# Tag Info

3

Let R be the ratio of number of Na atoms to number of Fe atoms. Then, for the first compound, we have $$\pu{R = }\frac{\pu{112300 ppm } /\ \pu{22.98977 g mol^{-1} } }{\pu{357200 ppm } / \ \pu{55.845 g mol^{-1} } } = \pu{0.7637 }$$ Since there are two Fe atoms in the compound, the Na number for the first compound is simply 1.527, i.e., twice the value of R ...

1

It seems to me, according to the comments on the question, that there are NO additional reactions or products to worry about. However, because of the foam, some of the original reactants can be tossed up, and find their way into the later parts of the apparatus as a haze/fog/cloud that is not completely inert. Is that more-or-less accurate? I'm trying to ...

6

It is true that acid-base indicators ($\mathrm{pH}$-indicators) are either weak acid or weak base. What I understood reading your question is that you have wrong impression about bases. To clear your view, not all bases contain $\ce{OH-}$ ions. Specially, most organic bases are weak and contain electronegative ion (e.g., $\ce{N}$ or $\ce{O}$) with at least ...

2

The indicators are never in the form "InOH" with a covalence between "In" and OH, producing $\ce{OH^-}$ ions in water. If a molecule contains one OH group attached to a Carbon atom, it would be an acid, an alcohol, en enol or a phenol. These sorts of molecules are never releasing $\ce{OH^-}$ ions in water. On the contrary, they are weak acids.

5

A basic indicator is usually a Broensted-Lawry base ( accepting protons ) rather than Arrhenius base ( releasing hydroxide ions ): $$\ce{B + H2O <=> BH+ + OH-}$$ or $$\ce{BH+ + H2O <=> B + H3O+}$$ depending on if the indicator is used in its base form or conjugate acid form (the latter is usually more stable and more soluble). Regardless of ...

1

The original equation cannot be solved using mathematics because it gives the coefficient of $\ce{XeO3}$ zero. The reason for that is exact reason given by Ivan Neretin (see else where). Yet, I'd like to show a way of solving a balancing problem of true chemical equation by using mathematics. First, assume the compound is $\ce{XeF4}$, which undergoes ...

5

Actually there is no unique solution. You have four linearly independent equations and six compounds. In a typical redox reaction you need one less equation than the number of compounds. When the difference is greater than one, you have a combination of multiple reactions. We can get unique numbers only by assuming that just one oxidation product is ...

4

I like how Ed V has explained the mathematical solution logically. I'd like to show another way of solving the problem. The equation is as written by OP: $$\ce{a XeF4 + b H2O -> c XeO3 + d Xe + e O2 + f HF \tag 1}$$ So atom balance of equation $(1)$ yields: $$\text{Balancing } \ce{F}: \quad 4a = f \tag 2$$ $$\text{Balancing } \ce{H}: \quad 2b = ... 7 As per the OP's request, here is a solution. The equation is:$$\ce{a XeF4 + b H2O -> c XeO3 + d Xe + e O2 + f HF \tag 1}$$So atom balance yields:$$4a=1f \tag 22b=1f \tag 31a=1c + 1d \tag 41b = 3c + 2e \tag 5$$The equations relate the six unknown coefficients a through f. But each coefficient is an integer ... 3 Based on available theoretical considerations and a available literature, a tetrahedral geometry appears to be a good assumption for \ce{Cr(NO)4}. The NO ligand can be rendered as a three-electron donor, if we take it as uncharged. It is actually similar to CO, really, in that it interacts through its \pi^* orbitals as well as its \sigma orbital; ... 4 Electrolysis or electrolytic refining is a technique used for extraction as well as purification of metals obtained by other refining methods. In the electrolytic refining process, a block or a strip of impure metal is used as the anode, and a thin sheet of pure metal is used as the cathode. Both the cathode and anode are dipped in an electrolytic cell ... 1 Maurice notes that alkali metal diatomic molecules are at best a very minor constituent. But alkali metals do fill their s valence shells in other ways. Most alkali metals are known to form alkalides, in which an electron is transferred from one alkali metal to another (or from barium to an alkali metal, in a more recent example). To effect this ... 4 After doing some digging I've found evidence to support that tin based inter-metallic compounds should be named 'stannide' such as in the paper by Fickenscher et al. [1]. Reference Fickenscher, T.; Rodewald, U. C.; Niehaus, O.; Gerke, B.; Haverkamp, S.; Eckert, H.; Pöttgen, R. The Stannides \ce{RE3Au6Sn5} (\ce{RE} = \ce{La}, \ce{Ce}, \ce{Pr}, \ce{Nd}, \... 7 Concise Inorganic Chemistry1 has the following to say about this tetra(cyclopentadienyl) titanium complex: Tetra(Cyclopentadienyl) compounds such as \ce{Ti(C5H5)4} may be made from \ce{TiCl4} and \ce{Na(C5H5)}. The formula may be written as \ce{[Ti(\eta_5-C5H5)2(\eta_1-C5H5)2]} where two cyclopentadienyl rings are attached by 5 C-atoms (\pi - ... 0 @Jan While your answer is technically correct and is the method that would most likely be used in an upper division/graduate inorganic or physical chemistry course, I think the OP's question was from the perspective of general chemistry where things are kept a bit more simplified. For example, in the general chemistry textbook I use by Petrucci, et.al., they ... 0 Your friend is right, it 'can' produce actually an extraordinary combustion/thermite reaction. In fact, for example, one can expect extreme temperatures when mixing fine metal powder (for example Aluminum dust) with nano-CuO. Per this June 2019 source from Science Direct, to quote from the abstract: Nanothermites (metal oxide/metal) can offer ... 2 By the question asked, I assume OP is in high school of freshman in college. Thus, I'd use this general description of combustion as a starting point: Combustion reactions are common and very important. Combustion means burning, usually in oxygen but sometimes with other oxidants such as fluorine. A combustion reaction happens quickly, producing heat, ... 4 Ultimately those structures are wrong. Most of the community of chemistry educators teaches that formal charge reduction creates hypervalency. All modern research and models shows this not to be the case. You can't hybridize a d orbital and even get the proper tetrahedral geometry here. Reducing formal charges to 0 makes no real world sense, when sulfur ... 2 Since the volume is given and it's a constant, the initial amount of ammonia n_0(\ce{NH3}) can be found from its initial concentration c_0(\ce{NH3}):$$n_0(\ce{NH3}) = c_0(\ce{NH3})\times V\tag{1}$$To find c_0(\ce{NH3}), an ICE table might indeed come in handy; however, yours needs corrections. First, I suggest to rewrite it according to the process ... 2 From the abstract of the article by Abu et al. [1]: Here we conclusively demonstrate that humans are, nevertheless, able to distinguish \ce{D2O} from \ce{H2O} by taste. Indeed, highly purified heavy water has a distinctly sweeter taste than same-purity normal water and adds to perceived sweetness of sweeteners. In contrast, mice do not prefer \ce{D2O}... 4 This question can easily be addressed considering reduction potential of each compound. Concentrated nitric acid is a powerful oxidizing acid, which has a high reduction potential:$$ \begin{align} \ce{NO3- + 4H+ + 3e- &<=> NO + 2H2O} &\quad E^\circ &= \pu{0.957 V} \tag1\\ \ce{NO3- + 3H+ + 2e- &<=> HNO2 + H2O} &\quad E^\circ &...

5

It is a chemical change The aspect of this that is probably confusing you is that both the product and the result in the case of an element are still a form of the element. But elements can exist in different chemical forms and moving between them is still a chemical process. A relevant example is the reaction occurring mostly in the upper atmosphere that ...

-2

$\ce{HNO_3}$ does not react with $\ce{Cl_2}$, as $\ce{Cl_2}$ is produced in the famous "aqua regia", which is a mixture of nitric and chlorhydric acid, having reacted according to :$$\ce{HNO_3 + 3 HCl -> NOCl + Cl_2 + 2 H_2O}$$ This equation is given in Cotton and Wilkinson, Chapter 5-13. In Chapter 12-8, it is stated that NOCl is never pure. It is ...

3

You have to consider the oxidation state of the xenon. Higher oxidation state makes the central atom a stronger Lewis acid in general, xenon is just one case. So we should consider each oxidation state separately. Look at the ordering when we do that: +8: $\ce{XeO2F4}>\ce{XeO3F2}>\ce{XeO4}$ +6: $\ce{XeF6}>\ce{XeO2F2}>\ce{XeOF4}>\ce{XeO3}... 1 The impossibility of a reaction producing simultaneously$\ce{Na_2O_2, H_2O_2}$and$\ce{CO_2}$can be proved by the following arguments, drawn from the Merck Index, under the reference "$8728$. Sodium Peroxide". 1)$\ce{Na_2O_2}$is produced by burning metallic sodium in a current of air, from which carbon dioxide has been removed. 2)$\ce{Na_2O_2}$... 8 This depends a lot on the compound in question. In general, a bidentate acetate would lead to a four-membered ring with the central metal; since the$\ce{O-C-O}$angle is relatively fixed at approximately$120^\circ$and since we expect the resulting compound to be at least somewhat symmetric, it would typically lead to unusually small bond angles or a$\ce{...

3

Basically, during the electrolysis process of molten nickel(II) iodide ($\ce{NiI2}$), $\ce{I– (l)}$ ions would be oxidized to $\ce{I2 (g)}$ at the anode, and $\ce{Ni^2+ (l)}$ ions would be reduced to $\ce{Ni (s)}$ at the cathode. My state assignments were based on following facts assuming the temperature of electrolytic cell is kept below $\pu{900 ^\circ C}$:...

2

Sad to see such glaring textbook errors. Nitrous oxide solubility in water decreases with increasing concentrations of NaOH. It does not react with NaOH to "form" NaNO. Search this paper on Google and show it to your teacher. J. F. Nunn, Respiratory Measurements in the Presence of Nitrous Oxide, Brit. J. Anaesth., 1958, 30, 254

3

In the $\pu{500 cm^3}$ solution there are is already $\ce{F^-}$ dissolved coming from the $\ce{NaF}$ $\pu{0.10 mol dm^{-3}}$ that was already there. This means that the solubility of $\ce{CaF2}$ is reduced due to the common ion effect. NB $\ce{NaF}$ can be considered completely dissociated because it has a very high $K_\mathrm{sp};$ therefore the initial ...

1

Gc3941d has mixed the solubilities of $\ce{NaF}$ and $\ce{CaF_2}$. Let's start the calculation from the beginning. The concentration of fluoride ion is : [$\ce{F^-}$] = $0.1$ M. The concentration of Calcium may be calculated from the solubility product $\ce{K_{sp}}$ and [$\ce{F^-}$] according to : :$$\ce{[Ca^{2+}] = K_{sp} /[F^-]^2} = 1.46·10^{-10}/(0.1)^2 ... 2 I upvoted the question because it is actually very insightful. We may think of simply removing the MgO layer, and voila, there is bare magnesium. I think not. If you took bare Mg (no oxide at all) in a vacuum, and dropped powdered MgO onto the Mg, then exposed the metal to an argon atmosphere, I think you could blow off the oxide powder because there would ... 1 Unfortunately, the OP's Nernst equation is incorrect for several reasons. The balanced net redox equation is$$\ce{5 Sn^{4+} (aq) + 2 Mn^{2+} (aq) + 8 H2O <=> 5 Sn^{2+} (aq) + 2 MnO4^- (aq) + 16 H+ (aq) \tag 1}$$and n = 10 electrons for the reaction as written. The reaction quotient, Q, is$$\ce{Q = \frac{[Sn^{2+}]^5 [MnO4^-]^2 [H+]^16}{[Sn^{4+}]^5 [...

3

I have found the answer: In the following case: $$\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}$$ Both the constituent entities are neutral and hence will exist independently, and thus will not form isomers of the original compound.

1

I suppose you would cool it back before measurement, otherwise it would be measuring of temperature dependency of honey refraction index. There would be change, it is the way how bees evaporate the excessive water to make honey stable. Obviously, it would be function of both temperature and time. As honey is very viscous, loosing water would be very slow in ...

3

As MaxW indicated in the comment, perovskites are a huge family of natural and synthetic compounds. Following is a breif description of the group of compounds named perovskites: The original Perovskite is a mineral of formula $\ce{CaTiO3}$. It was first discovered in 1839 by the Prussian minaralogist Gustav Rose in the Ural Mountains, and is named after ...

2

Answering this question requires a preliminary discussion of the solubility product constants of three silver halides and the formation constants of three silver complexes. First, note that $\ce{AgCl}$, $\ce{AgBr}$, and $\ce{AgI}$ are all insoluble in water, but insoluble is a relative term in the end. Their respective solubility equilibria and solubility ...

0

According to ASM ,(8 th ed. , vol 2), 5 to 20 % nitric can be used to clean aluminum for spot welding It is not intended to passivate or anodize. It was one of very many cleaning options.

1

The HMS column is for the Hermann–Mauguin Symbol. You can see the line for 55700 has "P4mm" which is the proper crystal symmetry to be of the Perovskite structure. That line is also notated "Perovskite#PbTiO3#B". So of the formulations listed that is the only one that has the Perovskite crystal structure.

1

Actually, the answer per this source: 'Reaction of aluminium with diluted nitric acid containing dissolved sodium chloride', is apparently no with respect to both dilute and concentrated nitric acid, except... to quote: Metallic aluminium was found not to react with either concentrated or diluted nitric acid. Providing the diluted acid contains dissolved ...

5

Py and Cl both ligands are monodentate . Therefore, the coordination number of Pt in both the cases are 4 and Pt in general shows +2 oxidation state for 4 coordination number. Again py is neutral ligand whereas Cl is anionic (-1). So, $\ce{[Pt(py)4]^2+}$ is positively charged and $\ce{[PtCl4]^2-}$ is negatively charged. In IUPAC nomenclature, for a ...

3

Mercury cyanide behaves like Mercury chloride $\ce{HgCl_2}$. Both are soluble in water, and both do not dissociate in water. These compounds are not salts. The bonds $\ce{Hg-Cl}$ or $\ce{Hg-CN}$ are more covalent than ionic. When the atom Hg is included in compounds it often behaves as a non-metal. This is due to a relativistic effect. In all elements ...

3

$$\ce{B2H6 + 6H2O -> 2H3BO3 + 6H2 \\ SiH4 + 4H2O -> Si(OH)4 + 4H2 \\ 2Na + 2H2O -> 2NaOH + H2 }$$ The rest do not react with water under ordinary conditions. $\ce{Al, Fe, B}$ react only with steam or when they are red hot. White $\ce{P}$ undergoes easy hydrolysis when coordinated with ruthenium, as illustrated by Vaira et al. [1]. $\ce{CH4}$ does ...

0

For simple ionic salts, the enthalpy of dissolution is given by $$\Delta H_{\text{disso}} = \Delta H_{\text{lattice}} + \Delta H_{\text{solvation}}$$ where $\Delta H_{\text{lattice}}$ is the lattice dissociation enthalpy (positive) and $\Delta H_{\text{solvation}}$ is the solvation enthalpy (negative). For exothermic or endothermic dissolution, we have ...

2

A 14 page article (https://www.jstage.jst.go.jp/article/jieij1917/6/4/6_4_398/_article) describes the blackening of incandescent light bulbs as being totally due to evaporation of the filament. Platinum had been used as far back as 1840, and Edison's first patent was for a carbon filament (1879), but by this time (1922), tungsten was the metal used in almost ...

0

There is no easy general pattern. The dissolution can be formally divided into 2 steps: breaking the ion lattice into free ions in vacuum hydration these "naked" ions by water. Both lattice forming and hydration enthalpies are negative and their absolute value increases with increasing ion charge and decreasing ion radius. The relation of their values ...

3

The simple answer for your query: the lone pair is practically immobile, and can't orient itself to chelate. To understand this, we first have to understand the structure of DABCO. The rings force a conformational rigidity on both the nitrogens, and therefore the fluxional process of nitrogen inversion is inhibited in these structures. Now, as the accepted ...

5

My copy of Concise Inorganic Chemistry, 5th edition (J.D.Lee, Wiley publishers)1 has nothing contrary to say about hydrazine or 1,4-diazabicyclo[2.2. 2]octane (henceforth referred to as DABCO). It is most likely that the author of your adaptation has used his liberty to add these facts in your copy. This then raises the question: is the author correct? ...

9

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 The preface of the book says: The approach :These reactions represent important facets of the elements involved but are typically presented as no ...

3

For multi-electron atoms where the spin-orbit coupling is weak, it can be presumed that the orbital angular momenta of the individual electrons add to form a resultant orbital angular momentum $\bf{L}$. Likewise, the individual spin angular momenta are presumed to couple to produce a resultant spin angular momentum $\bf{S}$ . Then $\bf{L}$ and $\bf{S}$ ...

3

The chromate and dichromate ions exist in equilibrium as follows: $$\ce{2CrO4^2- + 2H+ <=> Cr2O7^2- + H2O}$$ In a basic medium, $\ce{[H+]}$ is less, which favours the formation of reactant $\ce{CrO4^2-}$ whereas in an acidic medium, $\ce{[H+]}$ is high, which favours the formation of product $\ce{Cr2O7^2-}$ Sources: Wikipedia

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