ÌâÄ¿ÄÚÈÝ
£¨2008?ÇàÆÖÇø¶þÄ££©ÔÚ×öÖкͷ´Ó¦ÊµÑéʱ£¬ÎÒ½«Ï¡ÑÎËáµÎÈëÇâÑõ»¯ÄÆÈÜÒºÖУ¬ÒâÍâ¿´µ½ÆøÅݲúÉú£®Ð¡»ªÌáÐÑ£ºÊDz»ÊÇÄôíÁËÒ©Æ·£¿ÎÒ²éÑéºóÈ·ÈÏҩƷû´í£¬Ö»ÊÇÔÚÆ¿¿Ú·¢ÏÖÓа×É«·Ûĩ״ÎïÖÊ£®ÎÒÈÏΪÊÇÇâÑõ»¯ÄÆÈÜÒº±äÖÊÁË£®
£¨1£©ÇâÑõ»¯ÄÆÈÜÒº±äÖʵÄÔÒòÊÇ
£¨2£©ÀûÓÃÓëÉÏÊöʵÑ鲻ͬµÄÔÀí£¨²»ÓüÓÑÎËá·½·¨£©£¬ÎÒÓÖÉè¼ÆÁËÒ»¸öʵÑéÔÙ´ÎÈ·ÈϸÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ£®
д³ö£¨2£©ÖеĻ¯Ñ§·´Ó¦·½³Ìʽ
£¨3£©ÈçºÎÓøñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¿¼òÊö²Ù×÷²½Ö裺
£¨1£©ÇâÑõ»¯ÄÆÈÜÒº±äÖʵÄÔÒòÊÇ
2NaOH+CO2¨TNa2CO3+H2O
2NaOH+CO2¨TNa2CO3+H2O
£®£¨Ð´³öÓйػ¯Ñ§·½³Ìʽ£©£¨2£©ÀûÓÃÓëÉÏÊöʵÑ鲻ͬµÄÔÀí£¨²»ÓüÓÑÎËá·½·¨£©£¬ÎÒÓÖÉè¼ÆÁËÒ»¸öʵÑéÔÙ´ÎÈ·ÈϸÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ£®
ʵÑé²½Öè | ʵÑéÏÖÏó | ʵÑé½áÂÛ |
È¡ÉÙÁ¿ÇâÑõ»¯ÄÆÈÜÒºÓÚÊÔ¹ÜÖÐµÎ¼Ó ÂÈ»¯¸ÆÈÜÒº ÂÈ»¯¸ÆÈÜÒº £® |
ÈÜÒºÖвúÉú°×É«³Áµí ÈÜÒºÖвúÉú°×É«³Áµí |
¸ÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ£® |
Na2CO3+CaCl2=2NaCl+CaCO3¡ý
Na2CO3+CaCl2=2NaCl+CaCO3¡ý
£®£¨3£©ÈçºÎÓøñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¿¼òÊö²Ù×÷²½Ö裺
ÏòÈÜÒºÖеμÓÇâÑõ»¯¸ÆÈÜÒºÖ±ÖÁ²»ÔÙ²úÉú³ÁµíΪֹ£¬¹ýÂË£¬ËùµÃÂËҺΪÇâÑõ»¯ÄÆÈÜÒº£®
ÏòÈÜÒºÖеμÓÇâÑõ»¯¸ÆÈÜÒºÖ±ÖÁ²»ÔÙ²úÉú³ÁµíΪֹ£¬¹ýÂË£¬ËùµÃÂËҺΪÇâÑõ»¯ÄÆÈÜÒº£®
£®·ÖÎö£º£¨1£©¸ù¾ÝÇâÑõ»¯ÄÆÄÜÓë¿ÕÆøÖеĶþÑõ»¯Ì¼·¢ÉúÁË·´Ó¦À´·ÖÎö±äÖʵÄÔÒò£»
£¨2£©¸ù¾Ý̼ËáÄÆ¿ÉÒÔÓëÂÈ»¯¸Æ¡¢ÂÈ»¯±µÈÜÒºµÈ·´Ó¦ÄÜÐγɰ×É«³Áµí£¬¶øÇâÑõ»¯ÄƲ»ÄÜ·´Ó¦£¬À´¼ìÑéÇâÑõ»¯ÄƵıäÖÊ£»
£¨3£©ÓøñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¬¾ÍÊdzýÈ¥¸Ã±äÖʵÄÈÜÒºÖеÄ̼ËáÄÆ»ò°Ñ̼ËáÄÆת±äΪÇâÑõ»¯ÄÆÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£®
£¨2£©¸ù¾Ý̼ËáÄÆ¿ÉÒÔÓëÂÈ»¯¸Æ¡¢ÂÈ»¯±µÈÜÒºµÈ·´Ó¦ÄÜÐγɰ×É«³Áµí£¬¶øÇâÑõ»¯ÄƲ»ÄÜ·´Ó¦£¬À´¼ìÑéÇâÑõ»¯ÄƵıäÖÊ£»
£¨3£©ÓøñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¬¾ÍÊdzýÈ¥¸Ã±äÖʵÄÈÜÒºÖеÄ̼ËáÄÆ»ò°Ñ̼ËáÄÆת±äΪÇâÑõ»¯ÄÆÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£®
½â´ð£º½â£º£¨1£©ÇâÑõ»¯ÄÆÈÜÒºÄÜÎüÊÕ¿ÕÆøÖеĶþÑõ»¯Ì¼Éú³É̼ËáÄƺÍË®£¬Òò´Ë£¬Óë¿ÕÆø½Ó´¥µÄÇâÑõ»¯ÄÆÈÜÒº»á±äÖÊ£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2NaOH+CO2¨TNa2CO3+H2O£»
£¨2£©Ì¼ËáÑκÍËá·´Ó¦ÓÐÆøÌåÉú³É£¬ºÍº¬Ca2+£¨»òBa2+£©µÄÈÜÒº·´Ó¦»áÉú³É³Áµí£»ËùÒÔ£¬È·ÈϸÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖʵÄʵÑé²½Ö裺ȡÉÙÁ¿ÇâÑõ»¯ÄÆÈÜÒºÓÚÊÔ¹ÜÖеμÓÂÈ»¯¸ÆÈÜÒº£»ÊµÑéÏÖÏó£ºÈÜÒºÖвúÉú°×É«³Áµí£»¿ÉµÃ³ö½áÂÛ£º¸ÃÇâÑõ»¯ÄÆÒѱäÖÊ£»·´Ó¦µÄ·½³ÌʽÊÇ£ºNa2CO3+CaCl2=2NaCl+CaCO3¡ý£»
£¨3£©ÓñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¬¾ÍÊdzýȥ̼ËáÄÆ£¬ÏòÈÜÒºÖеμÓÇâÑõ»¯¸ÆÈÜÒºÖ±ÖÁ²»ÔÙ²úÉú³ÁµíΪֹ£¬¹ýÂË£¬ËùµÃÂËҺΪÇâÑõ»¯ÄÆÈÜÒº£®
¹Ê´ðΪ£º£¨1£©2NaOH+CO2¨TNa2CO3+H2O£»£¨2£©ÂÈ»¯¸ÆÈÜÒº£¬ÈÜÒºÖвúÉú°×É«³Áµí£¬Na2CO3+CaCl2=2NaCl+CaCO3¡ý£»£¨3£©ÏòÈÜÒºÖеμÓÇâÑõ»¯¸ÆÈÜÒºÖ±ÖÁ²»ÔÙ²úÉú³ÁµíΪֹ£¬¹ýÂË£¬ËùµÃÂËҺΪÇâÑõ»¯ÄÆÈÜÒº£®
£¨2£©Ì¼ËáÑκÍËá·´Ó¦ÓÐÆøÌåÉú³É£¬ºÍº¬Ca2+£¨»òBa2+£©µÄÈÜÒº·´Ó¦»áÉú³É³Áµí£»ËùÒÔ£¬È·ÈϸÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖʵÄʵÑé²½Ö裺ȡÉÙÁ¿ÇâÑõ»¯ÄÆÈÜÒºÓÚÊÔ¹ÜÖеμÓÂÈ»¯¸ÆÈÜÒº£»ÊµÑéÏÖÏó£ºÈÜÒºÖвúÉú°×É«³Áµí£»¿ÉµÃ³ö½áÂÛ£º¸ÃÇâÑõ»¯ÄÆÒѱäÖÊ£»·´Ó¦µÄ·½³ÌʽÊÇ£ºNa2CO3+CaCl2=2NaCl+CaCO3¡ý£»
£¨3£©ÓñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¬¾ÍÊdzýȥ̼ËáÄÆ£¬ÏòÈÜÒºÖеμÓÇâÑõ»¯¸ÆÈÜÒºÖ±ÖÁ²»ÔÙ²úÉú³ÁµíΪֹ£¬¹ýÂË£¬ËùµÃÂËҺΪÇâÑõ»¯ÄÆÈÜÒº£®
¹Ê´ðΪ£º£¨1£©2NaOH+CO2¨TNa2CO3+H2O£»£¨2£©ÂÈ»¯¸ÆÈÜÒº£¬ÈÜÒºÖвúÉú°×É«³Áµí£¬Na2CO3+CaCl2=2NaCl+CaCO3¡ý£»£¨3£©ÏòÈÜÒºÖеμÓÇâÑõ»¯¸ÆÈÜÒºÖ±ÖÁ²»ÔÙ²úÉú³ÁµíΪֹ£¬¹ýÂË£¬ËùµÃÂËҺΪÇâÑõ»¯ÄÆÈÜÒº£®
µãÆÀ£º±¾ÌâÊôÓÚʵÑé̽¾¿Ì⣬±¾ÌâÔÚ̽¾¿Ê±¾ßÓнϺõĿª·ÅÐÔ£¬Ñ¡ÔñµÄÊÔ¼Á²»Î¨Ò»£¬Ñ§Éú¿ÉÄÜÓв»Í¬µÄ´ð°¸£¬Ö»ÒªÉè¼ÆºÏÀí£¬ÏÖÏóÃ÷ÏÔ£¬½áÂÛÕýÈ·¶¼ÊÇÕýÈ·´ð°¸£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿