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不同组份出现的不同现象...
不同组份出现的不同现象
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展开全部反应结果:1、完全沉淀（PH=6~7）反应式：2NaOH+CuSO4=Cu(OH)2↓+Na2SO4此时，硫酸铜与氢氧化钠物质的量之比近似1：2。所得反应产物为蓝色絮奖沉淀，久置后上层溶液无色，即生成氢氧化铜沉淀与水。2、硫酸铜过量（PH值<6）反应式：4CuSO4+6NaOH=Cu4(OH)6SO4+3Na2SO4在试管内装1/3体积的硫酸铜后，滴入4~5滴氢氧化钠溶液，生成兰绿色块状沉淀。溶液久置后呈蓝色，经查证，兰绿色沉淀即为碱式硫酸铜。3、氢氧化钠过量（PH>7）反应式：4NaOH+CuSO4=Na2[Ca(OH)4]↓+Na2SO4我们在装了1/3试管的氢氧化钠后再滴入3~4滴硫酸铜溶液。久置后，上呈无色，而沉淀层为一种蓝白杂泪的糊状沉淀，此为铜酸钠。生成原因及生成物性质1、氢氧化铜[Cu(OH)2]这是实验最简单的结果，生成原因是一个铜离子正好与两个氢氧化铜。其为一种蓝色固体。难溶于水，加热易分解，微显两性。2、碱式硫酸铜。[Cu4(OH)6SO4]在实验中，氢氧化钠处于不足，故氢氧离子相对铜离子也就不足，使氢氧根离子无法完全结合所有铜离子，故它代替一部分硫酸根离子，生成碱式硫酸铜。这是一种兰绿色固体，受热可分解（生成硫酸铜和氢氧化铜），但受热时间比氢氧化钠分解的时间还要长，生成蓝色与黑色并存浊液[Cu4(OH)6SO4=CuSO4+3Cu(OH)2，Cu(OH)2=CuO+H2O]并且，碱式硫酸铜可溶于无机酸（实验用盐酸和稀硫酸），有机酸（实验用苯酚和乙酸），碱（实验用氢氧化钠和氢氧化钙）等。它还会与氨水反应，生成钢蓝色溶液，有氨味[Cu4(OH)6SO4+8NH3=3Cu(NH3)2(OH)2+Cu(NH3)2SO4]3、铜酸钠[Na2Cu(OH)4]它的形成原因相当复杂，删繁就简地就是氢氧与铜离子结合成氢氧化铜后，由于溶液中氢所氧根偏多，而氢氧化铜又微显两性（氢氧化铜可写为Cu(OH)2及H2CuO2）故又结合两个氢氧根离子，再与钠离子结合生成铜酸钠。这是一种相当不稳定的蓝白色物质，难溶于水，它只有在碱性条件下才能存在，溶液稀释后，它即分解成氢氧化铜和氢氧化钠[Na2Cu(OH)4=2NaOH+CU(OH)2↓]，在久置之后，它会风化分解[Na2Cu(OH)4=2NaOH+CU(OH)2↓]。它会与酸反应，例如与稀盐酸反应生成氧气与钠盐，与浓盐酸反应生成氯气和氧气。已赞过已踩过你对这个回答的评价是？评论
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0:-1;if(n!==r)for(var i=0;i0&&function(t,e,n,r){var i=document.getElementsByClassName(t);if(i.length>0)for(var o=0;o展开全部当pH显中性时，生成蓝色氢氧化铜沉淀。当pH小于6，即硫酸铜过量，它便生成碱式硫酸铜。而当氢氧化纳过量时，即生成不稳定的蓝白相间的铜酸钠沉淀。展开全部NaOH不足时生成蓝色沉淀NaOH过量时会生生络合物,易溶于水,它也就是高三生物实验中常用到的斐林试剂与双缩脲试剂,不过这两种试剂中NaOH多出的比例不同.
展开全部生成蓝色絮状沉淀。展开全部我想想反应后生成蓝色絮状沉淀 也就是Cu(OH)2
Cu(OH)2属于不溶性的金属氢氧化物,受热分解为对应价态的氧化物和水
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