New answers tagged inorganic-chemistry
1
It is actually the two single-paired electrons are more likely to be promoted than the two paired electrons.
Promoting the electrons always cost energy whether the electrons in question are paired or not. In the classical examples of hybridization (where they might actually promote a paired electron), this energy cost is compensated by the formation of ...
0
If you ask for an "easy" way to determine if a ligand is a pi acceptor/donor or none, look at the spectrochemical series. A quick-and-dirty rule (which means that it is correct most of the time but not always) is that weak ligands (iodide, bromide, hydroxide etc) are pi-donor ligands. The medium ligands (water, ammonia etc) are pi-neutral, and the ...
0
My take is that the terms "inorganic benzene" and "inorganic graphite" are meant to relay thoughts of both structural and chemical likeness to organic counterparts; however, graphite is not an organic compound since it is not a compound. It is an allotrope of carbon, just like diamond, buckminsterfullerene and graphene.
The IUPAC ...
0
Molecularity applies to single elementary reaction steps and is seldom higher than 2.
The reaction order is the "effective molecularity" of the whole set of linked reactions, if it had been a single step reaction. It need not to be integer and has just indirect relation to molecularities of elementary steps.
The reaction order is either determined ...
3
In principle, the oxidation of gold is the half-reaction $$\ce{Au -> Au^+ + e^-}$$But as this oxidation is done in a cyanide medium, $\ce{Au^+}$ is included in a complex ion : $$\ce{Au + 2 CN^- -> Au(CN)_2^- + e^-}$$ The second half-equation is $$\ce{2H2O + O2 + 4 e^- -> 4 OH^-}$$ So the final equation describing the dissolution of gold in an ...
1
I think you have got the calculation right. However, I would also consider the coupling between F and H because coupling through 3 bonds is also usually visible. So, each of your 8 lines will be split into four by the H-atoms.
That will be the major part of the spectrum. You would also get weak signals with a different splitting pattern due to the other ...
0
In short, $\ce{HNO3}$ is an oxidizing acid, so the destruction of the oxide layer competes with formation of new oxide. Something like this:
$\ce{Al2O3 + 6H+ -> 2Al^3+ + 3H2O}$
$\ce{2Al + 2Al^3+ + 6NO3- -> 2Al2O3 + 6NO2}$ (<-- maybe incorrect)
From Wikipedia:
Although chromium (Cr), iron (Fe), and aluminium (Al) readily dissolve in dilute nitric ...
1
Lithium ion batteries are pretty poorly recycled as is and so when I was looking through literature reviews I found only one mention of lithium ion polymer batteries being recycled. The paper that was using this method was also recycling lithium ion batteries the same way though so it wasn't clear if anything different had to be done.
The main thing that ...
4
I would not call graphite organic, but there is no clear-cut way of defining organic and inorganic compounds.
To your question, the pairs of compounds are isoelectronic. That means that if we assume that molecular orbitals (MOs) arise from atomic orbitals, the corresponding MOs in the two compounds have the same occupancy. In these cases it arises because ...
1
During titration of small amounts of acids, the molar amount of the indicator in 1-2 drops of $\pu{1 \%}$ indicator solution may not be negligible compared to the acid molar amount, affecting the result.
So for that cases, $\pu{0.1 \%}$ solution is used, to be able to dose smaller indicator amounts.
As the phenolphalein molar mass is about $M=\pu{318 g/mol}$,...
2
There are two little clues you can use.
First, mercury(I) isn’t actually present as $\ce{Hg+}$; the cationic species is actually $\ce{Hg2^2+}$ with a $\ce{Hg-Hg}$ bond. Thus, if this were a mercury(I) compound, the sum formula would be $\ce{Hg2[Co(SCN)4]2}$.
This obviously only helps you if the correct formula was supplied as part of the question but the ...
9
As pointed out by Maurice and myself, the argument provided in the answer by iad22agp is rather incorrect. Sulfuric acid is a dehydrating agent just because it is available in concentrated form whereas rest of the common acids like HCl(aq) are not. No, it is not a concentration effect. First of all, it is not possible to have HCl(aq) more concentrated than ...
2
How can potassium hydroxide be prepared from available potassium chloride (KCl) and Quick Lime (CaO)
Mixing $\ce{KCl + CaO}$, or $\ce{KCl + Ca(OH)2}$, will never produce pure $\ce{KOH}$ without $\ce{Ca(OH)2}$. And this Ca(OH)2 will prevent soap from being synthesized out of oil, as $\ce{(Ca(OH)2}$s destroy soap in case a little bit of soap has been synthesized. The only way of producing $\ce{KOH}$ out of $\ce{CaO}$ or $\ce{Ca(OH)2}$ is to mix it with ...
6
A salt like $\ce{Na2SO4}$ is essential in electrolysis. It provides ions $\ce{Na+}$ and $\ce{SO4^{2-}}$ which are attracted by the electrodes in the solution and migrate to them. When they arrive near the electrodes, they are not discharged. But they neutralize the charges of the ions that are produced out of water being destroyed at these electrodes. Let's ...
0
How can potassium hydroxide be prepared from available potassium chloride (KCl) and Quick Lime (CaO)
isn't potassium carbonate available as fertilizer? it's a lot more common. Then it would be a simple matter of mixing CaO with water and adding k2co3. Eventually you'd be left with solid caco3 and koh in solution
2
Assuming that "term" in your question refers to chemical nomenclature, the authoritative source of information would be the current edition of IUPAC Red Book [1]. From [1, p. 70]:
IR-5.3.2.2 Monoatomic cations
The name of a monoatomic cation is that of the element with an appropriate charge number
appended in parentheses.
[…]
$\ce{I+}$ iodine(1+)
...
1
Be aware that in the solubility comparison context, solubility product constants can be directly compared for compounds with the same number of ions created from the formula, where a greater solubility product means a greater solubility.
For compounds with different ion counts, one has to compare ( molar ) solubilities in $\pu{[mol/L]}$ , calculated from ...
3
TL;DR: The mechanism is different for fluorine and chlorine atoms. Fluorine atoms react and forms stable HF molecules while chlorine atoms turns into a radical by the action of UV which helps in the destruction of ozone.
Long answer:
Ozone depleting substance(ODS) are gases that take part in ozone depletion process. Most of the ODS are primarily ...
-1
Here is a depiction of how 'chlorine' destroys ozone per an education source:
CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged ...
4
The solubility of $\ce{AgCl}$ is equal to $\sqrt{K_\mathrm{s}} = \pu{1.3E-5 M}.$
The solubility $s$ of $\ce{Ag2CO3}$ is such that $K_\mathrm{s} = 4s^3.$ So that its solubility $s$ is equal to $s = \pu{1.2E-4 M}.$ This is ten times more than the solubility of $\ce{AgCl}.$
For gravimetric purposes, $\ce{AgCl}$ is a better choice.
1
You can use reducing agents to reduce the $\ce{SeO3^2-}$ salt to elemental selenium. This paper1 discusses the use of iron(II) salts in acidic medium (phosphoric - hydrochloric acid) as reducing agent. The reaction proceeds at r.t.
You can use other reducing agents like hydrochloric acid, sulfur dioxide, hydroxylamine hydrochloride, hydrazine hydrochloride. ...
1
The relationship between ionisation energy and emission spectra is complex
You make the assumption (implicitly) that the colour you see in emission spectra is from the single complete ionisation of an electron from the highest energy electron orbital in lithium. But the colour you see is not that simple for two reasons.
One is that you only see visible light ...
4
I don't think it is a good idea to connect ionization energy with the color of flame emission, especially in a Bunsen burner.
Ionization energy means that you have separated the electron out of the nucleus attractive field. However, electronic transitions, corresponding to visible wavelengths, take place while the electron is still bound to the nucleus. Thus ...
4
From this question on Physics SE, there is a linked audio track in Soundcloud here, which is what you are looking for. The OP references a jar of hair-stying gel which, upon being tapped with a mallet, produces a "reverberating" sound instead an otherwise expected "short 'tok' sound."
Note the the ringing gels are common ingredients in ...
3
You may have confused barium peroxide with barium metal. The metal can indeed displace hydrogen from hypochlorous acid, not to mention (for a metal as reactive as barium) from the water in which the acid is dissolved; but metal peroxides will produce oxygen and water from the acid. So your missing product is water, not hydrogen.
Tricky half-reactions
If we ...
6
Your reasoning is correct. Perhaps the answer key is mistaken.
For part (a), in the reaction
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
water is donating a proton $(\ce{H+}$ ion) and hence is behaving as a Brønsted acid. Since all Brønsted acids are Lewis acids, water is behaving as a Lewis acid.
For part (d) as well, the half reactions are:
$$
\begin{align}
\...
0
I am not being rude, but honestly I think you should review the meaning of the spectrochemical series and how the transition metal complexes are colored.
A ligand being strong or weak according to the spectrochemical series does not correlate to the complex being stable or not, but its ability to split the d orbitals of the metal. In the spectrochemical ...
4
The fallacy is the assumption that "inner $d$ orbitals" become "available". Generally they are not, in the Group 1 and Group 2 metals. There are rare occasions where some evidence of $d$-orbital bonding is found for heavier G2 elements, but the impact is small and not widespread.
In Groups 1 and 2, where there are inner $d$ orbitals ...
-1
I guess that 1.081 refers to the distance length (in Angstrom) separating the C and N atoms. You obtain this value by applying by using the values listed in the table; this is quite tedious, takes some time and require quite good knowledge about crystallography. Otherwise (and much more easy), you can use the crystallographic data listed in the table to ...
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