Oblicz ile gramów ołowiu znajduje się w 0,15 g antydetonatora o wzorze sumarycznym Pb(C₂H₅)₄
m Pb(C2H5)4 = 323uA Pb = 207u323u - 0,15g207u - xx = 0,1g Pb
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
m Pb(C2H5)4 = 323u
A Pb = 207u
323u - 0,15g
207u - x
x = 0,1g Pb