ÌâÄ¿ÄÚÈÝ

×øÂäÓÚãå´¨³Ç½¼µÄËÄ´¨ÓÀ·áÖ½Òµ¹É·ÝÓÐÏÞ¹«Ë¾ÊÇËÄ´¨¹æÄ£×î´óµÄÁÖ½¬Ö½²úÒµ¼¯ÍÅ£®ÔÚ×÷ΪԭÁϵÄÖñƬ¡¢Ä¾²Ä¡¢µ¾²ÝÖмÓÈëÏ¡NaOHÈÜÒº£¬½«·ÇÏËά³É·ÝÈܽ⣬¾­¹ý¹ýÂË·ÖÀëµÃµ½´¿ÏËάֽ½¬£®½áºÏÄãѧµ½µÄ»¯Ñ§ÖªÊ¶»Ø´ðÏÂÃæÓйØÎÊÌ⣮
£¨1£©Ö½µÄÖ÷Òª³É·ÖÊÇÏËÎ¬ËØ£¬Ö÷ÒªÓÉC¡¢H¡¢OÔªËØ×é³É£¬ÔòÏËÎ¬ËØÊôÓÚ
 
£®£¨Ñ¡Ìî¡°ÎÞ»úÎ¡¢¡°ÓлúÎ£©
£¨2£©µãȼһÕÅÖ½£¬ÓÃÒ»¸ö¸ÉÀäµÄÉÕ±­ÕÖÔÚ»ðÑæÉÏ·½£¬¿´µ½ÉÕ±­±ÚÓÐСˮÖéÉú³É£¬Ö¤Ã÷Ö½Öк¬ÓÐ
 
ÔªËØ£®
£¨3£©Ö½½¬³ØÖÜΧ¿ÕÆøÖг£ÃÖÂþÓж¾ÆøÌå¶þÑõ»¯Áò£¬¸ÃÆøÌåÒ×ÓëË®·´Ó¦Éú³ÉÑÇÁòËᣨH2SO3£©£¬¶þÑõ»¯ÁòÆøÌåÈܽâÓÚˮʱÖ÷Òª·¢Éú
 
±ä»¯£®
£¨4£©Ï¡NaOHÈÜÒº³¨·Å¿ÕÆøÖлá±äÖÊ£¬Ð´³öÆä»¯Ñ§·´Ó¦·½³Ìʽ£º
 
£®
¿¼µã£ºÓлúÎïÓëÎÞ»úÎïµÄÇø±ð,¼îµÄ»¯Ñ§ÐÔÖÊ,»¯Ñ§±ä»¯ºÍÎïÀí±ä»¯µÄÅбð,ÖÊÁ¿Êغ㶨Âɼ°ÆäÓ¦ÓÃ,Êéд»¯Ñ§·½³Ìʽ¡¢ÎÄ×Ö±í´ïʽ¡¢µçÀë·½³Ìʽ
רÌ⣺ÎïÖʵı仯ÓëÐÔÖÊ,»¯Ñ§ÓÃÓïºÍÖÊÁ¿Êغ㶨ÂÉ,ÎïÖʵķÖÀà
·ÖÎö£º£¨1£©¸ù¾ÝÓлúÎïµÄ¸ÅÄî·ÖÎö£»
£¨2£©¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉ·ÖÎö£»
£¨3£©¸ù¾Ý»¯Ñ§±ä»¯µÄ¸ÅÄî·ÖÎö£»
£¨4£©ÇâÑõ»¯ÄƱäÖʵÄÔ­Òò·ÖÎö£®
½â´ð£º½â£º£¨1£©ÓлúÎïÊÇÖ¸º¬ÓÐÌ¼ÔªËØµÄ»¯ºÏÎÏËÎ¬ËØµÄ×é³É·ûºÏÓлúÎïµÄ¸ÅÄ
£¨2£©Ö½È¼ÉÕÉú³ÉË®£¬Ë®ÖеÄÇâÔªËØÒ»¶¨À´×ÔÖ½£¬¹ÊÖ¤Ã÷Ö½Öк¬ÓÐÇâÔªËØ£»
£¨3£©¶þÑõ»¯ÁòÈܽâÓÚË®Éú³ÉÑÇÁòËᣬÉú³ÉÁËÐÂÎïÖÊ£¬¹ÊÊôÓÚ»¯Ñ§±ä»¯£»
£¨4£©ÇâÑõ»¯ÄƱäÖÊÊÇÒòΪÇâÑõ»¯ÄÆºÍ¿ÕÆøÖжþÑõ»¯Ì¼·´Ó¦Éú³ÉÁË̼ËáÄÆºÍË®£¬·´Ó¦µÄ·½³ÌʽΪ£º2NaOH+CO2=Na2CO3+H2O£®
¹Ê´ð°¸Îª£ºÓлúÎÇ⣻»¯Ñ§£»2NaOH+CO2=Na2CO3+H2O£®
µãÆÀ£ºÊìÁ·ÕÆÎտα¾ÖеĻù±¾¸ÅÄ³£¼û»¯Ñ§ÎïÖʵÄÐÔÖÊ£¬¿ÉÇáËɽâ´ð´ËÌâ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø