摘要:1 mol/L.阳极得到336 mL气体中.含0.01 mol Cl2和0.005 mol O2.转移电 子的物质的量为:0.01 mol×2+0.005 mol×4=0.04 mol. 此过程中阴极刚好析出全部铜: n(CuSO4)=n(Cu)==0.02 mol. 则c(CuSO4)==0.1 mol/L. (2)t2时溶液中c(Na+)=0.1 mol/L.c(SO42-)=0.1 mol/L. 根据电荷守恒有:c(H+)=2×0.1 mol/L-0.1 mol/L=0.1 mol/L.即溶液的pH=1. 答案:=0.1 mol/L.c(CuSO4)=0.1 mol/L

网址:http://m.1010jiajiao.com/timu3_id_81424[举报]

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网