New answers tagged inorganic-chemistry
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One error is the presence of an acid source which results in moving the below equilibrium (written in the reverse of the formation reaction) to the right which occurs slowly with water for many peroxides:
$\ce{ CaO2 + H2O <=> CaO + H2O2}$
Even if you performed the indicated reaction above with H2O2 acting on CaO, you still should check the pH of the ...
2
Here,
The oxidation number of chlorine in the perchlorate radical is +7 whereas, in the product side, chlorine has an oxidation number of minus 1 in AlCl3. As the oxidation number is reduced by 8 here (gaining 8 electrons), chlorine is, indeed, reduced.
Nitrogen has an oxidation state of minus 3 in NH4ClO4 whereas, it has an oxidation state of 0 in N2 ...
0
In $\ce{BF3}$ there is backbonding from $\ce{F},$ which increases the double bond character of the $\ce{B-F}$ bond. There is no backbonding in $\ce{BF4-}.$ This causes shorter bond length in $\ce{BF3}$ compared to $\ce{BF4-}.$
1
Wikipedia states that the industrial preparation of calcium peroxide is done with calcium hydroxide, not the acetate or a neutral salt. Compare this choice with those used commercially for magnesium peroxide and lithium peroxide.
You want to do these syntheses with strong bases because the peroxide itself is most stable in basic conditions. Neutral or ...
1
What you seem to be asking about is a two phase system, a solid phase in equilibrium with ions in solution. For such a system there are two reactions occurring at the same rate. Thus some of the solid is constantly dissolving and some of the ions from solution depositing at the same rate.
There is a named process, Oswald ripening, that describes the ...
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Brute force would be really good if it were also clever. An article on using a large spark generator (Ref 1) discussed controlled experiments and used to data to estimate the amount of nitrogen oxides produced by lightning. I suppose cleverness could be added if a proper surface (i.e., catalyst) could be found, or if a proper reaction could be designed.
The ...
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You can use PVD or CVD method. You will get a nano coating of Ni oxides. In PVD chamber you can heat Ni steel with $\ce{N2 + O2}$ or $\ce{Ar + O2}.$
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I know there exist industrial methods for that already, but im poking
around to see if maybe there's a way to do it under atmospheric
pressure without tailored molecules like enzymes or metalorgnic
complexes
Though Haber process which was used to industrially produce ammonia was well-known, some researchers have also attempted to split nitrogen molecules ...
1
The article provided by Karl has answered the question with electroplating method, which is not preferred but this is the only answer I have.
The method is based on MIL-P-18317. The electroplating solution consists of $\ce{Ni2SO4, Ni2(NH4)2SO4, ZnSO4}$ and $\ce{NaSCN}$
The black nickle coating are $\ce{NiS}$ and $\ce{ZnS}$.
$\ce{NiS}$ and $\ce{ZnS}$ are ...
4
Coordinate bonds are covalent bonds where both bond electrons stem from only one of the two bond partners. They are formed when a Lewis base donates two electrons into accepting orbitals of the Lewis acid. That's not what you're looking at in this case, where both $\ce{P}$ and $\ce{O}$ contribute one electron to the common $\sigma$ bond. The interesting part ...
1
It is impossible to be definitive with such a simple picture.
But, if we assume the colour code is black for carbon and that hydrogens have been omitted, the simplest answer is (to use a non-systematic name) tri-isopropyl methane. See this (very basic) description for systematic names.
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Probably a matter of availability/cost. Magnesium oxide is easily obtained by calcining the hydroxide or carbonate and can be kept in its oxide state, whereas calcium and heavier alkaline earth metals the anhydrous oxide us much harder to obtain, partly because the hydroxides require higher temperatures to valvoline in the first place. Thus for the heavier ...
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Why does Cu react with HNO3? The short answer is that copper is a transition metal (like Fe, Mn, Cr,..) that engages in REDOX reaction with an oxidizing agent. The latter here is HNO3 and the reaction is usually accompanied by radical formations.
Some possible chemistry which may help.
First, from "Fenton chemistry in biology and medicine" by Josef ...
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$\ce{Cu}$ reacts with $\ce{HNO3}$, though $\ce{Cu}$ cannot replace $\ce{H}$ due to its less reactivity.
The reaction that takes place is not similar to a metal displacement reaction $\ce{(CuSO4 + Zn -> ZnSO4 +Cu)}$ which is where what you state applies (taking the electropositivity series into consideration).
According to conditions, there are two ...
2
This should be a comment, but I decided to post it as an answer, since from my point of view, the present answers fail to shed light on the absolute simplicity of the truth.
We can state from the present answers that there will be no precipitates, but not because the concentration is smaller than an atom per liter, we don't know the volume used since it was ...
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Your intuition is spot on : the issue is with the definition of atomic radius, which according to that same Wikipedia page "is not a well-defined physical entity, for which there are various non-equivalent definitions". This ambiguity finds its roots in the quantum mechanical description of the atom.
If you use the column with the "calculated&...
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I would leave a comment, but the basics of the answer are as follows. It's hard to give an answer that's complete because the chemistry is very diverse for the different compounds.
What you're basically seeing is the transition corresponding to breaking the metal-sulfur bond (i.e. the bonding orbital/antibonding orbital) or electrons moving between the metal ...
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Learning the unknown sample is called analysis and is a subject of analytical chemistry.
But your question is probably not about this.
You mean "can i guess something without knowing chemistry at all?"
And answer is yes!
Look at the food salt.
Do you know it without any knowledge of chemistry? Yes, you do!
But how much?
You know what it looks like,
...
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The band gap between conduction band and the valence band plays a role in color of these compounds. The good example is $\ce{SnS2}$, which is bronze color (Image $\bf{A}$; $\ce{SnS2}$ is used in decorative coating where it is known as mosaic gold) and have a band gap of $\pu{2.18-2.44 eV}$ (Image $\bf{B}$; Ref.1). When an electron in the conduction band ...
1
TL;DR Yes, the reaction is feasible at $\pu{60 ^\circ C}$. Just extending my comment into an answer.
Goethite and ferrihydrite(not ferrihydrate) are two different things. They are polymorphs of iron oxyhydroxide ($\ce{FeOOH}$). Goethite is $\ce{α-FeO(OH)}$ while ferrihydrite is amorphous/nanocrystalline hydrated iron oxide, officially formulated as $\ce{...
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The reasoning you utilized for comparing the bond angles of $\ce{NH3}$, $\ce{NF3}$, and $\ce{NCl3}$, while sensible, is simplistic and works best only for a few cases where weighing the factors of steric repulsion against bond pair-bond pair repulsions is feasible. It seems that you have encountered this very problem in the case of comparing the bond angles ...
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The reasoning provided in the book conforms with the computational data and hence can be presumed correct. As the $\ce{C-X}$ bond lengths of increases, the position of electrons moves away from the $\ce{CO}$ (carbonyl electrons), and hence they are repelled less. Another proven method for the claim can be using the bent's rule provided by Yusuf Hasan. To ...
1
Hydrolysis of magnesium boride has previously been attempted1. Two hydrolysis products were observed. The main hydrolysis product was found to be $\ce{Mg3B2(OH)6}$ but at $\pu{-10 ^\circ C}$, a different product is formed whose composition was found to be $\ce{H3B2(MgOH)3}$ which immediately converts back to the main product if the temperature were slightly ...
1
It is always easy to balance redox equations by considering two half reactions. The equation to be balanced is:
$$\ce{Sn + HNO3 -> Sn(NO3)2 + NH4NO3 + H2O}$$
Thus, the two relevant half reactions to this redox equation are:
$$\ce{Sn (s) -> Sn^2+ (aq)} \\
\ce{NO3- (aq) -> NH4+ (aq)}$$
First balance elements other than $\ce{O}$ and $\ce{H}$. In this ...
-2
While balancing,
$$\ce{Sn + HNO_{3} \longrightarrow Sn(NO_{3})_{2} + NH_{4}NO_{3} + H_{2}O}$$
first balance oxidising agent and reducing agent only, we will get $\ce{4Sn}$ and $\ce{1HNO_{3}}$ , after balancing them, write number of extra moles of acid obtained while balancing other atoms. That will give us the answer.
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I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements
To start off, please go through my answer to this question: Why is chromate stable in basic medium and dichromate stable in acidic medium? to read about the ...
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Tin lined cans are commonly used in processing light colored fruit as the tin is preferentially oxidized by residual oxygen left in the product and the headspace of the can, headspace gasses can be reduced by steam flushing immediately before seaming. If the oxygen levels are high enough a black line will be found at the juncture of the product and the ...
1
First of all, your reaction is wrong. $\ce{HPO3}$ does not exist as a single entity. It is actually a cylic molecule of formula $\ce{(HPO3)_n}$. The simplest such compound is $\ce{(HPO3)3}$ which is called trimetaphosphoric acid. See this answer for more details.
So, what happens is that phosphorus pentoxide is very powerful dehydrating agent and can rip ...
3
The overall reaction is:
$$\ce{H2SO4 + Na2Cr2O7 → 2 CrO3 + Na2SO4 + H2O}$$
How to separate chromium trioxide and sodium sulfate from solution?
Since chromium trioxide is soluble, it exist as chromic acid in the solution. So, in the reaction pot, there are compounds like $\ce{H2CrO4, H2Cr2O7}$ along with our target compound $\ce{CrO3}$, all in equilibrium. ...
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Decomposition of lead(IV) oxide at different temperatures forms different types of oxides. At around $\pu{600 ^\circ C}$, lead(II) oxide forms. Wikipedia adopts the scheme of lead(IV) oxide decomposition in air summarized by White and Roy [1]:
$$\ce{PbO2 ->[\pu{293 ^\circ C}] Pb12O19 ->[\pu{351 ^\circ C}] Pb12O17 ->[\pu{375 ^\circ C}] Pb3O4 ->[\...
5
Lead dioxide can be heated to 600 °C and it will thermally decompose into PbO and O2. This likely creates lead fumes but this may be one way to accomplish the task.
Optionally, you could dissolve the lead dioxide in acid then reduce it from a salt.
Welcome to this community, but please use care. Be careful with the DIY approach to doing chemistry when it ...
4
I am afraid that no one knows why! The superficial rationalizations such as HSAB and ionic size etc are good for decorating general chemistry textbooks.
As scientists, we should accept the fact that all questions which a human mind can generate do not have answers. Unfortunately, as science students, we are trained that every question must have an answer at ...
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Why $\ce{HgS}$ and not $\ce{HgO}$?
Because of HSAB. F'x has already answered a previous question of why mercury has propensity towards thiols(or in general sulfur). Quoting the relevant information:
In Pearson's HSAB theory (hard and soft acids and bases), the reason
the $\ce{S-Hg}$ bond is be stronger than the $\ce{O-Hg}$ can be
explained because $\ce{S^2-...
4
Before going into details, let me state Wikipedia:
$\ce{^{208}PoF6}$ was probably successfully synthesised via the same
reaction (passing fluorine gas over $\ce{^{208}Po}$) in 1960 where a
volatile polonium fluoride was produced, but it was not fully
characterized before it underwent radiolysis and decomposed to
polonium tetrafluoride.
Even attempts were ...
2
I decided to do some more research for this question as I had faced it before. Turns out the proper name of this ligand is diethylenetriaminepentaacetate ($\ce{DTPA^5-}$). It's an expanded version of EDTA and is primarily used in MRI scanning where it forms complexes with the Gadolinium ion. It forms complexes with a coordination number of six and seven with ...
0
The grade for conservation is rather irrelevant. But is very relevant for the analysis, so it should be p.a. grade ( per analysis, try Sigma or Merck ) or similar. Farmaceutical grade could be good, but there should be run blank tests for pure water samples ( or pure urine ). Or making parallel analysis for old and new fluoride batches.
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In fact How I can limit the reaction of H2C2O4 and KOH?
If you are to react potassium hydroxide with oxalic acid in 1:1 mole ratio, you will get potassium hydrogenoxalate ($\ce{KHC2O4}$). Its hydrate form exist at a specific temperature range. This salt contains the hydrogenoxalate anion and hence known as acid potassium oxalate. But if you add an excess of ...
2
Hydrolysis is a reaction with water, producing two new ions. Doubly charged ions, like $\ce{Ca^{2+}}$ or $\ce{Mg^{2+}}$ may react with water and produce complex ions like :$$\ce{Ca^{2+} + H2O -> [Ca(OH)]^+ + H+}$$ $$\ce{Mg^{2+} + H2O -> [Mg(OH)]^+ + H+}$$ These complex ions are well known in chemistry. Their existences explain why poorly soluble ...
0
It is surprising that Silicon halides behave differently with water according to the choice of the halogen fluor or chlorine. The following reactions are known to happen : $$\ce{SiO_2 + 4 HF -> SiF_4 + 2 H_2O} ......... (1)$$
$$\ce{SiCl_4 + 2 H_2O -> SiO_2 + 4 HCl}........ (2)$$ These equations are exactly the opposite from one another. Why does ...
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This can be attributed to the formal charges of fulminate ion and cyanate ion. Structural stability of a compound sometimes depend on formal charge.
We observe that formal charges of fulminate ion is higher than cyanate ion for which fulminate ions are said to have a less favorable molecular configurations which leads to facile decomposition of the ...
1
As the other answer and comments imply, potassium superoxide is not a good reagent for absorbing carbon monoxide. Cuprous chloride solution works better, as in this study[1] where the salt is dissolved in an ammonia-bearing solution.
Reference:
R. V. Gholap and R. V. Chaudhari, "Absorption of carbon monoxide with reversible reaction in cuprous ...
4
In this question, the equations given are:
$$
\begin{align}
\tag{1} \ce{Ag+(aq) + e- &-> Ag} &\qquad E^\circ = \pu{+0.80 V}\\
\tag{2} \ce{Ag^2+(aq) + e- &-> Ag+} &\qquad E^\circ = \pu{+2.00 V}
\end{align}
$$
Now, since we know that $\Delta G$ is additive, we can use this property to proceed. (This is also the proof for why $E_\mathrm{...
5
It is convenient to solve problems like that with a Latimer diagram, which is a great tool for predicting conditions for the reactions of disproportionation and synproportionation.
A generic Latimer diagram
$$\ce{A ->[$E_1$] B ->[$E_2$] C}$$
posesses the following properties:
If $E_2 > E_1,$ then $\ce{B}$ is thermodynamically unstable and ...
3
This carbon they have reported (considering the amount present and without knowing the sample preparation) is likely adventitious carbon.
This carbon is on every surface and so perhaps the glass was not cleaned well before the measurement. Even then it is likely some carbon will still be present.
Adventitious carbon (binding energy 284.8 or 285.0 eV) is ...
1
Her is a source: THE MECHANISM OF IRON CATALYSIS IN CERTAIN OXIDATIONS that confirms Maurice's reported observations:
It is well known that ferrous iron is comparatively stable in acid solution and that it is rapidly oxidized to the ferric state by the oxygen of the air in alkaline solution.
On the question of why, I quote another work: Fe2+ adsorption on ...
1
First, let me state the standard formula of hydroxyapaptite:
The term "apatite" applies to a group of compounds with a general
formula in the form $\ce{M10(XO4)6Z2}$, where $\ce{M^2+}$ is a metal
and species $\ce{XO4^3-}$ and $\ce{Z-}$ are anions. The particular
name of each apatite depends on the elements or radicals M, X and Z.
In these terms, ...
2
It now reads:
The effect can also be clearly seen in the dissociation constants of the oxoacids of chlorine. The effect is much larger than could be explained by the negative charge being shared among a larger number of oxygen atoms, which would lead to a difference in pKa of log10(1⁄4) = –0.6 between hypochlorous acid and perchloric acid. As the oxidation ...
4
Please do not edit the Wikipedia article, as your wording is incorrect, and the existing wording, while awkward, is correct.
What the article is trying to say is that the electron density on each oxygen is less as the oxidation state of the Cl increases. We can understand this by imagining progressive oxidation starting with $\ce{HClO}$. Here, the Cl has ...
1
@ Paras Khosla. Your first equation is correct, but not your second one : $\ce{Cu_2O}$ reacts with $\ce{C}$ without getting decomposed previously into $\ce{Cu + O_2}$. $\ce{Cu_2O}$ is more stable than $\ce{CuO}$ above $1000$°C
Furthermore, $\ce{Cu_2O}$ does not react easily with Carbon in the solid state. The contact between the reacting solids is not good. ...
3
The $K_\mathrm{sp}$ of $\ce{Fe(OH)3}$ is varied from source to source, but a reliable source gives the value of $2.79 \times 10^{-39}$ at $\pu{25 ^\circ C}$. We'll use this value throughout the calculations.
Suppose, $s$ amount of $\ce{Fe(OH)3}$ dissolves some in water according to its $K_\mathrm{sp}$, but assume water is not ionized:
$$\ce{Fe(OH)3(s) <=&...
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