往四份相同的不饱和NaOH溶液中分别加入一定量的NaOH、Na202、Na2O、Na问题往四份相同的不饱和NaOH溶液中分别加入一定量的NaOH、Na202、Na2O、Na,保持温度不变,使溶液均恰好饱和.比较所加四种物

往四份相同的不饱和NaOH溶液中分别加入一定量的NaOH、Na202、Na2O、Na问题往四份相同的不饱和NaOH溶液中分别加入一定量的NaOH、Na202、Na2O、Na,保持温度不变,使溶液均恰好饱和.比较所加四种物
往四份相同的不饱和NaOH溶液中分别加入一定量的NaOH、Na202、Na2O、Na问题
往四份相同的不饱和NaOH溶液中分别加入一定量的NaOH、Na202、Na2O、Na,保持温度不变,使溶液均恰好饱和.比较所加四种物质的质量大小 物质的量大小
请说一下你的解法或思路

往四份相同的不饱和NaOH溶液中分别加入一定量的NaOH、Na202、Na2O、Na问题往四份相同的不饱和NaOH溶液中分别加入一定量的NaOH、Na202、Na2O、Na,保持温度不变,使溶液均恰好饱和.比较所加四种物
答案是：NaOH

2Na+2H2O===2NaOH + H2
产生1molNaOH消耗1mol水，需要加23克钠，消耗水溶解NaOH较少
Na2O + H2O == 2NaOH
产生1molNaOH消耗0.5mol水，需要加31克Na2O克，消耗水溶解NaOH较少
2Na2O2 + 2H2O == 4NaOH + O2
产生1molNaOH消耗0.5mol水，需要加39克过...

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2Na+2H2O===2NaOH + H2
产生1molNaOH消耗1mol水，需要加23克钠，消耗水溶解NaOH较少
Na2O + H2O == 2NaOH
产生1molNaOH消耗0.5mol水，需要加31克Na2O克，消耗水溶解NaOH较少
2Na2O2 + 2H2O == 4NaOH + O2
产生1molNaOH消耗0.5mol水，需要加39克过氧化钠，消耗水溶解NaOH较少
加入1mol氢氧化钠不消耗水，加入40克。
物质的量大小：NaOH大于Na大于Na2O=Na2O2
所加四种物质的质量大小：NaOH大于Na2O2大于Na2O大于Na

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本题比较麻烦，可以这样考虑，扣除掉不饱和NaOH溶液中饱和的那部分，剩下来就是水，然后分别加入一定量的NaOH、Na202、Na2O、Na，得到NaOH饱和溶液，换句话说，就是一定量的NaOH、Na202、Na2O、Na溶于水以后形成相同浓度的NaOH溶液。
设水的质量为a，加入的NaOH、Na202、Na2O、Na物质的量分别是x、y、z、w，溶于水以后产生的NaOH物质的量分别是x、2...

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本题比较麻烦，可以这样考虑，扣除掉不饱和NaOH溶液中饱和的那部分，剩下来就是水，然后分别加入一定量的NaOH、Na202、Na2O、Na，得到NaOH饱和溶液，换句话说，就是一定量的NaOH、Na202、Na2O、Na溶于水以后形成相同浓度的NaOH溶液。
设水的质量为a，加入的NaOH、Na202、Na2O、Na物质的量分别是x、y、z、w，溶于水以后产生的NaOH物质的量分别是x、2y、2z、w。溶液的总质量分别是a+40x、a+80y、a+80z、a+23w-1（根据化学反应方程式而来，不多说了）
所以根据质量分数相等得到：
40x/(a+40x)=80y/(a+62y)=80z/(a+62z)=40w/(a+23w-1)
即
x/(a+40x)=2y/(a+68y)=2z/(a+68z)=w/(a+23w-1)
由：x/(a+40x)=2y/(a+62y)→→a(x-2y)=18xy＞0→→x＞2y
由：2y/(a+62y)=2z/(a+62z)→→y=z
由：2z/(a+62x)=w/(a+23w-1)→→a(2z-w)=16wz+2z＞0→→2z＞w
所以物质的量的关系为：x＞2y=2z＞w
设质量分别为m1、m2、m3、m4
有关系：m1/40＞m2/39=m3/31＞m4/23

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Coco Chanel used to say that A woman with good shoes is never ugly,and They are the last touch of elegance. And she have done more for womens fashion than many other designers put together. Nowadays, ...

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