ÌâÄ¿ÄÚÈÝ

Сº£Í¬Ñ§ÔÚʵÑéÊÒÀï·¢ÏÖһƿƿ¸Çû¸ÇºÃµÄÇâÑõ»¯ÄÆÈÜÒº£®
£¨1£©ËûºÍѧϰС×éµÄͬѧһÆð¾ö¶¨¶ÔÕâÆ¿ÈÜÒºÊÇ·ñ±äÖʽøÐÐ̽¾¿£¬ÇëÄã°ïËûÃÇÍê³ÉʵÑ鱨¸æ£®
Ìá³öÎÊÌâ ÕâÆ¿ÇâÑõ»¯ÄÆÈÜÒºÊÇ·ñ±äÖÊ£¿
²ÂÏëÓë¼ÙÉè ÕâÆ¿ÇâÑõ»¯ÄÆÈÜÒºÒѾ­±äÖÊ£®
±äÖÊÀíÓÉÊÇ£º______£¬Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º______£®
ʵÑé̽¾¿ ²Ù×÷²½Ö裺ȡ³öÉÙÁ¿¸ÃÈÜÒºÓÚÒ»Ö§ÊÔ¹ÜÖУ¬ÔÙÍùÊÔ¹ÜÖеμÓ______£®
ʵÑéÏÖÏ󣺹۲쵽ÓÐ______£®
·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º______£®
½áÂÛ ÕâÆ¿ÇâÑõ»¯ÄÆÈÜÒºÒѾ­±äÖÊ£®
·´Ë¼ ÇâÑõ»¯ÄƱØÐë______±£´æ£®
£¨2£©ÈôÒª³ýÈ¥ÕâÆ¿ÇâÑõ»¯ÄÆÈÜÒºÖеÄÔÓÖÊ£¬Ê¹Ëü±ä³É´¿¾»µÄÇâÑõ»¯ÄÆÈÜÒº£¬Ð¡º£ËûÃÇӦѡÓõÄÊÔ¼ÁÊÇ______£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º______£®
£¨1£©ÓÉÓÚÇâÑõ»¯ÄÆÄÜÓë¿ÕÆøÖеĶþÑõ»¯Ì¼·´Ó¦Éú³ÉÁË̼ËáÄÆ£¬Ì¼ËáÄÆÄÜÓëÏ¡ÑÎËá·´Ó¦Éú³É¶þÑõ»¯Ì¼ÆøÌ壬ËùÒÔ£¬Ì½¾¿ÊµÑéÈçÏ£º
Ìá³öÎÊÌâ ÕâÆ¿ÇâÑõ»¯ÄÆÈÜÒºÊÇ·ñ±äÖÊ£¿
²ÂÏëÓë¼ÙÉè ÕâÆ¿ÇâÑõ»¯ÄÆÈÜÒºÒѾ­±äÖÊ£®
±äÖÊÀíÓÉÊÇ£ºÇâÑõ»¯ÄÆÔÚ¿ÕÆøÖÐÄÜÓë¶þÑõ»¯Ì¼·´Ó¦Éú³É̼ËáÄƺÍË®£¬Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2NaOH+CO2=Na2CO3+H2O£®
ʵÑé̽¾¿ ²Ù×÷²½Ö裺ȡ³öÉÙÁ¿¸ÃÈÜÒºÓÚÒ»Ö§ÊÔ¹ÜÖУ¬ÔÙÍùÊÔ¹ÜÖеμÓÏ¡ÑÎËᣮ
ʵÑéÏÖÏ󣺹۲쵽ÓÐÆøÅݲúÉú£®
·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºNa2CO3+2HCl=2NaCl+H2O+CO2¡ü£®
½áÂÛ ÕâÆ¿ÇâÑõ»¯ÄÆÈÜÒºÒѾ­±äÖÊ£®
·´Ë¼ ÇâÑõ»¯ÄƱØÐëÃÜ·â±£´æ£®
£¨2£©ÓÉÓÚ̼ËáÄÆÄÜÓëÇâÑõ»¯¸Æ·´Ó¦Éú³É̼Ëá¸ÆºÍÇâÑõ»¯ÄÆ£®ËùÒÔ£¬Ê¹Ëü±ä³É´¿¾»µÄÇâÑõ»¯ÄÆÈÜÒº£¬Ð¡º£ËûÃÇӦѡÓõÄÊÔ¼ÁÊÇʯ»ÒË®£¬·´Ó¦µÄ·½³ÌʽÊÇ£ºNa2CO3+Ca£¨OH£©2=2NaOH+CaCO3¡ý£®
¹Ê´ðΪ£º£¨1£©¼ûÉÏ±í£®£¨2£©Ê¯»ÒË®£¬Na2CO3+Ca£¨OH£©2=2NaOH+CaCO3¡ý£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø