# Tag Info

## New answers tagged stability

0

No. It does not. In order to understand this, you need to have this basic knowledge of the kinetics of multistep reactions In chemical kinetics, there are two ways to deal with multiple-step mechanisms Rate determining step method- Here one specific step is the slowest. So we consider all steps after this step to be equally fast. The rate-determining ...

1

I think the only real option for this is a clear plastic sheet that's had a chemical resistant thin film applied to the faces. One commercially available example is Makrolon AR2, it's polycarbonate with some proprietary coating that's listed as having >24hrs resistance to acetone and a few other chemicals.

-2

It is due to the difference in their second and third ionisation enthalpies. Due to the large difference in the second and third ionisation enthalpies arising from the disturbal of the stable half filled electronic configuration, +2 oxidation state is more favourable.

0

Hyperconjugation involves the electrons from an adjacent bond to "donate" electrons to an empty p orbital. Now if we consider a filled p orbital, or a carbo-anion, then there will be a repulsive force according to Coulomb's law. This will destabilize the molecule in question. However, resonance involves the delocalization of electrons in a molecule; whether ...

1

In carbanion the carbon carrying negative charge contains one $\mathrm{sp^3}$ hybrid orbital which contains a pair of electrons and therefore there is no possibility that it will undergo bonding interaction with the neighbouring $\ce{C-H}$ bond, due to repulsion between the electrons.

2

According to Palstics International's Chemical Resistance Chart, the A-rated plastics (no solvent attack) towards acetone are: ECTFE (Halar®): transparent films available Fluorosint® PTFE: white HDPE: transparent films available Nylon®, Type 6/6: white PP: clear sheets available PPS: opaque white PTFE: white Among these, polypropylene appears to be the ...

0

Other answers gave focused on the thermodynamics involved. But if we are focused on nucleophilic activity, which has a kinetic component, then indeed sulfur bases are more nucleophilic. This source explains that sulfur is less electronegative than oxygen so its electrons are "more available", which I take to mean more polarizable. More polarizability ...

1

$\ce{RSH}$ is a better acid than $\ce{ROH}$ (as the $\mathrm{p}K_\mathrm{a}$ of $\ce{RSH}$ is lower than the $\mathrm{p}K_\mathrm{a}$ of $\ce{ROH}$, shown in a previous answer). This means that $\ce{RSH}$ dissociates into $\ce{RS-}$ and $\ce{H+}$ more, i.e. the equilibrium below lies more to the right. $\ce{RSH + H2O <=> RS- + H3O+}$ To compare ...

3

There are several things which are needed to be known to be able to deal with this question. Firstly there is the 18 Valence Electron (18VE) rule (description at libretexts.org). The iron pentacarbonyl is an 18VE complex. It is coordinatively saturated. A low valent transition metal complex with strong ligand field ligands is considered to be coordinatively ...

1

While solubility, thermal stability and density are pretty hard to explain as @maurice said. But, there is good logic for the solubility of group 2 hydroxides. The lattice energy of a solid is inversely proportional to the radius ratio (as lower the radius ratio greater the packing efficiency). So, hydroxide being a small ion, makes a more stable aqueous ...

2

Nobody is able to foresee the solubility of a product. There are some experimental rules, but they all have exceptions, that nobody is able to explain. Just have a look on the Calcium salts made with the halogens (F, Cl, Br, I). There is a nice analogy among Cl, Br and I, but not F. Look ! The Calcium chloride $\ce{CaCl2}$, bromide $\ce{CaBr2}$ and iodide \$\...

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