Oblicz skład procentowy kwasu węglowego H2CO3 Proszę ze wzorem na dziś

mH₂CO₃ - 2 x 1u + 12u +3 x 16u = 62u   mH = 2u mC = 12u mO = 48u   %H = 2u/62u x 100% = 3,23% %C = 12u/62u x 100% = 19,35% %O = 48u/62u x 100% = 77,42%     ---------------------------------------------------------------------------------------------------- Litterarum radices amarae sunt, fructus iucundiores   Pozdrawiam :)  

m H2CO3 = 2*1u + 12u + 3*16u = 62u m 2H = 2*1u = 2u m C = 12u m3O = 3*16u= 48u %H = 2/62*100% = 3,2% %C = 12/62*100% = 19,4% %O = 48/62*100% = 77,4%