高一化学--用氯化铵,氯化钾,硫酸钾配制营养液
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(NH4)2SO4三种物质来配制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,0.12mol B0.66mol,0.5mol,0.24mol
C.0.64mol,0.5mol,0.24mol D.0.64mol,0.02mol,0.24mol
人气:241 ℃ 时间:2020-04-10 05:03:02
解答
1L溶液中NH4CL,KCL,K2SO4的浓度分别是0.5mol/L,0.16mol/L,0.24mol/L,因此他们的物质的量分别是0.5mol,0.16mol,0.24mol,所以,SO4的物质的量是0.24mol,NH4的物质的量是0.5mol,CL的物质的量是0.5+0.16=0.66mol,K的物质...
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