选择AD
解析:
A:C12H22O11+H2O→C6H12O6(葡萄糖)+C6H12O6(果糖)
B:C12H22O11+H2O→2C6H12O6(葡萄糖)
C:CH3COOC2H5+H2O→CH3COOH+C2H5OH
D:HCOOC2H5+H2O→HCOOH+C2H5OH
D
甲酸和乙醇
HCOOCH2CH3 + H2O =(可逆) HCOOH + CH3CH2OH
d
稀硫酸
chooch2ch3=======hcooh+ch3ch2oh
C和D
C: CH3-CO-O-CH2-CH3 + H2O = CH3-CO-OH + CH3-CH2-OH
D: HCO-O-CH2-CH3 + H2O = HCO-OH + CH3-CH2-OH
A
A C12H22O11(蔗糖)+ H2O —H+ → C6H12O6(果糖)+ C6H12O6(葡萄糖)
B C12H22O11(麦芽糖)+ H2O —H+ → 2C6H12O6(葡萄糖)
C CH3COOC2H5 + H2O —H+ → CH3COOH + C2H5OH
D HCOOC2H5 + H2O —H+ → HCOOH + C2H5OH
C D 比较明显 生成的两种物质式量不同
B 生成的是一种物质。。。。
而A 是“式量相同”的“两种”物质