ÌâÄ¿ÄÚÈÝ

ͨ¹ýÒ»ÄêµÄ»¯Ñ§Ñ§Ï°£¬ÄãÒѾ­ÕÆÎÕÁËһЩʵÑéÊÒÖÆÈ¡ÆøÌåµÄÏà¹Ø¹æÂÉ£¬ÒÔÏÂÊÇÀÏʦÌṩ¸ø´ó¼ÒµÄһЩװÖã¬ÇëÄã»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ʵÑéÊÒÓöþÑõ»¯Ã̺͹ýÑõ»¯ÇâÈÜÒºÖÆÈ¡ÑõÆøµÄ»¯Ñ§·½³ÌʽÊÇ________________£¬ÈôÒª½ÏºÃµØ¿ØÖƲúÉúÑõÆøµÄËÙ¶È£¬Ó¦Ñ¡Óõķ¢Éú×°ÖÃÊÇ__________(Ìî×Öĸ±àºÅ)¡£

(2)ʵÑéÊÒÓôóÀíʯÓëÏ¡ÑÎËáÖÆ±¸CO2µÄ»¯Ñ§·½³ÌʽΪ_____________£¬ ÊÕ¼¯×°ÖÃ×îºÃÑ¡Ôñ_______(Ìî×Öĸ±àºÅ)¡£ÎªÁËÊÕ¼¯¸ÉÔïµÄ¶þÑõ»¯Ì¼ÆøÌ壬¿É°ÑÆøÌåͨ¹ýÏÂͼµÄ×°ÖÃÖнøÐиÉÔïÔÙÊÕ¼¯¡£ÔòÏÂͼƿÖеÄÈÜÒºÊÇ____________(дÃû³Æ)£¬×°ÖÃÖеĶà¿×ÇòÅݵÄ×÷ÓÃÊÇ____________¡£

2H2O2 2H2O + O2 ¡ü B CaCO3+2HCl=CaCl2+2H2O+CO2¡ü D ŨÁòËá Ôö´ó·´Ó¦ÎïµÄ½Ó´¥Ãæ»ý ¡¾½âÎö¡¿±¾ÌâÖ÷Òª¿¼²éÁËÒÇÆ÷µÄÃû³Æ¡¢ÆøÌåµÄÖÆÈ¡×°ÖúÍÊÕ¼¯×°ÖõÄÑ¡Ôñ£¬Í¬Ê±Ò²¿¼²éÁË»¯Ñ§·½³ÌʽµÄÊéд£¬×ÛºÏÐԱȽÏÇ¿¡£ÆøÌåµÄÖÆÈ¡×°ÖõÄÑ¡ÔñÓë·´Ó¦ÎïµÄ״̬ºÍ·´Ó¦µÄÌõ¼þÓйأ»ÆøÌåµÄÊÕ¼¯×°ÖõÄÑ¡ÔñÓëÆøÌåµÄÃܶȺÍÈܽâÐÔÓйء£ £¨1£©ÊµÑéÊÒÓöþÑõ»¯Ã̺͹ýÑõ»¯ÇâÈÜÒºÖÆÈ¡ÑõÆøÍ¬Ê±Éú...
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø