2 years ago

S + 02 C + 02 Cu0 + H2SO4

Chemistry

1 answer:

vlabodo [156]2 years ago

6 0

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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