haoguci2007 haoguci2007

29-10-2021
Chemistry

contestada

S + 02
C + 02
Cu0 + H2SO4

Respuesta :

claremisner claremisner

29-10-2021

Answer:

S + 02= S0 - 4 e- → SIV (oxidation); 2 O0 + 4 e- → 2 O-II (reduction)

C + 02= C

Cu0 + H2SO4= CuO + H2SO4 → CuSO4 + H2O

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