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OH. so they want us to memorize the spectrum..that sounds weird. but anyway thanks
thanks..plz see if you can help me out with this one..I know.. but we have to learn it in quantum physics too [continuos and line spectra].
This will give you are rough estimate:
V I B G Y O R
400 - 700 nm
Graphs? Which paper and variant are we talking about?oh really ? i took a lot of time to draw the graphs !!!
look at the E0 values from booklet the more positive E0 cation discharge at cathode ...............http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_43.pdf
Q3c..plz tell me how do we dedude the products??
SOMEONE PLZ ANS MY QUESTION.........http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_4.pdf
tell me q1 a(ii) plz
E=RED-OXI
SHOULDN'T IT BE 1.36-.77
umm so what abt fluorine..its E0 value isnt given in the data booklet.look at the E0 values from booklet the more positive E0 cation discharge at cathode ...............
The degenerate d-orbitals split because of the existent repulsion between the electrons. This helps to form transition-metal-complexes. While forming a complex with a ligand the d-orbitals interact with the surrounding electron cloud, causing it to become non-degenerate (variable energies). This causes photons to be absorbed by the electrons (which then undergo state transitions) to emit radiations with corresponding frequencies (f) to the energy (E) of the photons [where E=hf].
P.S. Electrons in d-orbitals are repelled from the electrons in the ligands, based on electrostatic interactions. This causes two of the d-orbitals (dz2 & dx2-y2) [which have a stronger (repulsive) interaction with the ligand] to be at higher energy than the other three (dxy, dyz, dxz).
u say electrons absorb photon. where exactly do theese photons cum from ? and where does the radiation they emit go to ? . thnx for the knowledge btw
look, for finding heat of formation when enthalpy changes of combustion are given, the formula to use is pretty simple:
heat of combustion of reactant elements - heat of combustion of product
so if u substitute the values given u'll get the answer like this:
2(-393.7)+ 2(285.9) - (-1411) = +51.8 KJ/mol
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