某小组同学为探究分子的特征,进行了以下三个实验,请你参与其中,并回答下列问题:探究实验1:两位同学

2025-04-03 03:58:58
推荐回答(1个)
回答(1):


探究实验1:因为分子之间有间隔,酒精分子和水分子的大小不同,最后会出现混合后的体积比混合前的总体积小的情况.
探究实验2:分别在盛有同量的热水和冷水的两个烧杯中,加入一粒品红,观察到热水中的品红比冷水中的扩散更快,说明温度升高,分子的运动速率加快;
探究实验3:熟记:“正氧负氢,氢2氧1”来处理:
①用带火星木条接近液面下降较少的玻璃管尖嘴处,推测为氧气,故慢慢打开活塞,观察到的现象是木条复燃;
②用燃着木条接近液面下降较多的玻璃管尖嘴处,推测为氢气,故慢慢打开活塞,观察到的现象是气体被点燃,火焰呈淡蓝色,有爆鸣声;
③因为水通电生成氢气和氧气,水的电解是化学变化,因为有新物质氢气和氧气生成,故可以写出水通电发生变化的文字表达式 水

通电
氢气+氧气;
④该实验理论上产生氢气与氧气的体积之比应为2:1.但实际观察到氢气与氧气的体积之比略大于2:1,猜想是 氢气比氧气在水中溶解更少,或氧气比氢气在水中溶解多,或电解器漏气等;
⑤该实验证实了分子的特征是分子在化学变化中可以再分,而原子不能再分.
故答案为:
探究实验1:小于;分子间存在间隔;
探究实验2:热水与冷水的量相同,加入品红的量及颗粒大小相同;相同时间内热水中品红周围出现红晕要比冷水中品红周围出现红晕大;
实验探究3:①木条复燃; ②气体被点燃,火焰呈淡蓝色,有爆鸣声;③水
通电
氢气+氧气; 
④氢气比氧气在水中溶解更少,或氧气比氢气在水中溶解多,或电解器漏气等;⑤分子在化学变化中可以再分.

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