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0

The reactions you proposed are pure fantasy. They will never happen. Na2SiO3 does not react with either H2O2 or Hydrogen gas or CH2 (which does not exist, by the way). The only possibility would be to have silica precipitate by bubbling CO2 in the solution. This will produce a precipitate of Si(OH)4. But this precipitate is so gelatinous that it cannot be ...

2

The mistake you made is that you assumed mass concentrations of the ions from the stoichiometric relation, which is only applicable to the amount of substance, and, as a consequence, amount concentration. You have to convert mass concentration $γ$ to the molar concentration c first: $$c(\ce{Pb(OH)2}) = \frac{γ(\ce{Pb(OH)2})}{M(\ce{Pb(OH)2})} = \frac{\pu{0.... 1 There are better ways to do it. Copper (II) can be reduced to copper (I) with a variety of reagents, especially in the presence of a soft-base halide ion such as chloride which precipitates or complexes with the copper (I) (which is a good way to prevent "overshooting" to metallic copper). Some specific approaches can be seen here. 0 I tell a true story about myself. In my senior year of college, I took as an elective, a course in Pre-engineering Physics. At that point in my studies, I had already taken a lot of math (Advanced Calculus, Linear Algebra, Statistics, Theory of Differential Equations,..., and even graduate-level credit in Numerical Analysis, Numerical Methods,...). To my ... 4 Wikipedia reports that calculations show orthocarbonic acid spontaneously decomposes to the ordinary (meta)carbonic acid \ce{H2CO3} plus water. However, orthocarbonate esters \ce{C(OR)4} do exist. The orthoacid may, however, become stable at high pressure [1, with a nontechnical summary given in 2] and possibly exist within Uranus and Neptune. It is ... 2 Neutral oxides do not manifest either acidic either alkalic behaviour ( but eventually very weak in extremely alkalic/acidic environment). Amphoteric oxides manifest, more or less balanced, both behaviours in reasonably strong acidic/alkalic environments. 2 The value \pu{1.3653 mol L^{−1} atm^{−1}} is the solubility constant (or Henry's law solubility constant), not the solubility. The solubility is defined as the maximum possible concentration (the saturation concentration) of a solute under given solution conditions (e.g. temperature and pressure), whereas the solubility constant H^{cp} defines how solute ... 0 I recall a paper where heating the hydrate in a stream of \ce{HCl} gas avoided the formation of a basic salt. This may work for hexahydrate \ce{AlCl6⋅6H2O} also. References https://books.google.com/books?id=Pef47TK5NfkC&pg=PA1147&lpg=PA1147&dq=dehydration+with+a+stream+of+HCl&source=bl&ots=W8NJaeEg0E&sig=... 1 Metal nitrides can be produced from the amide (\ce{NH2−}) ion as the \ce{N3−} source (Metal nitrido complex). Also, heating water vapor to high temperature:$$\ce{H2O ->[\Delta] .H + .OH}\ce{.H + .H -> H2}$$In the presence of atomic nitrogen from heated nitrogen:$$\ce{.N + H2 -> .NH2}$$Cobalt is an active transition metal which can ... 1 I think this property has to do with the ease with which Cobalt makes complexes with NH3 and amines. The variety and the stability of the complexes containing \ce{Co^3+}, surrounded by NH3, ethylenediamine, H2O, and/or Cl, is unparalleled. Iron Fe does not make complexes with NH3 in usual conditions. Nickel, Copper and Zinc do make one or two complexes ... -1 Of course, the Lithium atom is smaller than Sodium. But the Sodium nucleus attracts its outer electron much stronger than Lithium, at it has 11 protons, and Lithium only 3. So both effects (size and charge) have to be taken into account. Apparently the size effect is not dominant. 2 I think obtaining anhydrous aluminium(III) chloride from hexahydrate \ce{AlCl6 · 6 H2O} isn't worth it, both economically and time-wise. Thionyl chloride might work, but, as you said, it's an unpleasant reactant to work with, which is also likely more expensive than dry \ce{AlCl3} (or is on par, depending on the purity of aluminium trichloride). A ... 2 Wikipedia pretty much has it all already. There are numerous magnesium chloride hydrates \ce{MgCl2 · x H2O} (x = 1, 2, 4, 6, 8, 12), however in the temperature range from 0 °C to 115 °C the most stable one is hexahydrate \ce{MgCl2 · 6 H2O}. Whether \ce{MgCl2 · 6 H2O} decomposes to hydroxo-salt or only dehydrates depends on the conditions. When ... 1 The problem could be solved with simultaneous equations, but the following are the wrong equations.$$K_{\mathrm{sp,}\ \ce{AgCl}} = [\ce{Ag+}][\ce{Cl-}]K_{\mathrm{sp,}\ \ce{AgBr}} = [\ce{Ag+}][\ce{Br-}]$$The right equations to use would be:$$K_{\mathrm{sp,}\ \ce{AgCl}} \ge [\ce{Ag+}][\ce{Cl-}]K_{\mathrm{sp,}\ \ce{AgBr}} \ge [\ce{Ag+}][\ce{...

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All calcium chloride hydrates give away water of crystallization upon heating without forming a hydroxide, and the anhydrous $\ce{CaCl2}$ melts without decomposition [1, p. 162]: anhydrous salts melts at 772 °C, while the mono-, di-, tetra- and hexahydrates decompose at 260 °C, 175 °C, 45.5 °C and 30 °C, respectively; the anhydrous salt vaporizes at 1935 °...

2

Both chloride and bromide ions are present in 10-fold excess over silver ions. That means that the chloride and bromide concentration in solution will not drop by much (they will remain major species). The solubility product of AgCl is 200-times higher than that of AgBr. If both AgCl and AgBr precipitate, the chloride solution would be 200-times higher than ...

5

One candidate for reaction with the rubidium is silica, $\ce{SiO2}$. Reference [1] reports that black powders are formed when alkali metals are contacted with silica. Such powders are more stable in dry air than ordinary alkali metals. The reference suggests that electrons are imparted from the alkali metals into the silica forming an electride complex. ...

2

Copper(III) nitrate cannot be obtained from aqueous nitric acid, and likely doesn't exist. Reaction carried under strongly oxidative conditions between $\ce{Cu(NO3)2}$ and fuming $\ce{HNO3}$ yields in nitrosyl copper(II) trinitrate $\ce{[NO+][Cu(NO3)3−]},$ [1, 2] sometimes written as adduct $\ce{Cu(NO3)2 · N2O4},$ which is contradictory to the crystal ...

0

Actually Fluorine has a high oxidizing power because its charge density is high due to its very small size, and it is also common for the hydration enthalpy that is if the charge density is high hydration enthalpy is high. So here for Fluorine there exist a greater oxidizing power and a higher hydration enthalpy because of its higher charge density.

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You have to evaluate shortcomings as compared to something. You can make a list and say "but InChI doesn't represent X" where X is ranging around mixtures Markush groups polymers from repeat units reactions toasters The point is: so what? All representations have some limitations. But limitations aren't shortcomings unless you have a different ...

1

Apparently you believe that $\ce{NaOH}$ may be an independent molecule. It is not in this case. $\ce{NaOH}$ does not exist in a solution. $\ce{NaOH}$ does not exist in the solid state either. $\ce{NaOH}$ does not exist in the solid state as a molecular compound. In the solid state, it is made of a huge pile of $\ce{Na+}$ and $\ce{OH-}$ ions, exactly like a ...

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Generally, personal protective equipment (PPE) is the last measure. First the employer has to check whether the risks to safety and health can be avoided (e.g. by replacing the toxic substances with substances that are not toxic). After that, the risks must be limited by technical means or collective protection or by measures, methods and procedures of work ...

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First you consider the state of hybridisation .all are SP3 .resonance is only possible with P orbitals and d orbitals too .here lone pairs of nitrogen in all are in SP3 back bonding is possible between 2p-2p ; 2p-3p ; 2p -3d . So as this cases are no longer valid we have to consider the electronegativity as you told in question .but remember it is my point ...

3

I think it generally means that the average oxidation state of the metallic element in the oxide is higher or lower. You can also say that the mass percentage of oxygen in the oxide is higher or lower.

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Aside from the reactive gas phase monochloride, magnesium is well known in the +1 oxidation state. The $\ce{Mg^+}$ ions, like similar ions of mercury, form diatomic pairs $\ce{Mg2^{2+}}$ in which each magnesium atom is covalently bound to the other with one valence, and the other valence is used for ionic bonding to the anionic ligand. Calcium has been ...

3

The reaction you wrote down is wrong on two counts. The reactant is not a hypothetical tetraaqua complex and the product is not a hypothetical tetraammin complex. The correct reaction is as shown below: $$\ce{[Cu(H2O)6]^2+ (aq) + 4 NH3 (aq) <=> [Cu(NH3)4(H2O)2]^2+ (aq) + 4 H2O (l)}\tag{1}$$ Note that I have used an equilibrium arrow here: the ...

0

We can find the average molar mass with the density formula involving pressure and temperature $\overline M=\frac{\rho TR}{p}$. We also know that, in this case, $\overline M=\frac{n_1M_1+n_2M_2}{n_1+n_2}$. Since we only "care" about $M_1$ we can write the average molar mass as $\overline M=\frac{n_1}{n_1+n_2}M_1+(1-\frac{n_1}{n_1+n_2})M_2$ and then find $\... 6 Unlike lithium tetrafluoroborate$\ce{LiBF4},$sodium tetrachloroaluminate$\ce{NaAlCl4}$doesn't decompose upon melting, and doesn't show any signs of decomposition or phase transition up to approx.$\pu{800 °C}$[1, p. 688]: FIGURE 3. Calculated$(\ce{NaCl + AlCl3})$phase diagram at$\pu{0.1 MPa}$(dotted lines are liquid–liquid miscibility gap ... 3 Both of these reactions are thermal decomposition reactions, but the difference basically occurs in the polarizing power of the cations$\ce{Ca^2+}$and$\ce{Mg^2+}$. The thermal decomposition of$\ce{Ca(HCO3)2}$is just the reverse reaction of the lime water test for the detection of$\ce{CO2}$, where the carbonate combines with$\ce{CO2}$and$\ce{H2O}$... 0 According to J.D.Lee's "Concise Inorganic Chemistry", the extra electrons in the metals of the 2nd half of the transition series, like Fe, Ni etc.. are excited to the p level .. and then the unpaired electrons in the d, s and p levels participate in metallic bonding .. so Ni has the max. unpaired electrons, followed by Co and then Fe ... and hence strength ... 2 To complement the previous answer: Atomic absorption lines are very sharp, with high extinction coefficient, low detection limit and are specific for the given element. Molecular absorption is band-wise, diffuse, with much lower ext. coefficient, higher detection limit and last but not least, the selectivity is lost. 2 Atomic absorption spectroscopy is a quantitative measure of the elements in the analyte. We atomise the analyte so that the absorption of light corresponds to the characteristics of the elements. Otherwise how we can predict the fundamental elements when these are still part of some molecules? We measure the absorbance and refer to the Beer-Lambert law. 0 Metal hydroxides are basic in water. Non-metal oxides are acids in water. The limit between metals and non-metals is a sort of staircase going through the periodic table from the middle of the first line to the lower corner at the right-hand side. Aluminium touches this staircase. This is why Aluminium behaves like a metal and like a non-metal, and why ... 0 When dissolving for exemple baryum nitrate in water, there is no baryum nitrate any more in the solution. The solution is a mixture of Baryum ions and of independent Nitrate ions. So there is no "selectivity of these nitrates" as you say. Your question has no meaning. The only order of reaction is the order of the solubility products of the corresponding ... 0 Gold looks pretty soluble in molten zinc. The melting temperature increases up to about 625 C with 50% gold.Charcoal on the surface of the melt will partly protect the zinc ; I have seen charcoal used on lead but not zinc. 1 Indeed if you do the cupelation, the lead will form litharge and some parts of the litharge will volatilize. But usually the cupel is made out of bone matter - this is very porous. While the molten lead and the gold will stay in the cupel, the litharge will be absorbed by the cupel like a sponge. If all lead is converted to litharge and all litharge is ... 1 Adding to the first answer: NiCr wires are used because they are cheaper as platinum wires and still have a good chemical resistance and high melting point. Even tough platinum is much superior to NiCr. 6 Metal chlorides are usually much easier volatilized than oxides or sulfates. They melt or sublime at temperatures much lower than 800°C. So they are volatilized in any flame like Bunsen burners. Platinum is not attacked by hot chlorides or acids, and is not oxidized in hot air. So it can be used for a great number of consecutive tests in the flame. It will ... 1 Shouldn't it clearly be the opposite, as increasing temperature favors the reverse reaction? Yes, they made a mistake. For all the other scenarios, they paired a figure of the rate changes with a figure of the matching concentration changes. For this scenario (increase in temperature) they matched it with a correct figure of concentration changes when the ... 3 Effective nuclear charge$Z_{\text{eff}}$is given by Slater's Rules. According to these rules, for$\ce{C}$, effective nuclear charge experienced by an outer shell electron is: $$Z_{\text{eff}} = 6 - 0.85\times 2 - 0.35\times 3 = 3.25$$ For$\ce{C+}$, effective nuclear charge is $$Z_{\text{eff}} = 6 - 0.85\times 2 - 0.35\times 2=3.6$$ Since the effective ... 2 According to this paper, the attacking species is$\ce{OH-}$rather than$\ce{H2O}$on$\ce{Cl2}$The paper also says that: The proposed mechanism also provides a good explanation for the high initial rate of reaction. At the instant the strong chlorine solution is mixed with the distilled water the hydroxyl ion concentration is considerably higher than ... 2 Heating hydrate copper sulfate at 105 degrees does nothing. The chemically bound water takes a higher temperature to remove all water molecules (water is lost in steps as a function of temperature). This low temperature removes only two water molecules out of five. I think the major role of copper sulfate, besides being a gentle dessicant, is to oxidize ... 5 A strategy different from the one described by Yusuf Hasan would be to look at the neutral isoelectronic molecule. The oxygen atom is isoelectronic to the fluorine cation, so$\ce{OF+}$is isoelectronic to$\ce{O2}$. It is difficult to write a nice Lewis structure for$\ce{O2}$, see https://chemistry.stackexchange.com/a/15061. Experiments show that dioxygen ... 7 A good way to think about stuff like this, is to imagine a possible synthesis for the given intermediate. One way of making this species would be to cleave the H-OF bond of HOF against it's electronegativity, that is, heterolytically cleaving the bond to make H- and [OF]+ . As we are going against electronegativity, you get the idea that the intermediates ... 2 You are right: 1 cubic meter of water contains$\pu{1E-4}$moles of$\ce{H+}$ions. But one mole is$\pu{6E23}$atoms or ions. So,$\pu{1E-4 mol}$contains$\pu{1E-4}\times\pu{6E23} = \pu{6E19 ions}.$Here we are. 6 There is no need (or possibility, really, in terms of standard lab capabilities) to oxidize sodium(I). In fact, one method relies on sodium(I) reduction to metal as a method of eliminating unwanted chloride. Method 1 Electrolysis of molten sodium chloride: $$\ce{2 NaCl(l) -> 2 Na(l) + Cl2(g)}$$ Oxidation of sodium metal to oxide by burning:$$\ce{4 Na + ... 1 Let me give an extreme example of chemical analysis. It is an anecdote. Some analytical chemists were using a very sensitive technique to analyze metals in blood. It is called neutron activation analysis. What they saw was that a very small quantity of molybdenum (Mo) was present in small quantities persistently. Nobody expected that. It turned out after a ... 1 Actually an organometallic scandium(I) compound is known, from P.L. Arnold at al. (1996). As with organometallic compounds generally, the metal is covalently bonded, and the use of$3d$orbitals as valence orbitals (there are 14 shared valence electrons associated with each scandium atom) in this and similar compounds favors including scandium as a ... -1 probably it is related to its filled$\mathrm{d}$orbitals. Filled or nearly filled orbital elements counted as soft acids, and$\ce{Ag+}$has totally filled$\mathrm{d}$orbitals, and it can donate to ligands, making ligand-$\pi\$ bonding, which causes stronger interaction.

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There are two ways to add two miscible solvents to each other. One is the bartender skill, whereby you attempt to create two different layers of solvents and then try to keep the flask motionless to prevent mixing. The other way is to mix them all along, whether by shaking, careless pouring, or a stirring bar. I will concentrate on the second as that is, in ...

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