Chemistry
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4 months agoNew answer posted
4 months agoContributor-Level 9
With weak field ligands Mn (II) will be of high spin and with strong field ligands it will be of low spin.
Ni (II) tetrahedral complexes will be generally of high spin due to sp³ hybridisation. Mn (II) is of light pink color in aqueous solution.
New answer posted
4 months agoContributor-Level 9
On moving Left to Right along a period.
Atomic Radius → decreases.
Electronegativity → Increases.
Electron gain enthalpy → Increases.
Ionisation Enthalpy → Increases
New answer posted
4 months agoContributor-Level 9
The vapour pressure of solution will be less than vapour pressure of pure solvent, so some vapour molecules will get condensed to maintain new equilibrium.
New answer posted
4 months agoContributor-Level 9
Bredig's Arc Method is used for preparation of colloidal sol's of less reactive metal like Au, Ag, Pt.
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4 months agoContributor-Level 9
Since spin only magnetic moment is 4.90BM so number of unpaired electrons must be 4, so If the complex is octahedral, then it has to be high spin complex with configuration t? g? e²g¹ in that case
CFSE = 4* (-0.4Δ? ) + 2*0.6Δ? = -0.4Δ?
If the complex is tetrahedral then its electronic configuration will be e? ², t? g² and CFSE will be = 3* (-0.6Δ? ) + 3* (0.4Δ? ) = -0.6Δ?
New answer posted
4 months agoContributor-Level 9
For ideal gas
PM = dRT
d = [PM]/R * 1/T
So graph between dV ST is not straight line
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4 months agoContributor-Level 9
In this acid base Titrating there is no use of Bunsen burner and measuring cylinder other laboratory equipments will be required for getting the end point of titration.
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