Tables or formula provided there at the bottom of page 5: ABV = (ϱ(ABW)/789.24)*ABW where ϱ(ABW) is R-K function decribed on page 12 of ABW at 20°C

Not a formula, but for anyone looking for an up to date calculator that gives reasonable-sounding results (a lot of the links on the other comments and answer are dead) this page seems to do it: https://www.essentialdistilling.com/resources/alcohol-dilution-and-proofing-calculator It also has many other useful calculators or related tasks.

I have had the same question recently. And there is no simple answer. Tables (like ones from International critical tables of numerical data, physics, chemistry and technology or this) are still the 1st hand source. Formulas used are just approximating this data. Check formula here on page 8 (to get the temperature-corrected ABV you have to invert it by solving non-linear function). Alternatively check this recent study. To convert between ABV (alcohol by volume, %) and density (specific gravity SG, g/cm3) you may use this approximation:
$$\mathrm{SG} = -0.002\cdot\mathrm{ABV} + 1.0012$$
PS. There is an unconfirmed formula at vinolab:

The hydrometer temperature correction for SG is performed with this expression:
$$\text{corrected-reading} = r \times \frac{(1.00130346 - (0.000134722124 \cdot t) + (0.00000204052596 \cdot t^2)\ - (0.00000000232820948 \cdot t^3))} {(1.00130346 - (0.000134722124 \cdot c) + (0.00000204052596 \cdot c^2)\ - (0.00000000232820948 \cdot c^3))}$$
where: $r$ = reading and $c$ = calibration temperature. This expression is based on °F, so the temperatures are first converted.

With some sample values it gave me results close to the table' ones (I have converted ABV to SG, amended as per temperature and formula above and finally converted back to ABV).

I think it wouldn't hurt if you include the formulas/data from each report you are referring to. URLs tend to rot over time. Also, the last formula taken from the report written by a student leaving John Doe from the template on the front page doesn't look convincing. Also, a tip: you can add #page=<number> (e.g. #page=8 ) at the end of the PDF's URL so that others open the document on that very page.

Formulas there are quite complex for my markdown skill. I have checked John Doe's formula against this data. It gives 0.8% discrepancy at worst.


text:
Why can't a balance that can weigh to the nearest milligram be able to weigh 23.8 milligrams of something?
I'm having trouble understanding the answer to this question:

Most chemistry laboratories have balances that can weigh to the nearest milligram, would it be possible to weigh $5.64\times10^{18}$ molecules of octadecane, $\ce{C18H38}$, on such a balance?

I worked through the equation and my end result was $\pu{0.00238 g}$ which is equal to $\pu{2.38 mg}$ so I thought that the answer was that it could, but the book states that the answer is that it cannot. Why can't it? Am I misunderstanding how weighing to the nearest $\pu{mg}$ works?

significant figures. Weighing $5.6\cdot10^{18}$ molecules is not the same as weighing $5.64\cdot10^{18}$ molecules.

So  I made a rounding error?

The balance can only weigh 23 or 24 mg, not 23.8 mg.

Ah, I think I get it. So, it only measures in whole milligrams. Thanks for your help.

Accepting what is the "wrong" answer bugs me. Although I hope the point about significant figures has been made adequately, let me go into more detail.
QUESTION

Most chemistry laboratories have balances that can weigh to the nearest milligram, would it be possible to weigh $5.64×10^{18}$ molecules of octadecane, $\ce{C18H38}$, on such a balance?

Li Zhi correctly points out that octadecane has a molecular weight of 254.5 g/mole and that $5.64×10^{18}$ molecules would weigh 2.38 milligrams considering the significant figures.
Li Zhi makes the correct assessment when he says "It certainly cannot weigh that amount with adequate precision for most purposes I can think of."
Li Zhi is also correct when he says that "A balance with an ability to weigh to the nearest mg would, in a perfect world give you a mass of 2 or possibly 3 mg." But this is accuracy not precision.
Thus a balance has two factors to consider. Accuracy and precision. A balance which weighs to 1 mg has neither the accuracy nor the precision to weigh out 2.38 milligrams.
Accuracy: On average 2 mg will be 0.38 mg too low and 3 mg will be 0.62 mg too high. So either weight is biased.
Now if a chemist weighs out 2 mg 62% of the time and 3 mg 38% of the time, then on average the chemist weighs out 2.38 mg. (See how stupid this is?)
Let's look at the reverse too. On average 2 mg will contain $4.73\times10^{18}$ molecules and 3 mg will on average contain $7.10\times10^{18}$ molecules.
Precision: $5.64×10^{18}$ molecules of octadecane implies +/- 1 part in 564. If we assume the balance rounds, which is reasonable, then 2 mg is precise to 1 part in 4, not nearly enough.
The point here overall is that chemists make assumptions all the time to simply problems using significant figures. Using significant figures has to become innate to a chemist so that one can cut through the Gordian knot offered by so many chemistry problems.

Couldn't agree less, but check my math.
First $\ce{C18H38}$ has a molecular weight of $254.5~\mathrm{g/mol}$.
Second $\mathrm{5.64\times 10^{18} ÷ 6.02214 \times 10^{23} = 9.3654 \times 10^{-6}}$ and so is $2.4\ \mathrm{mg}$.
A balance with an ability to weigh to the nearest mg would, in a perfect world give you a mass of 2 or possibly 3 mg. This wouldn't be very useful, but it wouldn't give you zero information, either. So, since I don't understand the context in which your book asked the question, I'd have to say it doesn't seem to be completely correct if it states "it cannot" weigh that amount. It certainly cannot weigh that amount with adequate precision for most purposes I can think of. But if, say, you were trying to determine whether there was enough of a sample to use in a mass spec., or to calculate the yield of a reactor-on-a-chip, then it would suffice.

Your math is correct. The OP should have gotten 2.38 mg not 23.8. // We'll just have to agree to disagree on how significant figures impact the answer.

Thanks for the correction, I had a bit of a mix-up with my decimals when copying from my work to the screen it seems.

"Would it be possible to weight this amount?" should, I think, be interpreted as "Would it be possible to weigh out this amount?" It's not asking if the balance can give you an answer for the weight of a sample; rather, it's asking if the balance can be used to tell you how much stuff to take out of hte bottle to make 2.4mg.

I would like to expand upon the other answer a bit and claim that the balance indeed cannot weight 2.38 mg.
First of all, the question cannot be interpreted literally. The question must be interpreted as either:

Given this sample of $C_{18}H_{38}$ and this balance, can you conclude that it has $5.64 \times 10^{18}$ molecules.

or

Can you use this balance to make a sample that has $5.64 \times 10^{18}$ molecules.

Now, $5.64 \times 10^{18}$ has 3 significant figures. What it actually means is that you want between $(5.64\pm 0.01) \times 10^{18}$ or $(5.64 \pm 0.005) \times 10^{18}$ molecules. This converts to $2.38$ mg (I am simply assuming you did your math correctly), again with 3 significant digits.
Now, all measuring devices are living creature made and imperfect $^\text{[citation needed]}$. A measuring device that is accurate to the nearest unit could be off by up to that unit (or half, depending on who you ask.) For example, if you use a ruler accurate up to mm, then a measurement you make on that ruler can be off up to 1 or 0.5 mm, depending on who you ask. If you measure a line, and the line ends in between the 6mm notch and the 7mm notch, it does not make sense to say that the line is 6.5mm. The 6mm notch and the 7mm notch could be off up to a millimeter, after all. Maybe it depends on the season.
As such, you cannot use a measuring device that is accurate up to mg to conclude that something has 2.38 mg, or measure out 2.38 mg of something.
As an addendum, I would like to note that the question asks about a balance.

A balance's being accurate to the nearest mg could mean a couple of things.

One side is up to 1 mg heavier than the other side.
There is some friction in the balance, and it simply would not move, unless there is a difference of more than 1 mg between the two sides.
Where you put the items on the plate, and what the initial position of the balance is, could have an effect on the weighing, but no more than 1 mg worth.

A "balance" doesn't have to be of that form.  The question is probably about a microblance (google link) which would be a frictionless design.  Spring balances (wikipedia) may not be common in chemistry labs but are another use of the term.

+1 but the term "balance" is used to refer to any weighing scale.

FYI: The "balance" with the digital readout that you see in a chem lab probably does work on the same principle as the balance shown above, except that instead of comparing the unknown weight to a set of known masses, it compares the unknown weight to a force supplied by an electromagnet.  The readout is proportional to the amount of current that must flow in the magnet in order to exactly balance the unknown weight on the pan.

"The 6mm notch and the 7mm notch could be off up to a centimeter" Surely, you mean "millimeter", right? I really hope my millimeter precision ruler isn't off by a centimeter!

The significant figure factor is about precision not accuracy.


text:
Equilibrium Position
So if you had the equilibrium $\ce{A <--> B}$ and you increase the concentration of A. I understand that equilibrium will shift to the right. To say this (I believe) is to say that the $\ce{A -> B}$ reaction is favoured. But when at equilibrium again will the position of equilibrium be back in its original position, as I believe the position of equilibrium to be defined as the amount of A relative to B.
If this is true, does it follow that a shift in equilibrium does not necessarily mean a permanent change in the position?
If it was a $\ce{A + B <--> C}$ equilibrium and you increase the concentration of A the equilibrium will shift to the right. How would you determine if the position had changed at equilibrium?
Thanks very much.

Not knowing what the chemicals in your equation are or what kind of properties they have, I can't tell you how you would measure their concentrations.
Nonetheless, given:
$$\ce{A + B <--> C}$$
If you add an additional, preferably known, quantity of A, then the concentration of B should decrease and the concentration of C should increase.
This is actually just the real meaning behind your statement that "the equilibrium will shift to the right".  In this case, with two reactants, obviously more of the second one (B) must be consumed by the addition of more A in order to produce more C (shift to the right).
Again, how you would measure the concentrations of B and C depends on which compounds B and C actually represent.


text:
When are Kp and Kc values equal?

The picture shows the correct answers to this question.
I'm not sure why these answers are correct. $K_p$ only counts with gases, while $K_c$ only counts with aqueous solution + gases. I assumed that any reactions with only gases, or a mix of gases and solids, will have an equal $K_p$ and $K_c$ value, but that is clearly not the case.

Wrong definitions. Kc is uses molar concentrations, whereas Kp uses the partial pressures of the gasses

I was referring to the physical states that are involved in Kc and Kp.

They are equal when the sum of the stoichiometric coefficients of the products equals the sum of the S.C. for the reactants (assuming that you are treating the gas as ideal which I assume you are).

Disclaimer: The following information has been extracted from my own Chemistry lecture notes which was built on contributions from many other general chemistry textbooks.  It is a stereotypical example of relating $K_c$ and $K_p$.
$K_c$ and $K_p$ relationship
For a simple reaction involving gas phase substances
$$\ce{aA(g) <=> bB(g)}$$
The equilibrium constant can be written in terms of molar concentrations or partial pressures as
$$K_c = \frac{[\ce{B}]^b}{[\ce{A}]^a} \qquad\text{or}\qquad K_p = \frac{(P_\ce{B})^b}{(P_\ce{A})^a}$$
Assuming that the gas behaves ideally (a common and convenient assumption for general cases), we can write the ideal gas law for each gas such that
$$ PV = nRT \quad\to\quad P_\ce{A}V = n_\ce{A}RT \quad\to\quad P_\ce{A} = \left(\frac{n_\ce{A}}{V}\right)RT$$
$$ PV = nRT \quad\to\quad P_\ce{B}V = n_\ce{B}RT \quad\to\quad P_\ce{B} = \left(\frac{n_\ce{B}}{V}\right)RT$$
We note that
$$ \frac{n_A}{V} \equiv M $$
where $M$ is simply molarity (moles/unit volume).  Since $[\ce{X}] = M_x$, we get
$$P_\ce{A} = [\ce{A}]RT \qquad\text{and}\qquad P_\ce{B} = [\ce{B}]RT$$
Substitute these terms into the $K_p$ equation to give
$$K_p = \frac{(P_\ce{B})^b}{(P_\ce{A})^a} = \frac{([\ce{B}]RT)^b}{([\ce{A}]RT)^a}$$
and recognize that
$$\frac{[\ce{B}]^b}{[\ce{A}]^a} = K_c$$
Plug this relationship into the $K_p$ expression to give
$$K_p = K_c (RT)^{b-a} \quad\to\quad K_p = K_c (RT)^{\Delta n}$$
Where $\Delta n$ is simply the difference of the sum of the stoichiometric coefficients of the products and reactants from a balanced chemical equation like so
$$\Delta n = \sum (\mathrm{products}) - \sum (\mathrm{reactants})$$
It should be clear from this that the only way $K_c = K_p$ within the context of an ideal gas is when $\Delta n = 0$.
$$\begin{align*}
K_p = K_c(RT)^{\Delta n} \quad\to\quad K_p &= K_c(RT)^{0} \\
&= K_c (1) \\
&= K_c
\end{align*}$$
By visual inspection of your image, it is clear why these are the correct answers based upon the derived relationship offered here.

Very elegant proofing. Just a wandering question, is there any relationship been Kc, Kp and oxidation states of the reactants and products considered for the Kc, Kp? I wanted to shoot it as a question, but then I came across this. Looking at OP's chosen reaction stoichiometries, I get the feeling there is.

Kc=Kp when the mols of gas on the left side of the equation is equal to the number of mols of gas on the right side of the equation.

This is already mentioned in another answer which also elaborates on why this is the case and demonstrates the proof.


text:
Condensation Rate is Increased with a Decrease in Temperature?
I'm having trouble understanding why the rate of condensation is increased with a decrease in temperature. My reasoning is as follows:
If...
The rate of condensation (R) depends on the concentration of gas molecules (reactants), and it's defined as:
R = k[Gas]
And...
Any rate is proportionate with respect to temperature; as is k...
Then why would a decrease in temperature increase the rate of condensation (R)?
My deductive reasoning for this is as follows:

The problem in your question is the over-simplified statement that the rate of condensation is simply proportional to the concentration of the gas.  That is only part of the very complex picture here.
The primary key to condensation rate is the degree of supersaturation of the water vapor at the temperature of the object on which the water is condensing.  Under conditions of sub-saturation, i.e. the ambient water vapor concentration (vapor pressure) is less that the vapor pressure of water at the temperature of the object, no bulk (excluding molecular layers of chemisorbed water) condensation occurs.  This of course make your equation R = k[Gas] completely false under these conditions.
Once the partial pressure of water vapor exceeds the saturation vapor pressure at the temperature of the object, i.e. the relative humidity exceeds 100% (by decreasing the temperature), then bulk condensation of water can occur.  Because our atmosphere has an excess of water condensation nuclei, relative humidities of greater than 102% rarely occur.  So, at a given temperature, the condensation rate is roughly proportional to the partial pressure of water for relative humidities of 100-102% at the surface of the object.  Note that the relative humidity increases with decreased temperature. Under these specific conditions, your equation is approximately correct.  I say approximately because there are other factors like air movement (wind velocity), and kinetic limitations like the rate at which water diffuses across the nm or so layer of air at the surface of the object at which the wind velocity approaches zero.
Further complicating the issue is the temperature dependence of the water vapor concentration at 100% humidity.  In other words, there is much more water vapor in the air at $\mathrm{30^oC}$ and 100% humidity vs. $\mathrm{5^oC}$ and 100% humidity.  This means that the amount of water vapor available for condensation at any given degree of supersaturation, say 101% humidity, is a function of temperature.
Because of the complexity of this issue my answer has been largely qualitative in nature.  For a thorough discussion of this issue (albeit even more complex as it includes discussions of water vapor-to-ice condensation), see this article, "Supersaturation of Water Vapor in Clouds", Journal of Atmospheric Science, December 15 2003, p. 2957.

The answer is more complicated than I thought. Is there a way of proving this mathematically? I included my deductive reasoning in my original post.

I think I see the problem. Note that at lower temperatures, delta T is a larger negative because you are talking about departure from equilibrium. Does this sound right? Your deductive reasoning looks sound, with a couple caveats. The only condition placed on your system is "the change in temperature is not extremely high", which I think is there to address changes in heat capacity with temperature, not any of the conditions I described above. Also, I think the answer is "3", where evaporation rate decreases more than condensation rate decreases. Each process should be slower at lower temps.

Condensation is the phase transition from gas to liquid.  A condensing gas needs to release energy to form a fluid.  A low temperature more readily accepts the heat from the condensing gas, enabling condensation.
Edit.  Your reasoning in the image seems to ignore the evaporation is endothermic and therefore Delta h will  change sign during condensation.

The temperature of a gas is really just the average kinetic energy of all of the molecules. When you have a container, each gaseous molecule will be exhibiting a different magnitude of molecular vibration. When the average kinetic energy (heat) of the molecules decrease to a certain point, they will condense - go from a gas to a liquid. At lower temperatures, a greater proportion of these molecules will be at an energy level that is low enough to enter the liquid phase.
At a higher temperature, a greater proportion of molecules will have the required energy to become, or stay, a gas.


text:
How much energy can be produced by burning sewage gas?
I am an eighth grader. I need to make a science project on the production of electricity from sewage. I know that almost sixty percent of sewage gas is Methane and most of the other constituents are combustible. I wanted to know how much energy can be produced by combustion of sewage gas and what amount of it can be converted to mechanical and thermal energy and how much of it is lost.
I was thinking of burning the gas to heat water to steam which in turn would turn the blades of a turbine, which would generate the electricity. I don't know exactly how that would work and how efficient that would be, but I am working on it.

First of all, welcome to SE.Chemistry!  Your question is excellent and demonstrates that you have taken some time to learn about your topic (which is obviously a very important issue for our growing world).  Do you know how combustion of the gas or the sewage itself would be converted to mechanical or thermal energy and why some of it would be lost?  If you had some idea of these things, this would be an excellent addition to your question.  If you can learn these things, or take your best guess, I think you should edit your question to include those things.  Again, good job and good luck!

You need to decide by what kind of machine you want to burn the sewage gas, and then check the technical specifications. The absolute heating value of methane should also be simple to find.

@airhuff, I have edited the question to include what I was thinking. I know that some of the energy would be lost because of the second law of thermodynamics, and that's about as much as I understood about that.


text:
Which functional group is more polar in a drug: primary amide or a carboxylic acid?
There is a lot of speculation as to which functional group is more polar, I've read a lot around this on the web and I can't decide which one is actually more polar? Can someone help because I need to know how to identify the most polar groups in drugs to create some drug monographs.

Under physiological conditions the carboxylic acid will be deprotonated, which puts a negative charge on it. That might (or might not; I don't know) affect the conclusion you draw.

Meh, acids and amides are so similar they are hard to differentiate. I think the negative charge at physiological pH is your only real chance of a definite indicator. I’m not too good at all the pharmacological stuff, though.

If the primary amide is hydrolyzed (via an amidase) in vivo, then it's a moot point.

@Michael Nguyen. Did you perhaps mean to compare a carboxylic acid to a primary amine, not an amide? Just curious.

It depends on your exact definition of polarity
The measured dipole moments of formamide (3.7 D)1 and formic acid (1.4 D)2 in the gas phase indicate that the amide is intrinsically more polar.
However, this neglects additional factors relevant to the discussion: ionization and solvation. Under physiologic conditions, carboxylic acids are generally deprotonated resulting in a formal charge separation with a -1 charge delocalized across two oxygens. Depending on what parameters you are concerned with, this may be a more profound effect than the difference in gas phase polarity.

A carboxylic acid would be more polar.  Amide bonds are essentially how proteins are bonded together, and are the less-polar condensation product of an acid and and amine.  One way to think of it is, amino acids are more polar, and are converted to proteins, which are less polar.  Polarity may change with pH, but is the case at least for the only pH that you'd care about, which is that of the human body.

The primary amide is more polar. The nitrogen and oxygen are conjugated, giving the C-N bond partial double bpnd characteristics, meaning it forms stronger H bonds. Also the difference in electronegativities between oxygen and nitrogen mean it is more polarisable. These characteristics are reflected in the boiling point and dielectric constants of amides vs carboxylic acids.


text:
Dangerous exothermic reactions at home
Exothermic reactions are essentially reactions that release energy as heat (in part, at least) to the surroundings, and may cause an explosion or combustion, depending on the amount of heat released and the specific conditions (see this story as an example of an unexpected exothermic reaction at home).
We mix up all kinds of stuff at home, mostly in the kitchen, and each "ingredient" may be composed of numerous distinct chemical compounds. This huge number of possible combinations might make it probable that a redox reaction, which may be exothermic and there's no telling how much exothermic, will happen from time to time when mixing two (or more) foodstuffs together. However, it is pretty rare to hear about significant exothermic reactions that may be called "a household accident", such as the one I brought earlier.
My question is: how is this so? Is there any energy threshold that we do not normally cross in the kitchen? Or are the concentrations not large enough? How come kitchen chemistry is not more violent than it actually is? What makes "experimental" food chemistry safe for billions of people without chemistry knowledge?

Cool question.  My first though is that we've been cooking things since the stone age, and any recipe containing a violently exothermic, or otherwise deadly, product or intermediate, would not have made its way into modern non-kitchen-blowing-up cookbooks ;)

@airhuff, that sounds quite right, as evolution and natural selection can usually explain surviving trends and behaviors throughout history. But I don't want you to get off that easily... your comment may be true for basic foodstuffs and cooking methods that have been known since the beginning of recorded history, but what about all those new food products that are constantly being developed, usually by physical and chemical processes meant for longer preservation, better taste, etc. These have not yet been subjected to evolution, but are still experimentally safe... Thanks for your comment!

New products and methods can indeed be more violent. Try to boil an egg in a microwave, for instance.

Point absolutely taken, both Don_S and @IvanNeretin.  The world of cooking has changed as drastically as the rest of the world in the past century.

We oxidize things like crazy in the kitchen, on stoves, in all kinds of ovens, pressure cookers (ohh, there's a good one!), etc. until we pull out some hot, partially oxidized product.  I realize that oxidation is not at all the sole purpose of heating when cooking, but I wonder if there is any cooking or preparation technique for which the goal is to reduce something...

I don't exactly get the point here. We're using combustion all the time as source of heat and yes it's dangerous... If you really want to get into food itself, its burnt all to often if you're not careful, or want to use flambe technique.

@Mithoron, it isn't really about combustion. I was thinking more about the mixing of several ingredients prior to any physical process they are subjected to such as heating up to the boiling point, freezing etc. I was pointing out that the large variety of available compounds makes it probable that some household materials to create dangerous chemical reactions (not only in the kitchen; I gave an example in my original question and another common dangerous cleaning practice involves mixing chlorine-based bleaches with an acid.

Life on Earth happens in pretty tight chemical limits, except for some exotic organisms we rarely encounter.
So, on the human side, the stuff we want to eat is not really very reactive. Carbohydrates, fats and proteins will give you no great fireworks no matter how you mix the three of them. The trace elements we need would be capable of some spectacular reactions, but we don't eat them in their pure form.
Which brings me to the "what is available" side. for almost all of human history, humans ate what they found in nature. And the stuff that reacts violently when mixed with organic matter is not really lying around to be used as an ingredient. Sure, some natural events are capable of producing pure oxygen, fluorine, or whatever, but they don't lie around for a long time waiting to be harvested, they react away. And plants and animals (=our food) tend to evolve in a way such that their bodies use the nitrogen and sulfur to build amino acids, not TNT and sulfuric acid.
There are a few substances in the kitchen which can give you a bit of a bang when mixed. If you want to impress a second grader, such classics as baking soda + vinegar, or mentos + cola will do nicely. But stuff which tends to pack more energy, and likes to free it in a flash, is not really food. You will notice that you need advanced processing methods to get all these ingredients - even though some of them (vinegar fermentation) were known in antique times, they are not something that grows on trees.
You can certainly find some edge cases - I keep pure NaOH on hand for pretzels, it is also used for some styles of noodle - but they are rare and concern ingredients which are not willy-nilly mixed into random dishes.


text:
If an atom of any element gained energy, does it always mean that its electrons gain energy too?
My book says ;
"The transitioning of electrons in atoms from one energy level to another is associated with changes in the energy of the atom."
But it hasn't made it clear whether or not the gaining of energy (of any form) by an atom will mean that the electrons are now at a higher energy level than before.

The "atom" includes the electron. The energy is stored in the interaction of electron and atomic core. Unless you put too much energy in the atom, when they split and fly off, each taking part of the energy with it.

@Karl never mind i actually thought too much about it and that is why i confused myself

No, as atoms can for instance gain (or lose) kinetic energy as a whole and this does not influence the (pure) electronic states of the atom. The electrons may be excited to a higher energy level when the energy that is applied to the atoms (e.g. light or collisions with fast electrons) matches the energy gap between the current level and an excited level. For molecules, energy can also be trnsferred to rotation and/or vibration of the nuclei.
On a side note, atoms can also have excited nuclei (carbon has for instance an excited nuclear state around 7.4 MeV that is very important for the formation of carbon in the Universe).

you probably mean $7.4 MeV$ ?


text:
Why is the probability of finding an electron in an orbital not 100 %?
Here is what I know and please correct me if I am wrong:
An orbit can have 2 electrons only but electrons do not exist in defined orbits. An orbital is like a hollow sphere where the 2 electrons are most likely to be found. In other words the 2 electrons of an orbit are not confined to that pathway and can be found anywhere in the orbital.
Now my question is:
Why isnt the probability of finding the two electrons of an orbital $100~\%$? Most websites say the probability is $95~\%$, where else could they be and why?

There can be zero, one or two electrons but no more in an atomic energy level. The spatial extent  at position r is called the wavefunction or orbital $\psi(r)$. The chance of the finding  electrons  around the nucleus is usually plotted as a boundary surface inside which  there is a  probability of, say, 95%  probability density. The electrons can in theory extend anywhere but  the chance of finding the electron at a large distance is so vanishingly small that we use the boundary surface as a useful guide to express where the electrons are to be found.

@porphyrin you mean "...no more in an orbital", right? Energy levels can have 8+ electrons (n=2,3...). May be my mistake

No, just max of two per energy level , that is to say each electron has to have a unique set of quantum numbers (Pauli principle) . However, some energy levels are found to be degenerate, which is why, for example  6 electrons fill p orbitals, but there are three of them, so max of two in each. Each energy level has its own unique wavefunction or orbital.

There is a nonzero probability of finding an electron anywhere except for at the nodes, where the probability is 0 by definition. An orbital might be better thought of as an infinitely large cloud that is more dense in some areas than others. Integrated over all space, there is also by definition a 100% chance of finding an electron. This should make sense intuitively (if if doesn't exist somewhere then where is it and what does the integral even mean?).
Orbitals really aren't so well-defined as the nice little shapes we represent them with (think cloud), but we can make them well-defined by arbitrarily deciding how much space we wish to enclose. It is often decided for the practicality of the representation to show a space such that the probability of finding an electron in it is 95%, though mathematically it's true that there is a chance of finding it outside the space.

The concept of an orbital directly derives from the concept of a free rotator in simple quantum mechanics. Before students learn about a free rotator, they typically learn about the harmonic (potentially including the aharmonic) oscillator and — first of all — the particle in a box. These are the three probably most fundamental quantum mechanical concepts which can be used as a first basis to understand everything else.
The particle in a box is a somewhat idealised case because the box is defined to have walls of infinite potential energy. Only in that case is the particle actually confined to the box. All other fundamental concepts in principle have a wave function contribution from $-\infty$ to $+\infty$ in all relevant coordinate axes. Which means that there is a nonzero probability of finding the particle anywhere in the coordinate system.
For the free rotator — which in principle gives us hydrogen orbitals — this means we need to define an arbitrary cutoff point if we want to display anything at all. Because 5 is a nice number, most of the time people choose a cutoff value of $95~\%$ and draw a line around the entire area (or volume) inside of which the electron has a $95~\%$ chance of appearing. Depending on your application, you might want to choose $90~\%$ (for less crowding) or $99~\%$ (to show more interactions) instead.
$100~\%$ cannot be used since that would put everything into an orbital and therefore the concept of an orbital would lose its usefulness. In general, an electron belonging to an atom somewhere in New Zealand has a nonzero probability of being found right here where I am sitting in Germany — and likewise a nonzero probability of being found at the other end of the Andromeda galaxy. (Note that while they are nonzero, they average out to zero at quite a large number of significant digits.)


text:
Chemical breakdown of sunflower seed to determine the allergy-causing substance
I have a severe anaphylactic reaction if I ingest a sunflower seed, and carry an Epi-Pen at all times. What is it in the seed kernel that could be so problematic and is it going to also be present in sunflower oil and in the sunflower organic derivatives used so much now in lotions, shampoo, and many other products today? I try to avoid any food or product that states that it contains sunflower in any form but it is becoming more difficult to avoid, especially when many crackers and such state "may contain sunflower or canola or other" oils. I even fear putting lotion on that might contain it!

Personal medical questions are off limits on the site. This is a life threatening problem and you should certainly consult an allergist to discuss this problem. Don't solicit advice from us nitwits.

@MaxW, this seems to me like purely a question of curiosity, and nothing in the question suggests that the OP intends to take any action based on any answer.  The fact that she's had severe anaphylactic reactions in the past and carries medication to control it suggests she's seen / is seeing a doctor about this.  At least that's my take.  If this is the case (Theresa W. Moore please comment on this) do you think the personal medical question policy applies? The definition here is pretty brief.  Are you sure this violates that policy?

@airhuff yes I'm sure. I certainly don't want to offer an opinion if the chemical that is in the seed is not in the oil so the oil is safe. She should absolutely have that conversation with her allergist not us.

@MaxW, yeah, I guess any chance the OP would act on our advice is definitely just not worth it.


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How to sketch reaction coordinate plot from mechanisms?
I am interested in learning about strategies for drawing reaction coordinate plots from the reaction mechanism. The sketches I am after don't need to be highly accurate or precise, but should reflect any notable intermediate states. For example, how would I draw the reaction coordinate diagram of a nucleophilic substitution of an arene showing features like the formation of the sigma complex, and the return to aromaticity via depronation?

Usually these are calculated by finding a transition state then the two intrinsic reaction coordinates to get to the transition state.  The results of these two files are knitted together to give an intrinsic reaction coordinate for that part of the reaction.  The result can be viewed using jmol (or similar) and the energy of each frame from can be plotted ... see https://chemapps.stolaf.edu/jmol/jmol.php?source=https://ndownloader.figshare.com/files/1868269 and right click and play to see what I mean.