New answers tagged

1

It might be a protein crystal. Proteins are large and the interstitial space is filled to a large extent with water. The volume of a protein crystal occupied by the solvent can be 50% or so. However the fraction of waters observed in crystals (relatively fixed) is often low, on the order of 10%. For instance, from a report on the crystal structure of crambin ...


3

Greg is quite correct: no clear question. But sometimes you don't know enough to define a question clearly. The implicit question is: "Is this mechanism helpful or even correct?" The second, clear, question, is why does rusting take time to happen. I think the answer to the first, implicit, question is that it is a start, but far from helpful and ...


2

A reaction's rate does depend upon the consumption of its reactants, and the manner in which the reactants interact. Let's consider the following reaction: $$\ce{A + B -> final products}$$ The rate of consumption of reactants $$\dfrac{-d[A]}{dt}= k[A][B]$$ and $$\dfrac{-d[B]}{dt}= k[A][B]$$ It seems that this reaction is a second order reaction since the ...


1

It depends on temperature. At higher temperatures, hypochlorite has tendency to disproportionate: $$\ce{3 ClO- -> 2 Cl- + ClO3-}$$ Otherwise it mostly releases atomic oxygen : $$\ce{ClO- -> Cl- + O}$$ In reality, it would be combination of both in various ratio, depending on relative reaction rates, all ending as mixture of sodium hydroxide, carbonate,...


0

$\ce{B2O3}$ - diboron trioxide (boron trioxide), prefix di means two and prefix tri means three. So, there are two boron atoms and three oxygen atoms in this compound. Boric anhydride. The word anhydride means anhydrous or waterless. This is the anhydride of boric acid ($\ce{H3BO3}$). $\ce{B2O3 + 3H2O -> 2H3BO3}$


6

I have no expertise, but Pourbaix does. Let's take a look at some of his diagrams which offer a hint to what is going on, with both silver and gold. Or more accurately, why gold is indeed less vulnerable to attack by hydrogen peroxide than silver. Begin with just the diagram for hydrogen and oxygen in aqueous solution. Ross[1] gives a diagram that ...


1

Potassium nitrate is transformed into potassium nitrite only when heated alone. And it is not really toxic, as it is added in ham to prevent its too early oxidation in air. When heated with charcoal, sulfur or any other oxydable substance, potassium nitrate is transformed into $\ce{K2CO3}$ or $\ce{K2S}$. For example, black powder is a mixture $\ce{KNO3 + C +...


0

I found this exact point discussed indirectly in JD LEE's inorganic chemistry 5th ed in page 110. In it, the author discusses the CO molecule system where for C it's on order and O it's another, the following point is noted in the last sentence to assume the order was of the lighter element (in this case, C). Then it is noted that if CO is is ionized to $CO^+...


2

I'm not a metallurgist or even not a good physical chemist. However, I'd like to explain what's happening in $\ce{AuCu}$ alloy with temperature and any expert can intervene. According to Ref.1, $\ce{Cu}$ and $\ce{Au}$ are both univalent group IB metals with the atomic size difference ~12%. The $\ce{AuCu}$ alloy forms the tetragonally distorted fcc lattice ...


1

If you are going to do anything differently than is normally done, you should understand what you are going to do and why it should work.(*) The essential (but often not sufficient) conditions for successful electroplating by any metal are: The used metal salts/compounds are soluble, so ions can migrate within the solution in applied electrostatic gradient. ...


1

No current needs to flow from an open circuit half cell. Well, no external current needs to flow. An equilibrium is established between the solution and the electrode. This equilibrium involves many transitions, back and forth, but is in equilibrium. There is a net current of zero. Now, how do you measure a voltage without getting a current? You back it up ...


3

Yes, it is possible to fully decompose cinnabar to elemental mercury and sulfur. What conditions need to be satisfied in the new method? Before reaching to decomposition temperature, cinnabar phase transitions to a different polymorph. Data from this paper1 indicate that α-HgS (trigonal) → β-HgS (cubic) phase transition occurs at 673 K and is completed at ...


3

Zero in science is not taken as rigorously as in mathematics. It is seldom exactly zero, but rather any value smaller than some significant or noticeable threshold. Scientific zero is in data processing context any value with its difference to mathematical zero being statistically not significant. In the question context, the zero current means a current ...


3

The answer is no ! If an electric currant is flowing through a piece of metal (platinum or anything else) the same number of electrons is getting in and getting out of this piece at both ends of the piece. No atoms are reduced or oxidized. The total number of electrons present in the metallic piece does not change. There is no oxidation and no reduction in ...


3

Reduction and oxidation are defined as gaining and donating electrons from the valence shell of an atom respectively. Whereas in the case of electrons flowing in the electrodes and connecting wires the electrons are passing from the interatomic voids and not being released by the wires and the electrode. There will be some electrons which will interact with ...


3

All group 13 elements (can technically) form nitrides from the direct combination of the elements (elemental reaction). Boron: See @Oscar's answer. Also, see references(1,2) Aluminum: There is a paper(3) which discussed about combustion of aluminum in a high-temperature and high pressure (up to 300 MPa) nitrogen atmosphere to form aluminum nitride. Gallium: ...


1

In H2O, as O is more electronegative than hydrogen, the resultant bond dipole is towards O, which means both the lone pair and bond pair dipole are acting in the same direction and the dipole moment of H2O is high. (1.84 D) On the other hand, In the case of F2O, the bond dipole is acting towards fluorine, so in F2O the lone pair and bond pair dipole are ...


22

The simple answer is that metals which form sufficiently insoluble oxide layers stifle the very process which caused the oxide layer in the first place. It's not just a matter of energy of formation of the oxide, but whether the first reaction product formed will be soluble or insoluble, then whether it will be adherent or non-adherent. In the case of the ...


2

This was answered in the comments, but these second equilibrium expression should not actually be a $K_b$ value. Rather, it should be $1/K_a = 55600$. Using this value in the same calculations, we find that $x=5.5805\times 10^{-4}$ and $y = 5.5795\times 10^{-4}$, so that $\ce{pH} = -\log(x-y) = 7.00$.


0

I think: $$\frac{1}{d^{2}} =\frac{\frac{h^{2}\sin^{2} \alpha }{a^{2}} +\frac{k^{2}\sin^{2} \beta }{b^{2}} +\frac{l^{2}\sin^{2} \gamma }{c^{2}} +\frac{2hk}{ab}(\cos \alpha \cos \beta -\cos \gamma ) +\frac{2kl}{bc}(\cos \beta \cos \gamma -\cos \alpha ) +\frac{2lh}{ac}(\cos \gamma \cos \alpha -\cos \beta )}{1+2\cos \alpha \cos \beta \cos \gamma -\cos^{2} \alpha ...


4

As Ed V points out, the conditions of formation of struvite have many variables and the struvite crystals produced are variable. On the one hand, this could be a pleasant exercise of playing with atomic models, but on the other, it may be unproductive unless you were to use computer graphing to model various configurations. Perhaps you could start with a ...


1

As suggested by @porpyrin, in the second calculation you're assuming the force constant $k$ is constant in both molecules so a ratio can be taken. Let $\tilde{\nu_H}$ and $\tilde{\nu_D}$ be the hydrogen and deuterated versions respectively, $$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$ $$\frac{\tilde{\nu_H}}{\tilde{\nu_D}} = \frac{\sqrt{k/\mu_H}}{\...


1

No, adding acid to a ferrous sulfate solution will not prevent its being oxidized by bromine water. Fresh FeSO$_4$ solutions are light blue-green, almost colorless, but pick up O$_2$ readily: Fe$^3$$^+$ + e$^-$ --> Fe$^2$$^+$ +0.770 V and O$_2$ + 2 H$_2$O + 4e$^-$ --> 4 OH$^-$ +0.401 V (The half cell voltages are only indicative of what's ...


3

I don't know which equations you are using, but this is the one I was taught: $$E(v)=hc\tilde{\nu}\left( v+\frac{1}{2}\right) -hc\tilde{\nu}\chi\left(v+\frac{1}{2}\right)^2.\tag{1}$$ Here, $\tilde{\nu}$ is the wavenumber of the vibration; $v$ is the quantum number that represents the vibrational levels $(v= 0, 1, 2,\ldots);$ $\chi$ is the anharmonicity ...


3

In the case of boron, a reaction with nitrogen does occur under some eye-popping conditions: Combustion of boron powder in nitrogen plasma at 5500 °C yields ultrafine boron nitride used for lubricants and toners.[1] The characteristic molecular thermal energy $kT$ at this temperature is roughly half an electron volt, which is enough to at least partially ...


3

There are three kind of oxides/hydroxides: basic (e.g., $\ce{NaOH}$, metal hydroxides), acidic (e.g., $\ce{CO2}$, non-metal oxides), and amphoteric (e.g., $\ce{Al(OH)3}$, Group 13 and 14 hydroxides). Amphoteric Hydroxides act either as Bronsted-Lowry bases (accepting protons) or as Lewis acids (accepting an electron pair), depending on reaction conditions. ...


3

OP's main question: Why would substitution of $\ce{H}$ with $\ce{D}$ alter chemical shift? OP has correctly indicated that the $\ce{Sn-D}$ bond will be shorter/stronger than $\ce{Sn-H}$ due to the lower ZPE, which may cause the chemical shift difference. That is the major reason for isotope effect in chemical kinetics mainly due to bond strength (shorter ...


1

Assuming it is given that $\ce{X2O5}$ behaves like an ideal gas, the best approach to do this problem is manipulating $pv = nRT$ to get the equation, which fits the given data. Suppose $m$ is the mass of the gas, $M$ is the molar mass of the gas, and $\rho$ is the density of the gas. Then $n = \frac{m}{M}$: $$ pv = nRT = \frac{m}{M}RT \ \Rightarrow \ M = \...


3

A buffer is a mixture of a weak acid with the conjugate weak base. Is a mixture of $\ce{NH4OH}$ and $\ce{NH4NO3}$ a buffer solution? I have seen some examples with $\ce{NH4OH}$ and $\ce{NH4Cl}$ but never $\ce{NH4NO3}$. Both mixtures contain $\ce{NH4+}$, a weak acid. Where is the conjugate weak base? You can either write $\ce{NH4OH}$ more conventionally (or ...


-1

If you are not sure about the use of the parameter Q, go back to Nernst's law for each electrode, namely $$\ce{E_{M^{2+}/M} = \pu{ E°_{M^{2+}/M}} + \pu{0.0296 V} \times \log[M^{2+}]}$$ Here the redox potentials of the two half-cells are, according to Nernst's law :$$\ce{E_{Cu} = \pu{+ 0.339 V} + \pu{0.0296 V} \times \log0.10} = \pu{0.309 V}$$ $$\ce{E_{Fe} ...


2

Whether you say if $\ce{F2}$ hybridises or not you reach the same answer - the molecule is linear. As indicated in the comments, hybridisation isn't an accurate reflection of how bonding actually works, however it is a good enough approximation to determine bond angles for simple molecules like ammonia. Note: to add some detail on your comment of the lone ...


5

Reactions can be of order $0, 1 , 2$ or $3$ , but never more. No collision may happen between more than three particules simultaneously. If an equation is written with more than $4$ substances on the lefthand side, it means that the corresponding reaction is a sum of several elementary reactions. Mechanisms in inorganic reactions are hard to establish ...


0

The OP asked the question again, and proposed the following solution from his book. My textbook says to solve it in the following way: 1) Ignore the 0.10 M KI since it doesn't matter. 2) Since you have excess AgCl(s), you can calculate the Ag+ concentration using its 𝐾sp. 𝐾sp=1.77×10−10=𝑥^2 𝑥=1.3×10^−5 This is the Ag+ concentration 3) Substitute ...


0

Silver iodide is not hygroscopic. It is used to seed clouds, because when the atmosphere is saturated with water vapor, liquid water should start condensing. But to do this, it needs a seed, a place when the first molecules will start to condense. No water molecule feels that it should be the first one to show other molecules where to be condensed. But, as ...


2

I decided to give a help here since OP has given enough effort to solve this problem. First, OP needs to know the molarity of a solution is defined as the amount of solute (in this case $\ce{NaBr}$) in $\pu{mol}$ per $\pu{1.0 L}$ of solution. The first part of problem states: If $\pu{2.60 g}$ of $\ce{NaBr}$ is dissolved in enough water to make $\pu{160.0 mL}...


2

It comes to be $\pu{252 mL}$ precisely: $$V = \frac{\pu{2.6 g}}{\pu{103 g}} \times \frac{1}{0.1} \times \pu{1000 mL} = \pu{252.4 mL}$$ $\text{Molar Mass of NaBr} = \pu{103 g mol-1}$.


1

Hypothetical temperature of starch melting or even boiling is much higher than temperature of starch pyrolysis, not speaking about critical values. Additionally, starch is mixture of molecules of various lengths and branching. Therefore, melting, boiling nor critical parameters would not have been defined, even if starch had been thermally stable.


1

The anion $\ce{AuCl4−}$ is the most known complex of gold(III). It is very stable complex and have very high formation constant ($K_f$). That is the reason the metal chloride complex have a lower reduction potential than metal aqua ion. As pointed out in the other answer, this is a consequence of Nernst's law. Let's look at the two redox equations we ...


8

The authoritative source Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (Red Book) lists $\ce{H[AuCl4]}$ as an example of a salt in the subsection IR-4.4.3.4 Generalized salt formulae [1, pp. 61–62]. Further, introduction to the section IR-8 Inorganic Acids and Derivatives underlines that IUPAC nomenclature is established from composition ...


-1

This difference of redox potentials is a consequence of Nernst's law. The second equation is nothing else as the first one, where the concentration of the $\ce{Au^{3+}}$ in a solution of $\ce{AuCl4^{-}}$ is extremely low. This sort of comparison could be used to calculate the equilibrium constant of the equilibrium $$\ce{Au^{3+} + 4 Cl- <-> AuCl4^{-}}$$...


2

I think the experts here have muddled the problem by stating "no equilibriums," so I'll solve it. Given the reaction: $$\ce{KI(aq) + AgCl(s) <=>[excess AgCl(s)] KCl(aq) + AgI(s)}$$ there are equilibriums. There are always equilibriums in chemistry. In this case it is much better to think of the reaction as proceeding quantitatively. That is ...


3

What is meant by a "very strong" Be−O bond? If berylium's tendency to hold on to water ligands is unusually strong, is it due to its small ionic size? You got that absolutely right. $\ce{Be\bond{-}O}$ is a strong bond because of the small size of Be. Smaller cation size means a stronger pull on the $\ce{O}$ electrons, thus reducing the bond length ...


0

I'm assuming we can neglect differences due to radioactivity/radioisotopes in biological systems. Hydrogen vs Deuterium The kinetic isotope effect - the difference in reaction rates and reactivity for different isotopes of the same element - is well known for hydrogen and its two stable isotopes protium and deuterium. For example levels of deuteration above ...


0

The oxidation numbers cannot be predicted from the structure of the atom. They are experimentally determined. Not much chance you predict that while nitrogen can get all integer oxidation numbers from $−3$ to $+5$, its neighbor oxygen can only have $−2$, $−1$, $0$, $+1$, $+2$ in known compounds. And then come fractional oxidation states...


3

This question requires idea of the Linear Combination of Atomic Orbitals (LCAO). It involves the following equation: $\psi_n = \sum\limits_{i} c_{ni}\phi_{i} = c_{n1}\phi_{1} + c_{n2}\phi_{2} + ... c_{ni}\phi_{i}$ Here $\psi_n$ represents the wavefunction for the resulting molecular orbital, $\phi_{i}$ represents the atomic orbitals that contribute to the ...


1

First, you should be given $K_\mathrm{a}$ of the weak acid. If you know it, you can set up the expression for $K_\mathrm{a}$: $$\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}= K_\mathrm{a}$$ Let $x$ be the amount of the weak acid dissociate at equilibrium. Thus, at equilibrium, $[\ce{H+}] = [\ce{A-}] = x$ and $[\ce{HA}]$ will be its initial concentration minus $x$. ...


1

The ICH guidelines give some details about the presence of excipients, and the forms they may take (along with acceptable types/classes of excipients); “The excipients chosen, their concentration, and the characteristics that can influence the drug product performance (e.g., stability, bioavailability) or manufacturability should be discussed relative to the ...


-3

All kinds of weird molecules exist in space. The molecule is formally known as bis(diazo)methane according to chemdraw. By looking at the structure it contains a Diazomethane moiety. Resonance would also take place to introduce the charges at both nitrogen atoms. The structure doesn't seem to exist on any chemical database nor any official papers have been ...


0

The everything OP and orthocresol discussed in the comment session is in this reference book (Ref.1). I include its PDF file (Ref.1) for benefits of OP and future readers. An example of coupling is depicted in the following figure (from Ref.2): This spectrum is qualified as other inorganic molecules similar to $\ce{S=PF2H}$ molecule. What I want to show is ...


0

So a Coordination compound contains a complex ion and necessary counter ions. For your example,[Mn(H2O)6]$^{3+}$ is a complex ion while for example [Mn(H2O)6]Cl$_3$ would be the coordination compound, Cl being the counter ion to balance the charge.


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