Chemistry
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6 months agoContributor-Level 9
o Chlorophyll: Magnesium present in chlorophyll.
o Vitamin B? : Cobalt (cyanocobalamin).
o Anticancer drug: Platinum (Coordination compound of platinum).
o Grubbs catalyst: Ruthenium.
New answer posted
6 months agoContributor-Level 10
The addition of HBr to H? C−CH? −CH=CH? proceeds via carbocation formation.
Formation of a 1° Carbocation (A): H? C−CH?
Formation of a 2° Carbocation (B): H? C−CH? −C? H−CH?
The 2° carbocation (B) is more stable than the 1° carbocation (A). Therefore, the activation energy (Ea) for the formation of B is lower, and B is formed faster.
New answer posted
6 months agoContributor-Level 9
o Alcoholic potassium hydroxide (Alc KOH):- used for β - elimination.
o Pd / BaSO? : - Lindlar's Catalyst.
New answer posted
6 months agoContributor-Level 10
A diagonal relationship is observed in the periodic table between elements of period (2) and period (3).
Li and Mg
Be and Al
B and Si
Li and Na do not show a diagonal relationship.
New answer posted
6 months agoContributor-Level 9
When we add cryolite (Na? AlF? ) in the extraction of aluminium, the melting point of alumina decreases
New answer posted
6 months agoContributor-Level 10
For the coagulation of a negative sol, the species Ba²? has the highest flocculating power (referring to the Hardy-Schulze rule, where higher charge leads to greater coagulation power).
New answer posted
6 months agoContributor-Level 9
Quantum Numbers and Orbitals:
o Total Node = n - 1 (l => Azimuthal Q.N)
o Radial Node = n - l - 1
o Angular Node: l
o If Angular node = 0, then l = 0, i.e., S orbital.
o If Radial nodes = 2, then n - l - 1 = 2.
o Substituting l = 0 gives n - 0 - 1 = 2, so n = 3.
o Therefore, the orbital is 3s.
New answer posted
6 months agoContributor-Level 10
Haematite: Fe? O?
Bauxite: Al? O? ·xH? O
Magnetite: Fe? O?
Malachite: CuCO? ·Cu (OH)?
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