题目内容

在无土栽培中需用浓度为0.5mol/LNH4Cl、0.16mol/LKCl、0.24mol/LK2SO4的培养液,若用KCl、NH4Cl和(NH42SO4三种物质来配制1.00L上述营养液,所需三种盐的物质的量正确的是 
A.0.4mol、0.5mol、0.12molB.0.66mol、0.5mol、0.24mol
C.0.64mol、0.5mol、0.24molD.0.64mol、0.02mol、0.24mol
D
0.5mol/LNH4Cl、0.16mol/LKCl、0.24mol/LK2SO4的混合溶液,设溶液体积为1L,所含离子的物质的量为NH4:0.5mol,K:0.64mol,Cl:0.56mol,SO42:0.24mol,若用KCl、NH4Cl和(NH42SO4三种物质来配制1.00L上述营养液,需用0.64mol KCl、0.02molNH4Cl、0.24mol(NH42SO4,故选D
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