ÌâÄ¿ÄÚÈÝ

ÏÂÁл­ÓкáÏßµÄÎïÖÊ£¬ÔÚ·´Ó¦Öв»Äܱ»ÍêÈ«ÏûºÄµÄÊÇ£¨¡¡¡¡£©
¢Ù½«º¬ÓÐÉÙÁ¿H2O£¨g£©µÄH2ͨ¹ýÊ¢ÓÐ×ãÁ¿Na2O2µÄÈÝÆ÷Öв¢²»¶ÏÓõç»ð»¨µãȼ
¢Ú½«1molCuÖÃÓÚº¬2mol H2SO4µÄŨÁòËáÖмÓÈÈ
¢ÛÔÚÒ»¶¨ÌõʲÏÂ3Ìå»ýH2ºÍ1Ìå»ýN2ÔÚ´ß»¯¼Á´æÔÚÏ·´Ó¦
¢ÜÔÚÇ¿¹â³ÖÐøÕÕÉäÏ£¬Ïò¹ýÁ¿µÄCa£¨ClO£©2ÈÜÒºÖÐͨÈëÉÙÁ¿CO2£®
A£®¢Ù¢ÚB£®¢Ú¢Û¢ÜC£®¢Ú¢ÜD£®¢Û¢Ü
¢Ù½«º¬ÉÙÁ¿H2OµÄH2ÆøÌåͨÈëÊ¢ÓÐ×ãÁ¿Na2O2µÄÃܱÕÈÝÆ÷ÖУ¬²¢²»¶ÏÓõç»ð»¨Òýȼ£¬2H2O+2Na2O2=4NaOH+O2£¬2H2+O2=2H2O£¬ÇâÆøÄÜÍêÈ«ÏûºÄ£¬¹Ê¢Ù²»·ûºÏ£»
¢ÚŨÁòËáËæ×Å·´Ó¦µÄ½øÐбäΪϡÁòËᣬÒÀ¾Ý½ðÊô»î¶¯Ë³Ðò±í¿ÉÖª£¬Í­²»ÓëÏ¡ÁòËá·´Ó¦£¬¹ÊÍ­²»ÄÜÏûºÄÍêÈ«£¬¹Ê¢Ú·ûºÏ£»
¢ÛÔÚÒ»¶¨ÌõʲÏÂ3Ìå»ýH2ºÍ1Ìå»ýN2ÔÚ´ß»¯¼Á´æÔÚÏ·´Ó¦£¬×îÖÕ´ïµ½»¯Ñ§Æ½ºâ£¬²»ÄܽøÐг¹µ×£¬ËùÒÔÇâÆø²»Äܱ»ÍêÈ«ÏûºÄ£¬¹Ê¢Û·ûºÏ£»
¢ÜÔÚÇ¿¹â³ÖÐøÕÕÉäÏ£¬Ïò¹ýÁ¿µÄCa£¨ClO£©2µÄÐü×ÇÒºÖÐͨÈëÉÙÁ¿CO2£¬·¢Éú·´Ó¦Ca£¨ClO£©2+CO2+H2O=CaCO3¡ý+2HClO£¬¹âÕÕ2HClO=2HCl+O2¡ü£¬ËùÒÔCa£¨ClO£©2²»ÄÜÍêÈ«ÏûºÄ£¬¹ÊD·ûºÏ£»
¹ÊÑ¡B£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø