Can anybody please explain the Zn Mole fraction and mole percent based on
x = 0.0, 0.1, 0.2 for
Co(1-x)ZnxFe2O4 nanoparticles
Dear Willey John,
Thanks a lot. But in the CoZn0.9FeO4 ,please explain about mole percent of Zn Doping percent ( namely 0.9).
Regards,
Abbas Rahdar
Molar ratio can be found by this way:
Co/Zn = x / 1- x
For any x value, you should compute Co/Zn molar fraction.
Dear Satish Bolloju ,
Thanks.i want to synthesis Zn1−xCuxS nanoparticles that Cu is doping agent with dopant content ranging from x=0.00,0.05, 0.10 .
I prepare 0.1 M CuCl2 50 ml ,So for x=0.00 ,1.oo ,2.00 , how much CuCl2 need in ml,respectively?
please answer for 0.1 M CuCl2 50 ml and x=0.00 ,1.oo ,2.00 , how much CuCl2 need in ml,respectively.
Dear Fernando Acosta-Humánez,
for Zn1−xCuxS nanoparticles , you said Cu/Zn = x / 1-x (Cu is doping agent )and for Co(1-x)ZnxFe2O4 nanoparticles you said Co/Zn = x / 1- x (Zn is doping agent ) . please explain this difference?
Dear Mr. Abbas Rahdar,
Did you get your problem solved, please?
I am here to help you. Please let me know if you still need this help. I am looking forward to hearing from you.
Dear Sir, still it is not clear that how to calculate atomic percent doping and mole percent doping
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