# Tag Info

2

So you need $$K_p = \frac{p_{\ce{H2S}}^3}{p_{\ce{H2}}^3}$$ where $p_{\ce{H2S}}$ and $p_{\ce{H2}}$ are partial pressures of $\ce{H2S}$ and $\ce{H2}$ respectively. Partial pressure is nothing but the mole fraction of the substance times the total pressure. You already have mole fractions, so that part is done. Now notice that in the expression of $K_p$ , you ...

3

In multielectronic atoms we have a relatively large difference between $s$ and $d$ orbitals when they have the same $n$ quantum number, or in terms of the actual quantum mechanics when they have the same total number of nodes in the wavefunction (this being what we label as $n-1$). But in the case of transition metals the $s$ orbital that mixes in with the $... 4 To say that boron forms only covalent compounds is an oversimplification. It can be involved in metallic or even predominantly ionic bonding, especially when combined with elecyropositive elements like magnesium. Compounds such as magnesium diboride ($\ce{MgB2}$), discussed in the referenced answer, or calcium hexaboride ($\ce{CaB6}$) are best described as ... 1 Another explanation. It costs something to remove one election from an neutral atom, whatever the nature of this atom. It costs more to remove a second electron from an atom that has already lost one electron. But it is feasible. As a consequence, removing a third electron form an atom that has already lost two electrons may be too much, if there is another ... 2 There is no simple and general relationship between valency and oxidation number. Look at the example of Oxygen. It depends of experimental evidences. Oxygen has always valency$2$. But it has an oxidation number equal to$-2$in the vast majority of its compounds, like water$\ce{H2O}$or$\ce{CO2}$. It can also be at oxidation number$-1$in$\ce{H2O2}$, ... 6 TL, DR: They do not contradict. Although larger ions decrease the lattice energy by increasing ion distance, it is compensated by another factor that increases the lattice energy. And like Nisarg Bhavsar said, product formation does not solely depend on lattice energy. Other factors such as the electron affinity and ionization energy of the element should be ... 3 Let's say you are titrating a weak acid$\ce{H3A}$with a strong base. $$\ce{H3A + OH- -> H2A- + H2O}$$ $$\ce{H2A- + OH- -> HA^{2-} + H2O}$$ $$\ce{HA^{2-} + OH- -> A^{3-} + H2O}$$ When all$\ce{H3A}$is reacted with$\ce{OH-}$to form$\ce{H2A-}$, the first equivalence point is reached. Before the first equivalence point is reached, the$\ce{H3A - ...

3

The alpha particles used in this experiment have an energy of about $5.59\ \mathrm{MeV}$. This energy is too high; the alpha particle cannot simply capture two electrons and form a stable helium atom. By way of comparison, the first and second ionization energy of helium is $24.6\ \mathrm{eV}$ and $54.4\ \mathrm{eV}$, respectively. Nevertheless, alpha ...

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An alpha particle is basically a helium nucleus, with two positively charged protons and no electrons, the particle would react with any electrons it comes into contact with. However it likely depends on which electron it interacts with, and how close the particle comes that determine the outcome. As ScienceDirect describes here - In some cases, this force ...

1

I had this example worked out. You should be able to copy the method, it is a bit long but easier than using the equations as errors are easily made there. To find the symmetry of the vibrations of a C2v molecule such as H2O or SO2. (Note that the method only works this way for 180 deg rotation groups if not you need to use cos/ sine to find amounts.) The ...

0

Well ! Titration is an operation that has to be done with the highest precision. The necessary end-point volume has to be determined at ± 1 drop of reactant, using an apparatus (pH meter, conductimeter, colorimeter, etc.) or a colored indicator, changing color at the end-point. Such a titration can be done with an acid-base reaction. But it can be done with ...

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No. While neutralization is often used in titration, other kinds of reactions may also be adapted to this laboratory technique. Precipitation for the measurement of chloride ion concentration is a well-known example.

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The acids $\ce{HNO3,H2SO4}$ , and the ions $\ce{NO3-,SO4^2-}$ are involved in redox reactions . standard reduction potentials of these ions are $\ce{(1) SO4^2- +4H+ +2e- -> H2SO3 + H2O (E^0= -0.93V)}$ and $\ce{(2) NO3- +2H+ +e- ->NO2 + H2O (E^0= +0.83V)}$; Reduction potentials reveal that $\ce{NO3-}$ is more prone to be reduced i.e. nitrate ion is ...

6

According to NIST CCCBDB, the $\angle(\ce{F-C-F})$ bond angles of $\ce{CH2F2}$ and $\ce{CHF3}$ are ${108.421 ^\circ}$ and $108.099 ^\circ$ respectively as calculated using CCSD=FULL/aug-cc-pVTZ. Similarly, the $\angle(\ce{Cl-C-Cl})$ bond angles of $\ce{CH2Cl2}$ and $\ce{CHCl3}$ are ${112.329 ^\circ}$ and $109.992 ^\circ$ respectively as calculated using CCSD/...

0

Warning: This is just an effort not complete answer. I also request higher members to give answer / put bounty as this question is not having complete answers. Borazine is planar. But the borazine ring does not form a perfect hexagon. The bond angle is $117.1°$ at the boron atoms and $122.9°$ at the nitrogen atoms, giving the molecule distinct symmetry. ...

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It doesn't get harder to add an electron to a halogen atom as you go down the group! It just gets less easy (Ref 1). Big distinction, since the addition of an electron to the radical is exothermic. However, other than this imprecision in the graphic, it is generally true that reactivity drops as you go down the column, as electronegativity does also, so ...

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In symmetric trialkylamines, $\ce{NR3}$, the bond angle under consideration is $\ce{\angle (R'-N-R)}$, where $\ce{R' = R}$ In this answer, it has been proven using Coulson's Theorem, that for the bond angle between any two equivalent bonds, we can use the formula, $$\cos \theta = \frac{s}{s-1}$$ A brief explanation of Coulson's Theorem and how it is used to ...

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NaCl/ rocksalt lattice is a FCC structure The length of the cubic for FCC is taken as: a Volume of lattice is: a^3 Based on 1 atom +2 half atoms, the diagonal length is derived as Sqrt(2a^2 )= a sqrt 2 The diagonal length is made up of 2 atoms or total length is equals to 4r, where r is the radius of atom. Hence, r = (a sqrt 2) /4 OR a= 4r/ sqrt 2 APF of ...

2

A simple, intuitive answer based solely on high-school chemistry and available emperical data; 1s < 2s < 2p < 3s < 3d < 4s < 4d. For Be, all the orbitals up to 2s are filled, so the "new" elctron has to go into the 2p orbital. Since there is a large gap between the 2s/2p energy levels of Be(it is one of the only elements in the ...

4

The presence of $d$ orbitals can indeed provide a means for otherwise closed-shell atoms of alkaline-earth metals to accept electrons. Wu et al. [1] describe carbonyl complexes of Ca, Sr, and Ba in which the valence $d$ orbitals of these metal atoms are indeed engaged in (covalent) bonding. In principle, all atoms actually have $d$ orbitals. Getting them to ...

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I've recently done an NBO calculation on the staggered conformation of the ferrocene molecule, using the 3-21G basis set, using the embedded Gaussian engine on the free version of the WebMO application. This calculation is easily reproducible, so I am not going to post the exact details here- the important conculsion from the calculation is that there are ...

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Disclaimer: This is not going to be an answer to the question. This is just a collection of thoughts on it, which hopefully helps to form a better understanding. First and foremost, and this should go without saying: If you have experimental data, use it. If you do not have experimental data, go find it. No calculation is good enough to be used without ...

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There was a previous question in Chem.SE which discussed the reaction between iodine and ozone. The product was found to be tetraiodine nonoxide, $\ce{I4O9}$. This reaction was performed in gas phase and published in a paper1. The reaction was done in a flow system, with $\ce{N2/O2}$ mixture as carrier gas at a pressure of 100 kPa and temperature range of ...

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Ozone is not mentioned in the synthetic techniques reported by Wikipedia, citing Greenwood and Earnshaw[1]. Summary: Orthoperiodate ion, $\ce{IO6^{5-}}$, is obtained by alkaline oxidation of iodate either electrolytically on a lead dioxide anode or chemically with chlorine. The product may then be acidi-fied to $\ce{H5IO6}$. Alkaline conditions favor ...

3

From Atkins' Physical Chemistry, The electronegativity is a parameter introduced by Linus Pauling as a measure of the power of an atom to attract electrons to itself when it is part of a compound. Pauling used valence-bond arguments to suggest that an appropriate numerical scale of electronegativities could be defined in terms of bond dissociation energies, ...

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within hybrid theory, R groups carry non-bonding electrons (typically in s orbitals) beside the orbitals for bonding that they lend Nitrogen for use in hybridization, so now Nitrogen has more s orbitals to mix with their own p orbitals. It’s not so much this increase in s character (non-directional) rather than the decrease in p character (directional) that ...

3

According to the reference mentioned in the question (Ref.1): The term ‘trans-influence’, being a long-established concept of broad relevance in the realm of inorganic chemistry, was defined first in 1966 by Pidcock et al. as the ability of ligand L in a complex to weaken the metal–ligand bond trans to itself. This ground-state phenomenon should be ...

1

You are right. When adding a small amount of $\ce{HCl}$ to a solution of $\ce{Na2CO3}$ you produce some $\ce{NaHCO3}$. When adding more $\ce{HCl}$, you cannot avoid that some of this $\ce{NaHCO3}$ does react with the newly created $\ce{NaHCO3}$ producing some $\ce{CO2}$. But this $\ce{CO2}$ will immediately react with $\ce{Na2CO3}$ which is still in excess ...

0

There are two ways to react sodium carbonate (A) with hydrochloric acid (B): You can add A to B, or You can add B to A. If the ratios differ from 1:1, you can get different reaction products by adding them differently. For instance, if you dump (A to B) a solution of 0.1 mole of Na$_2$CO$_3$ into a solution of 0.1 mole of HCl at a moderate rate, with ...

0

$6.6$ is neutral for pretty much all purposes. So is $6.0$. So is $5.0$. You need to go way lower than that. Table vinegar may have $\rm{pH}\approx2.5\dots3.0$, and that's what you want to use to hear that gentle fizz of dissolving carbonates. Yes, this is a viable method. No, it will not corrode steel pipes to any significant extent, provided that you apply ...

0

Reducing the pH of tap water to $6.6$ or $6.9$ will not corrode pieces of Zinc or Iron metals. Even pH $5$ is not sufficient.

3

When considering bond angles of molecules of main group ($\ce{\angle A-X-B}$) there are several different factors to consider: The row of the the central atom $\ce{X}$ determines the energy difference between the valence s and p orbital, which determines the extent of mixing (or hybridisation if you wish) The sizes of A and B, because steric repulsion, A ...

5

There are a few papers on Zinc-Platinum Systems (e.g., Ref.1 and 2, which are also sited in Ref.3), but most of them are in German. However, there were few useful data published by Johnson and Dillon in their Research and Development Report (Ref.4): $$\bf{The \ structures \ of \ platinum-zinc \ intermetallic \ phases}\\ \begin{array}{l|l r r} \hline \text{... 0 One of the "various forms" of cobalt chloride is Drierite, a mixture of cobalt chloride and calcium sulfate which is a common dessicant. It is blue when active and pink when hydrated (Ref 1). "Drierite can be regenerated by spreading the granules in a tray and heating them in an oven at about 425°F (220°C) for 1 to 1.5 hours." A more ... 2 Some of the experimental details are missing, but if you recall$$\mathrm{Abs} = \varepsilon \cdot c \cdot d$$about the absorption depending on the molar extinction coefficient \varepsilon, the analyte concentration c, and the optical path length d, it could be due to a low concentration of the analyte. Independent of the former, given the \pi ... 3 \mathrm pπ-\mathrm dπ bonding is very weak and normally not considered to be very effective in bonding. Rather, the bonding can be explained via ionic bonding instead due to the large electronegativity differences. This can be seen in siloxanes and trisilylamine. According to J. Am. Chem. Soc. 1980, 102 (24), 7241–7244 Since our ab initio calculations ... 8 Entries mp-2815 and mp-1018809 differ in their lattice constants (c = 14.879\ Å vs. c = 13.983\ Å) reported, thus differ in unit cell volume and density (4.05 g/cm³ vs. 4.31 g/cm³) while sharing the same space group symmetry P6_3/mmc. But only the former is labeled as «stable», the later one is annotated with «Decomposes To \ce{MoS2}». It is ... -4 \ce{H-O-F} has a lower bond angle than both \ce{F-S-F} and \ce{F-O-F}. The reason is probably that the lower electron density around \ce{H} results in significantly less steric interaction between \ce{H} and the lone pairs of the \ce{F} atom. In both \ce{SF2} and\ce{ OF2}, the two \ce{F} atoms with high electron densities would tend to ... -4 I looked up the net and found the following order- HOF(97.2°)<SF2(98°)<OF2(103°) (Of course, the values might not be exact but this is what I found on the net.) Now, my POV- Yes, your analogy that bond angle of OF2> SF2 is correct. It's a direct result of Drago's Rule. The electron cloud in OF2 will be shifted much towards the Fluorine atom. Bond ... 7 Buttonwood has given an excellent answer with a nice demonstrational video. However, there is one point I'd like to emphasize. In ideal gases, the molar mass is directly proportional to the density:$$PV = nRT \ \Rightarrow \ \frac{n}{V} = \frac{P}{RT} \tag1$$If the molar mass of a gas is M and its mass is m, n = \frac{m}{M} and \frac{n}{V} = \frac{m}... 4 Aqua regia doesn't actually dissolve lead metal. It converts to insoluble lead salt. From here: When we are working with an acidic chloride solution, like HCl (with an oxidizer) or aqua regia, most of the lead will become an insoluble lead chloride which if it is left to settle can be removed by careful decanting of the solution and filtering, what lead ... 1 You can think of it as: polarity causes slight concentration of pi electrons on nitrogen atoms in contrast to boron in the resonance hybrid due to which cyclic delocalization does not occur "completely"; hence the aromatic character is hindered Further reference: https://www.researchgate.net/publication/... 9 If your sample is solid and crystalline, knowledge of the unit cell's dimension and symmetry (e.g., fcc) and density allows you to determine the molecular mass of your compound. In fact, in crystallography, you may determine the macroscopic density of your sample (e.g., find a liquid which i) wets your crystals and ii) lets your crystals float) in first ... 1 Let us understand and try to answer your question with a few simple points. First it is important to understand why exactly is borazine aromatic. Boron has an empty p-orbital and Nitrogen has a lone pair of electrons. According to requirements for pi-backbonding, there needs to be a sigma bond between both the atoms and requirement of one empty orbital for ... 3 TL;DR: Decomposition of nitrites always afford \ce{NO2} if you apply the correct reaction condition to it. Long answer To generalize the decomposition of metal nitrates (I concluded from the article above, and hereby using a generic monovalent metal M for the equations), we can say this: The first stage of decomposition is always the decomposition to ... 1 According to the CRC Handbook of Chemistry and Physics, melting point of \ce{CaF2} is \pu{1418 ^\circ C}, of \ce{Al2O3} (in its corundum form) is \pu{2053 ^\circ C}, and of \ce{Na3AlF6} is \pu{1009 ^\circ C} (lowest of the all three). Thus the cryolite should melt first. I have read mentions of the problem you asked in the book Nonaqueous ... 2 As usual, I suggest to start with a RICE table to get an overview, denoting initial partial pressures with p_0 and the change in partial pressure with x:$$ \begin{array}{lccccc} &\text{R} &\ce{&CH3OH(g) &+ &NOCl(g) &<=> &CH3ONO(g) &+ &HCl(g)}\\ &\text{I} && p_0(\ce{CH3OH}) && p_0(\ce{NOCl}) &...

2

Images are from Wikipedia. The first obvious thing that you can see is that N2O4 is planar but N2H4 is not. And the primary reason starts here. N2O4 can be viewed as two nitrogen dioxide molecules joined together. If you draw a Lewis structure of NO2, you will recognize two things. Firstly, it has one unpaired electron. So imagine when two NO2 molecules come ...

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This answer is similar to this answer on $\ce{[Co(NH3)5(H2O)]Cl3}$, Thanks to Loong (Faded Giant) for the references. According to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book), the systematic names of coordination entities are generated by using the following principles: IR-9.2.2.1 Sequences of ligands ...

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Metallic osmium is harmless but it reacts with oxygen at room temperature, forming volatile osmium tetroxide. Any other osmium compounds are also converted to the tetroxide if oxygen is present. This makes osmium tetroxide the main source of contact with the environment and human body. Exposure to osmium tetroxide cause harm to the skin, eyes, and ...

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