ÌâÄ¿ÄÚÈÝ
ijУ¿ÎÍâ»î¶¯Ð¡×éµÄͬѧÃÇÀûÓÃË®µÄµç½âʵÑé̽¾¿Ë®µÄ×é³É£¬ËûÃÇÁ¿È¡192.7mLË®£¨Ë®µÄÃܶÈΪ1.00g/cm3£©£¬²¢ÏòË®ÖмÓÈëÁË7.3gÇâÑõ»¯ÄƹÌÌ壬³ä·ÖÈܽâºó°´ÏÂͼËùʾµÄ×°ÖýøÐÐʵÑ飬½ÓͨµçÔ´£¬µ±A¹ÜÊÕ¼¯µ½22.3mLÆøÌ壨ÆøÌåÃܶÈΪ0.09g/L£©Ê±£¬Í£Ö¹ÊµÑé¡£Çë·ÖÎö»Ø´ð£º
£¨1£©Ë®Í¨µç·Ö½âµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ£º
£¨2£©ÊµÑéÖÐA¹ÜÉú³ÉµÄÆøÌåÖÊÁ¿ÊÇ £»
£¨3£©ÏòË®ÖмÓÈëÇâÑõ»¯ÄƹÌÌåµÄÄ¿µÄÊÇ
£»
£¨4£©Í£Ö¹ÊµÑéʱʣÓà¡°Ë®ÖС±ÇâÑõ»¯ÄƵÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿£¨ÁÐʽ¼ÆË㣩
£¨1£©Ë®Í¨µç·Ö½âµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ£º
£¨2£©ÊµÑéÖÐA¹ÜÉú³ÉµÄÆøÌåÖÊÁ¿ÊÇ £»
£¨3£©ÏòË®ÖмÓÈëÇâÑõ»¯ÄƹÌÌåµÄÄ¿µÄÊÇ
£»
£¨4£©Í£Ö¹ÊµÑéʱʣÓà¡°Ë®ÖС±ÇâÑõ»¯ÄƵÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿£¨ÁÐʽ¼ÆË㣩
(1)2H 2O 2H2¡ü+O2¡ü?
(2)22.3mL¡Á0.09g/L¡Á10-3=0.002g?
(3)ÔöÇ¿¡°Ë®¡±µÄµ¼µçÐÔ?
(4) ¡Á100%=3.65% £¨2·Ö£©
(2)22.3mL¡Á0.09g/L¡Á10-3=0.002g?
(3)ÔöÇ¿¡°Ë®¡±µÄµ¼µçÐÔ?
(4) ¡Á100%=3.65% £¨2·Ö£©
£¨1£©µç½âË®µÄ»¯Ñ§·½³ÌʽΪ2H 2O 2H2¡ü+O2¡ü£»£¨2£©m=¦ÑV=22.3mL¡Á0.09g/L¡Á10-3="0.002g;" £¨3£©´¿¾»µÄË®ÊDz»µ¼µçµÄ£¬¼ÓÈëNaOH¹ÌÌåÄ¿µÄ¾ÍÊÇÔöÇ¿¡°Ë®¡±µÄµ¼µçÐÔ£»£¨4£©µç½â²úÉú0.002gH2ÐèҪˮµÄÖÊÁ¿Îª 0.002g ¡Á9=0.018g£¬Ôò·´Ó¦ºóÈÜÒºµÄÖÊÁ¿=192.7mL¡Á1.00g/cm3+7.3g-0.018g=199.982g£¬ÔòNaOHµÄÖÊÁ¿·ÖÊýΪ
7.3g£¯199.982g¡Á100%="3.65%" ¡£
7.3g£¯199.982g¡Á100%="3.65%" ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿