Câu 24:      BTKLC H O x xx y: 0,15180 342 6,126 12 6    Y C H O x yn 0, 01: y 2 0, 03Ag12 22 11 m0, 01.342 3, 42(%)C H O12 22 11áp án: C

     BTKLC H O X XX Y

Câu 24:      

BTKL

C H O x xx y: 0,15180 342 6,12

    Y C H O x y

 0, 01: y 2 0, 03

 m0, 01.342 3, 42(%)

C H O

22 11

áp án: C