New answers tagged

-3

First of all it’s basicity is not 3 rather it’s PH value is 3. PH 3 doesn’t mean it will give off 3 H+ ions, who told you this, PH is found by taking the log that’s why we get such a small number so that we can talk about it easily. An acid is strong if it acts as acid in a reaction with the other acids, see to quote an example if you react Phosphoric acid ...


3

Your solution is correct up to the point you assumed that you can double the concentration of hydrochloric acid. Unfortunately, this is wrong and not at all how stoichiometry works. Let's focus on what's important. In the nutshell, we are dealing with a typical neutralization reaction: $$\ce{H3O+ + OH- <=> 2 H2O}$$ and note that $\ce{BaCl2}$ as a ...


0

On a microscopic level, you still have «a forward» reaction described by $\ce{AgNO3 + NaCl -> NaNO3 + AgCl}$ and a «backward reaction», which may be described by $\ce{AgCl + NaNO3 -> AgNO3 + NaCl}$. However, reaching the thermodynamic equilibrium of these two reactions is heavily influenced by the low solubility product of $\ce{AgCl}$ at about $\...


4

The inorganic crystal structure database (ICSD), by specifying that the structure contains only a single element, you can reduce the number of hits to 55. Structure examples are Cr, Po, Mn, Sb, Se, P, Ca, Li, As. Some are specific phases occurring at different temperature or pressures. So Po is not the ONLY metal to have a simple ...


3

Firstly a correction and general clarification: your first general assertion is not true. The metal surface acts as a heterogenous catalysts (supported by paper linked later), and the kinetics can be sufficiently modeled by the Langmuir-Hinshelwood model for a monomolecular reaction in which the rate determining step is the surface reaction $$ \ce{A_{ad} -&...


0

Your balance equation is correct except for not simplifying it. Simplified equation should be: $$\ce{2Al + 2NaOH + 6H2O ⟶ 2Na[Al(OH)4] + 3H2}$$ I also think you're getting a little bit confused about your equation. You probably think reacting $\ce{Al}$ and $\ce{NaOH}$ with same mole ratio (2:2) make them two limiting reagents. If that is the case, you are ...


-1

I think you're getting a bit confused with your equation. It's probably better to write two distinct reactions, one with $\ce{NaOH}$ and one with $\ce{Al}$. 35% of $\pu{500g}$ is $\pu{165g}$ $\ce{NaOH}$, and 2.5% of $\pu{500g}$ is $\pu{12.5g}$ $\ce{Al}$.


2

The smoky black stuff is probably not tantalum oxide. The only such oxide that's well characterized is $\ce{Ta2O5}$, which is white. Tantalum itself is a rare element, so if the jewelry contains tantalum at all it's probably a thin coating or cladding on a base metal substrate.


0

So basically, you're comparing a soft base and a hard base. In HSAB theory, a soft acid like Pd2+, would preferably react with a soft base like R-PH2. Soft acids and bases are easily polarized because of their size. Although, the possibility of the ligand acting as a bidentate can't be ignored, the P to Pd2+ interaction is thermodynamically favored than N to ...


1

A metal phosphine complex, dichlorobis(triphenylphosphine)nickel(II) exists in two isomeric forms (blue and red) based on how you prepared it (Wikipedia). The blue isomer, which is paramagnetic can be prepared by treating hydrated nickel chloride ($\ce{NiCl2.6H2O}$) with triphenylphosphine in alcohols or glacial acetic acid. When allowed to crystallize from ...


3

It can be prepared using hydrogen fluoride, in a double displacement reaction: $\ce{SbCl5 + 5 HF -> SbF5 + 5 HCl}$ However, hydrogen holds onto fluorine more tightly than does antimony in $\ce{SbF5}$. Consider the reverse of your proposed reaction, which is preferred energetically: $\ce{SbF5 + H2 -> SbF3 + 2HF}$


0

In $\ce {SbF3}$, antimony has 3+ oxidation number. Oxidizing it with $\ce {F2}$ yields $\ce {SbF5}$ and antimony has 5+ oxidation number. Then adding $\ce{H2F2}$ just results in transfer of one fluoride ion, $\ce {F-}$, to $\ce {SbF5}$, resulting in $\ce {SbF6-}$ and $\ce {H2F+}$. There is no evident redox reaction if antimony trifluoride is simply mixed ...


-2

Referring to the replies of Mr. Ivan Neretin, complexes that has a form of M(AB)2C2 posses 8 stereoisomers, 2 of which are optically inactive. The other 6 isomers can be interchanged to their respective enantiomers.


2

Disclaimer: I realize that I am not really answering the question of color change, but perhaps this perspective adds something nonetheless. To pinpoint the origin of a mineral's color is not an easy task, I think. Some minerals "inherently have color", in the sense that the pure mineral has a color. Other minerals rely on impurities in their crystal ...


2

First of all, why do you think your copper is not oxidized? Each copper has a charge of +II while $\ce{OH^-}$ and $\ce{PO_4^3-}$ add up to -4. For all further discussions we need to look deeper into the crystal structure. The compound crystallizes with two crystallographically destinguishable copper ions. One forms a $\ce{[Cu(PO_4)_4(OH)_2]^12-}$ ...


-1

Maybe Titanium also? TiCl2 + 2H2O → TiO2 + HCl + H2 2TiCl3 + 4H2O → 2TiO2 + 6HCl + H2 TiCl4 + 2 H2O → TiO2 + 4 HCl TiO2 is insoluble Not sure how it compares to silver but it may be better than lead and mercury.


0

I know that chromium(III) reduces to chromium(II), but I'm not sure about the reaction. Would it be $$\ce{CrCl3·6H2O + Zn -> Cr(H2O)6Cl2 + ZnCl + H2}?$$ Your concept is right and I am glad nobody taught you the concept of nascent hydrogen. It is an obsolete idea of the 1920-40s. Please don't use it anywhere. Wikipedia has a nice summary https://en....


4

When electrons come back down they emit the same wavelength of light and that's the colour we see. I think this is a major misconception held by students. The color of these transition metal complexes is due to the absorption of light in the visible region. Very crudely, imagine you have six colors in the rainbow, if you remove a certain color portion from ...


3

Aluminum chlorohydrate (ACH) is a group of specific aluminum salts having the general formula $\ce{Al_nCl_{(3n-m)}(OH)_m}$ (Wikipedia). It is used in cosmetics as an antiperspirant and as a coagulant in water purification. Specifically in wastewater treatment processes, ACH is used as a coagulant to remove dissolved organic matter and colloidal particles ...


1

The Cu2+ and Ni2+ complexes, while having the same ligands, have different amounts of d-electrons (8 for Ni and 9 for Cu), meaning their ∆o value is going to be different. For both complexes the t2g is completely filled meaning the ∆o only increases based on the electrons in the eg. This is why Cu2+ will absorb higher energy light (orange light) in order ...


1

Nitrates tend to lower the temperature on dissolving, having a positive enthalpy change of solution. For example, $\ce{NH4NO3}$ has an enthalpy of solution in water of ~26 kJ/mol at ~300 K, and it is used in self-cooling cold-packs for that reason. Sorry, I could not locate the enthalpy change of solution for $\ce{Fe(NO3)3}$, so this is not a definitive ...


2

Another option could be a ion selective electrode (ISE) selective for chlorides, but it would make sense rather for regular checks, as there would be high one time investment. The principle is the same as for the potential reference electrode ... $$\ce{Ag | AgCl | KCl}$$ ... but this time with mono or polycrystallic $\ce{AgCl}$ exposed to the solution. ...


-1

Basic strength of different hydrides of gp-15 decreases as we go down ↓ the group. i,e. NH↓³ > PH↓³ > AsH↓³ > SbH↓³ > BiH↓³ Reason:- As we go down the group ___ 1. The size of the central atom of different hydrides of gp- 15 increases with the increase in atomic number. Therfore, the electron density decreases due to which electron donating capacity will ...


1

I'm unsure the level which this question is being asked, but hopefully this answers your question. As you pointed out "why did Oxygen take 2 electrons from magnesium" well if you would allow me to be rather 'hand-wavy' about the whole thing the reason is due to a activation energy vs ionization energy and bond-dissociation energy. If you stick some fresh ...


3

The only well known method to check for (the absence of) chloride ions is precipitation as AgCl (solubility 2 mg/l at 20°C). Mercury(I) chloride is similarly badly soluble, but toxic (and unstable, disproportionation) and therefore bad practice. Lead(II) chloride is already far more soluble (4 g/l) than AgCl, and practically all other simple chlorides are ...


0

Corrosion is a more or less involuntary/undesired reaction. If you apply a current, the corrosion is defined by the current you apply, so you are controlling the system, not just observing it. In general, electrochemical forcing puts your system into a highly unnatural regime, but can certainly provide interesting information. Corrosion can be affected (in ...


3

Your compound has only one proper axis of rotation and it is $C_2$ as shown in the diagram. Therefore, it is the principle axis of rotation. By definition, $\sigma_\mathrm{h}$ is a plane of symmetry perpendicular to the principle axis of rotation. However, the plain you were talking about is paralleled to the principle axis of rotation. Thus, it is not a $\...


3

If a metal is made to react with sulfuric acid, it will form a sulfate salt. When you dissolve those sulfates in water, you would be studying the color of the aquated metal ion. Sulfate is just a counter ion in each case. Search crystal field theory. You are right, iron (II) sulfate in water would be light green, Zn would be colorless, Mn would be light ...


0

The basic types of chemical bonds are learned already at school: atomic bond (single bond, double bond, triple bond), ionic bonding, atomic bond with partial ionic character, metallic bonding. The first three mentioned types depend on the difference of electronegativity difference. Single bond, double bond and triple bond depend on the outer electrons (...


1

The dot represents an unpaired electron. It's written that way as a reduced Lewis dot diagram. The reduction works like this: Start with the typical dot-diagrams for Nitrogen and Oxygen atoms, i.e.$$ \begin{array}{ccc} \textbf{Nitrogen} & \qquad & \textbf{Oxygen} \\[-10px] {\Huge{\begin{array}{rcl} & \cdot \phantom{\cdot} \\[-50px] \...


4

Nitric oxide ($\ce{NO}$) is a free radical and hence why that dot is for. Explanation for the dot and the reason why its there have been well answered in other responses. Nevertheless, I decided to put the molecular orbital representation of $\ce{NO}$ as depicted below for your convenience (Ref.1): There are three electrons in antibonding orbitals and ...


5

It is not necessary, but optional, to express explicitly the radical status. In other cases, like alkyl radicals, the dot marking is mandatory, not to be confused e.g. with a functional group. For curiosity, the ground state of the oxygen molecule - triplet oxygen - is a biradical, with 2 unpaired electrons. What we write as $\ce{O=O}$ is singlet oxygen, ...


11

Usually the dot is put there to emphasize that the nitric oxide is a free radical that includes an unpaired electron. This is especially notable by comparison with $\ce{NO^+}$, which does not have the unpaired electron. Note that the nomenclature $\ce{·NO}$ should not be rendered as showing the unpaired electron on nitrogen. The unpaired electron is ...


7

According to this Wikipedia article, there has been military research into nano-thermite, which can contain molybdenum, bismuth or tungsten oxides. It would be possible to test for residues of those after the fire. Also, "nano-thermite" sounds awesome!


14

Typical military-grade thermite mixtures (e.g. used in incendiary hand grenades) contain a large amount of barium nitrate (e.g. 29.0 % in Thermate-TH3, see Patent No. US 6766744). It should be possible to detect the barium oxide that is left over after the fire, maybe first with a simple flame test on site followed by AAS (atomic absorption spectroscopy) in ...


0

You don't need to know the original amount of the substance because the $K_\mathrm{sp}$ will give you the saturated concentrations of the ions present. For $\ce{Co(OH)3},$ the $K_\mathrm{sp}$ is $\pu{1.6e-44}$ which means the saturated concentration of the solution is $\pu{4.93e-12 M}$ for $\ce{Co^3+}$ and $\pu{9.86e-12 M}$ for $\ce{OH-}.$ You can ...


2

Many nuclear power plants use so-called "passive autocatalytic recombiners" for the catalytic oxidization of hydrogen that could be released into the containment in the event of severe accidents. (Note that the affected units of the Fukushima-Daiichi plant did not have such equipment.) The catalytically active materials, typically platinum and/or palladium, ...


1

It seems to me that it will be difficult to get a uniform concentration of Zn in a non-uniform material like dried leaves, unless you measure only in bulk (this pile of dried leaves weighs X grams and contains 2000X micrograms of Zn). Things get more confusing if the insects don't eat all the dried-leaf structure (leaving stems, for example). I'm also not ...


3

This is a nice example: the standard reduction potentials are close, so the Nernst equation ultimately reveals what is favored to happen spontaneously, ignoring all real world complications. Consider a galvanic cell having a tin $(\ce{Sn})$ metal electrode in $x~\pu{M}$ aqueous $\ce{Sn(NO3)2}$ and a lead $(\ce{Pb})$ metal electrode in $y~\pu{M}$ aqueous $\...


1

Actually, if you balance given redox reaction strictly following the half-reactions method where you have to equate the number of transferred electrons, you won't end up with fractional coefficients: $$ \begin{align} \ce{\overset{+7}{Mn}O4- + 8 H+ + 5 e- &→ \overset{+2}{Mn}^2+ + 4 H2O} &|\cdot 2 \tag{red}\\ \ce{2 \overset{-1}{Cl}^- &→ \overset{0}...


2

The balanced equation seems correct. Multiplying by $2$ we get the following: $$\ce{2 MnO4- + 16 H+ + 10 Cl- → 2 Mn^2+ + 5 Cl2 + 8 H2O}$$ which is also correct.


1

If you want a general solution for complex mixtures, where simplified formulas do not apply, you may need to involve numerical mathematics. As solving the equation set may lead to polynomial equation of higher order you may like. Enumerate equations of the known. First, charge balance. Then, mass balances. Finally, dissociation equilibriums. Involve a ...


0

Can you please elaborate on that part? I know that the E value of Mno4- is +1.51, and the one for BrO3- is +1.48. What does that suggest?? that Mno4- will more likely to be reduced by Fe+ than Bro3-?? please explain The first and foremost hint, even before you calculate the electrode potentials is the solubility rule taught in general chemistry. ...


1

According to the Globally Harmonised System (GHS), hydrochloric acid with a concentration of c ≥ 25 % and sulfuric acid with a concentration of c ≥ 15 % are labelled "causes severe skin burns and eye damage" (H314) and are classified based on standard animal test data in skin corrosion Sub-category 1B ("Corrosive responses in at least one animal following ...


3

The reactions are ongoing this way: Relatively free electrons of potassium reduce water: $$\ce{2 e- + 2 H2O -> H2 + 2 OH-}\tag{1}$$ That leaves metal positively charged. Liquid ammonia, if exposed to alkali metal, reacts with electrons much slower than water, forming a dark blue solution of solvated electrons. As electrons progressively kick out ...


1

It's because of the mechanism of catalysis, they both react with hydrogen peroxide as follows. $$\ce{2KMnO4 + 3H2O2 => 3O2 + 2MnO2 + 2KOH + 2H2O}$$ $$\ce{KI + H2O2 => KIO + H2O}$$ $$\ce{KIO + H2O2 => KI + H2O + O2}$$ The most important bit is that in the first equation the peroxide is actually a reducing agent and the permanganate doesn't actually ...


0

The coordinate covalent bonds are still counted as 1 electron, since the pair is shared between both atoms. That means the Carbon will have a formal charge of -1 and Oxygen a formal charge of +1 on Carbon Monoxide. Carbon FC: $\ce{(4 valence electrons on neutral atom) - (1*2e- from lone pair + 3*1e- from covalent bonds) = -1}$ Oxygen FC: $\ce{(6 valence ...


4

Yes, concentrated strong acid can destroy litmus paper. Paper is mostly cellulose, which is a carbohydrate polymer. Concentrated sulfuric acid will pull the hydrogen and oxygen out of it as water, leaving carbon behind. The indicator dye that's supposed to change color probably won't fare too well, either. You can dilute the unknown by adding a drop or two ...


0

Polishing with Cerium oxide, electrolytic cleaning, or diluted acid solution (nitric, hydrochloric, orthophosphoric, or hot acetic acid) are recommended in the Q&A on ResearchGate.net: "How to remove native oxide growth on copper or nickel foil?".


0

If you want to ruin both reagents, then you can go ahead and mix them. Hydrogen peroxide oxidizes glycerol to various reagents. A possible reaction pathway can be depicted as follows (Ref.1): This reaction is catalyzed by transition metals and other substances (Ref.1-3). References: G. Wu, X. Wang, T. Jiang, Q. Lin, “Selective Oxidation of Glycerol with 3%...


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