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0

$\ce{SO4^2-}$ is actually going to make the pH basic because $\ce{HSO4-}$ is a weak acid (not $\ce{H2SO4}$, only the first hydrogen ionizes completely). By itself aqueous $\ce{Mg^2+}$ will not form magnesium hydroxide in water. But with the $\ce{OH-}$ from the $\ce{HSO4-}$ is will form a suspension with some dissolved and some not so dissolved, but still the ...


0

This is a diagram of the d orbitals of a generic d8 complex in a tetrahedral and square planar configuration. The tetrahedral complex would be expected for pi donor ligands (Cl-, OH-, etc.) where the pairing energy is greater than delta t, while the square planar complex would be expected for strong field pi* acceptor ligands (think CO, NO+, and CN-, which ...


1

Given the problem as stated, you did the problem correctly. However I would have done it a bit differently to make it easier to check. I like to do the problems in steps. I also dislike carrying a lot of fractions in intermediate calculations since I get confused easily. (This is basically M. Farooq's answer with some explanation.) Given $$\ce{2FeS2 + 11/...


1

Note the following relations from the equation you wrote: 1 mol $\ce{FeS2}$ = (11/4) mol $\ce{O2}$; 1 mol $\ce{FeS2}$ = (1/2) mol $\ce{Fe2O3}$; 1 mol $\ce{FeS2}$ = 2 mol $\ce{SO2}$. You already determined that $\ce{FeS2}$ is the limiting reactant because we have 600/120 mol of iron sulfide and 800/32 mol of $\ce{O2}$; This implies that 5 mol $\ce{FeS2}$ ...


0

When you mate connectors there is mechanical abrasion. Thus electronic connectors with silver contacts would not be resistant to silver sulfide formation because of this mechanical abrasion which would remove the silver oxide coating. That is why gold plating is often used on higher quality connectors. Gold is resistant to tarnishing of any sort and the gold ...


4

Silver sulfide is formed by Ag and hydrogen sulfide not sulfur dioxide. You need reducing conditions where the latter can be reduced to $$\ce{SO2 +reduction -> H2S}$$ Actually, sulfur dioxide chemisorbs on ultraclean silver surface, however heating can remove it. So this is reversible sorption, as suggested by Lassiter [1]. Just note that Auger ...


0

This is my first answer guys, so be nice :) In order to calculate the moles of an ideal gas from its the volume, you have to know that one mole of an ideal gas will occupy a volume of 22.4 litres at STP (Standard Temperature and Pressure). Now, the calculation should look like this: 90 litres of NH3 ÷ 22.4 litres of gas per mole = 4.0 moles of NH3 We ...


0

It is an odd rendering. I'm not sure what the actual mechanism is to create $\ce{BeH2}$ but the simplistic reaction would be $$\ce{Be + H2 <=> BeH2}$$ The MO for $\ce{H2}$, which is shown in the figure below is taken from Wikipedia. The right side of the diagram you showed neither represents a hydrogen molecule, nor two independent (and hence ...


1

Here is a simple minded answer. High melting point compounds tend to follow the octet rule - the number of valence electrons sums to 8. The Group IVB carbides (TiC, ZrC, HfC) have an 8 electron sum (4+4) and have very high melting points - for HfC it is approximately 7100F. The Group IIIB nitrides and phosphides have an 8 electron sum (3+5) and are the ...


2

It looks like that particular MO was created with the two possibilities of hydrogen combinations: Either the two 1s electrons are same phase or different phase. Combining those options, alongside the AOs for $\ce{Be}$ produces the MO for $\ce{BeH2}$ This picture would have been clearer if they had opted to use the MO of dihydrogen, (bonding AND antibonding) ...


0

Any unpaired electron is taken into consideration while trying to quantify spin only magnetic moment. valence electron could be unpaired (not always), and inner shell electrons are paired so they don’t contribute (since they are fully filled). That’s why usually outer shell electron is considered and no it’s not calculated for ions only.It is applicable for ...


12

Based on research inspired by andselisk's answer, chemists stored it in glass vessels coated in wax (similar to the receiver setup Scheele used to prove the silicon dioxide precipitate was from the glassware itself. The fourth paragraph down in this blog post on The Chronicle Flask touches on it (emphasis mine): Where do you put something that eats ...


32

In the original 1771 experiment, Scheele used a very simple setup consisting of a glass retort with a glass receiver (round-bottom flask). Yes, the glass was etched to some degree by the fumes, but it was not drastic enough to destroy the apparatus. From Anders Lennartson's The Chemical Works of Carl Wilhelm Scheele [1, p. 22]: 3.1 Publication 1. ...


3

Alternative method to MaxW method: Assume that an initial $\pu{10.0 mL}$ solution of $\pu{1.0e-10 M}$ in each of $\ce{Ba(CN)2}$ and $\ce{BaI2}$ is clear (homogeneous). That means $\ce{Ba(CN)2}$ and $\ce{BaI2}$ have dissociated completely. Thus concentrations of ions are as follows: $$\ce{[CN-]_i = [I-]_i} = \pu{2\cdot10^{-10} mol \! L^{-1}}$$ Suppose when ...


5

Consider a $\pu{10.0 mL}$ solution containing $\pu{1.0e-10 M}$ each of $\ce{Ba(CN)2}$ and $\ce{BaI2}$. If $\pu{3.5e-9 mol}$ of $\ce{AgNO3(s)}$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate? $K_\mathrm{sp}(\ce{AgCN}) = \pu{6.0e-17}$; $K_\mathrm{sp}(\ce{AgI}) = \pu{8.5e-17}$. Assuming that $\ce{Ba(CN)2}...


0

The crystal structure shows that the copper is six coordinated and has four oxygen atoms from water molecules in a plane and the other two octahedral positions are each occupied by one oxygen from an $\ce{SO4}$ group. The fifth water molecule has its oxygen bridging two $\ce{CuSO4}$ molecules, see the sketch, and is bonded to four oxygens, two from water ...


-2

The definition of a chemical reaction: We call changes that change bonding form of atoms, which results changes in the structure and properties of matter, is called a chemical reaction. So this reaction isn't a chemical reaction. Copper (II) Sulphite is still Copper (II) Sulphite and water was'nt in bond with Salt it was trapped in crystal system of salt P.S:...


0

When analytical chemists talk about hardness of water as ppm CaCO3, it is just a statement of "titratable" calcium / magnesium by ethylenediamminetetraacetate (EDTA). From your comments, you would like to replicate hard water perhaps for a project. Keep in mind that hardness are of two types temporary and permanent hardness. Permanent hardness can replicated ...


0

The questions asks you to differentiate Na2SO4 and Na2SO3. There are two main concepts involved here: H2SO3 is a very weak acid (better to write SO2 (aq)) and H2SO4 is an extremely strong acid. Certainly, HCl would not be able to displace H2SO4. Another point which you should remember is that salts of strong acids and strong bases are pH neutral. However, ...


1

The last statement …sodium sulfate and sodium sulfite are both basic and would turn pH indicator the same color is true only for $\ce{Na2SO4}$ as it's formed by both strong base and strong acid and won't noticeably affect pH. $\ce{Na2SO3}$, on the other hand, undergoes hydrolysis: $$ \begin{align} \ce{Na2SO3 + H2O &<=> NaOH + NaHSO3} \\ \ce{...


0

A basic salt is formed when a weak acid reacts with a strong base. $\ce{LiF}$ is formed from strong base $\ce{LiOH}$ and weak acid $\ce{HF}$ to give a basic salt. $\ce{HF}$ is not a strong acid.


0

It is an interesting question, however there is a misconception. You are giving too much attention to the word neutral. The word neutral is just a qualitative version. Numbers and equations are more accurate way to discuss science. You have to see how water's $\pu{pH}$ at 100 °C is calculated. How do you get those numbers? Lets us do it algebraically, so ...


0

The redox potential is a measure of tendency of redox system to exist rather in oxidized form ( negative potential values) or reduced form ( positive values ). Therefore the higher the standard redox potential is, the stronger oxidation effect and weaker reduction effect have the respective oxidized/ reduced forms of the redox system. And vice versa. ...


1

Have a look at the method given in this answer. In your example subject each vector on each atom to the operation in the point group, e.g rotation, reflection etc. Count 0 if the vector is moved else count 1 if it remains unchanged and -1 if it is unchanged but just points in the other direction. Now you have the reducible representation. Use the method ...


0

CO>CN->Ethylene Diamine>NH3>NCS-............Cl>....... this is the ligand strength order form the NCERT textbook so we have four strong ligand which will cause splitting of orbitals also called octahedral splitting and pairing takes place aginst the hunds rule making this diamagnetic and due to pairing it is low spin


0

This is not a full answer, but you should note that halogens occupy the lower end of the spectrochemical series, and they are (almost) never able to act as strong field ligands. The magnitude of crystal field splitting energy of $Co^{+3}$ is low, such that any ligands barring the halogens do act as strong field ligands. This makes it a low spin complex. ...


1

Let's subtract the two reactions from each other to simplify things a bit: $$\ce{2ClF + 2KBr -> 2KF + Cl2 + Br2}\ \ \ \ \ \ \ \ \ \ [1]$$ $$\ce{ClF + 2KBr -> KCl + KF + Br2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2]$$ $$\ce{ClF + KCl -> KF + Cl2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ [3] = [1]-[2]]$$ Now we can discuss [2] and [3] separately by splitting them up ...


6

The parameter $a$ from the van der Waals equation $$\left(p+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$$ is not to be taken only as the measure of strength of intermolecular bonding. Rather, it is overall measure of significance and frequency of intermolecular interaction. It may be based on purely elastic electrostatic interactions, where only the ...


0

Aside of influence on the crystallization rate, you have also consider the direct chemical influence. The sodium carbonate hydrolyzes in not enough alkaline solutions. $$\ce{Na2CO3 + H2O <=> NaHCO3 + NaOH}$$ Precipitation of carbonate is interfered by solubility of Bicarbonates. $$\ce{CaCO3 v + H2O + HCO3- <=> Ca(HCO3)2 + OH-} $$ Addition of ...


1

At low pHs there is little "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Most of the carbonate species are dissolved $\ce{CO2}$, $\ce{H2CO3}$ and $\ce{HCO3-}$. Thus the precipitate forms slowly and you get relatively large crystals. At high pHs there is a lot "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ ...


3

What is the crystal structure of bismuth oxyhydroxyphosphate (BOHP)? Not exactly sure what an answer looks like so here is everything. First off looking at the diffraction pattern from the paper, bismuth oxyhydroxyphosphate matches that of petitjeanite so that is definately the prototype structure. Using the information about diffraction, from the ICDD ...


2

It is difficult to calculate the energy levels of the orbitals beyond hydrogen. "The first experimental evidence for the filling of the 5f shell in actinides was obtained by McMillan and Abelson in 1940," from Wikipedia So the answer, as you suspected, is that experimental evidence from spectroscopy helped determine the electron configuration of the ...


0

If you actually passed current thru the cell, you would have been cathodically protecting the cathode - that is, preventing the dissolution of iron. On top of this "cathodic protection", however, you have copper deposition. And the copper could have been loosely deposited as black nano particles. At the anode, you are discharging SO4-- dianions, which then ...


4

Aluminum does assume oxidation states, see the list of elemental oxidation states in Wikipedia. Aluminum assumes a negative oxidation state when forming compounds with more electropositive elements like magnesium. The magnesium compounds referred to above include $\ce{Mg_{17}Al_{12}}$. In this compound there is enough magnesium to half-fill the valence $p$ ...


0

Metal fluoride thermal stability trends are largely dictated by lattice enthalpies; what matters is the relative size of the cation in consideration (the charge/radius ratio - a measure of polarizability - is also a key factor). Smaller cations can better hold small anions like fluoride. The larger the lattice enthalpy factor, the tougher thermal ...


1

You may follow the procedure in this reference. I included the abstract for your convenience: Abstract: The hydroamination of styrene with aniline catalyzed by phosphine-ligated palladium triflates exhibits a substantial $\ce{^{13}C}$ isotope effect at the benzylic carbon. This supports rate-determining nucleophilic attack of amine on a $\eta^3$-phenethyl ...


1

$\ce{NH4OH <=>NH3 + H2O}$, that is, they are in equilibrium. Since the solution is weakly ionized, (Kb is ~1.8 * 10-5 at 293 K), there is little $\ce{NH4OH}$ present at a given instant, so ammonia water is only weakly alkaline. One might say the answer $\ce{NH3 + H2O}$ is 99.998% correct, but $\ce{NH4OH}$ is still 0.002% correct, not entirely wrong. ...


1

It is postulated that NH4OH as a molecule does not exist in solution. A better way to write might be NH3(aq). As to the evolution of gaseous ammonia, this reaction would only proceed if you heat up (i) a moist paste of NH4Cl and Ca(OH)2 or (ii) strongly heat a solution, otherwise ammonia would remain dissolved in the solution.


4

Steels are nitrided in ammonia gas at 900 to 1050 F ; It forms a very hard , very thin ( < 0.01 ") hard layer of iron nitride . An addition of aluminum alloy to the steel ( 0.5 to 1.0 % ) enhances the nitriding. Nitriding is applied typically to cutting tools like drill bits and wearing surfaces. At higher temperatures "carbo-nitriding " is done where C ...


0

There is no general rule, just partial empirical rules for particular anions, which can be memorized from textbooks. There is theoretical reasoning why salts are soluble or not, concerning the energy of the crystal lattice, energy of ion hydration, entropy change etc. But easier way is to go for empirical rules and "chemical sense".


1

We should not extend the idea of electrode potentials for very high concentrations such as concentrated sulfuric acid which is on the order of 18 M. The interesting property of these acids (HNO3 or H2SO4) is that they behave as oxidizing agents or even as dehydrating agents (H2SO4) especially at high temperature and at high concentrations. This is not to be ...


2

You asked this historical question before (check Sci math history which is pretty good). For historical searches, Google scholar works the best. https://scholar.google.com/scholar?hl=en&as_sdt=0%2C44&q=lassaigne+test&btnG=&oq=lassaign The very first article is very useful on the history of sodium fusion test. "A lost centenary: Lassaigne's ...


1

Historical availability and procedural compliance. Why to use potassium or lithium if the procedure says sodium ? Plus, lithium is significantly more expensive and reacts with gaseous nitrogen. Potassium may be too reactive for manipulation and reaction with some samples.


18

Equilibrium will be far to the right, as a stable six membered ring is formed. The proton speaks for itself. References: Queen, A. The kinetics of the reaction of boric acid with salicylic acid. Can. J. Chem. 1977, 55 (16), 3035–3039 DOI: 10.1139/v77-421.


3

By googling "boric acid salicylic acid" I have found the salicylic acid acts like it was a diol toward the boric acid, but just 1:1. https://www.nrcresearchpress.com/doi/abs/10.1139/v77-421


4

1.20g of hydrated tin chloride decompose to form 1.01g of anhydrous tin chloride on heating. Calculate the value of x 1.20 Stannous chloride hydrate = 1.01 anhydrous stannous chloride So 1.20 - 1.01 = 0.19 grams of water Moles of anhydrous stannous chloride $$\dfrac{1.01}{189.60} = 0.0053270$$ Moles of water $$\dfrac{0.19}{18} = 0.01065$$ ratio of ...


2

There is no special unit analysis. It's a recommended practice to solve the problem algebraically first using proper notations for physical quantities, and plug the numeral values at the end minding the units — this way you reduce the chance of making the erroneous calculations and keep track of all units; as a bonus, you simplify handling significant ...


1

The Unit/dimension analysis is in this case rather "using a cannon against sparrows". The first spotted error is $\pu{500 g}$ of $\ce{HCl}$. Where did it come from ? Note that the molar mass of $\ce{HCl}$ is about $\pu{36.5 g / mol}$. Get the molar mass of $\ce{NaOH}$. Calculate the amount of mols of $\ce{NaOH}$. Calculate the equivalent amount of mols of ...


3

You have rightfully assumed the reaction takes place in either basic or neutral medium, otherwise $\ce{Cr(OH)3}$ wouldn't be the product. This automatically suggests $\ce{H+}$ are not used for balancing half reactions. Your solution is fine for the provided reaction given that chromium(III) hydroxide stays precipitated. Here I've basically formatted your ...


0

Add up the reactions (double the second one so there is sufficient hydroxide) to get a net reaction with your starting materials as reactants and your desired product as one of the products: $$\ce{FeSO4.7H2O(aq) + 2KOH(aq) + KNO3(aq) -> Fe3O4(s) + NO2-(aq) + ...}$$ Then, find the limiting reactant, calculate the theoretical yield from it, and compare it ...


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