ÌâÄ¿ÄÚÈÝ

£¨9·Ö£©Ä³Ñо¿Ð¡×éÓû¼ìÑé²ÝËá¾§ÌåÑùÆ··Ö½â²úÎ²¢²â¶¨ÑùÆ·ÖвÝËá¾§ÌåµÄÖÊÁ¿·ÖÊý£¨¼ÙÉèÔÓÖʲ»²ÎÓë·´Ó¦£©¡£²ÝËá¾§Ì壨H2C2O4¡¤2H2O£©µÄ²¿·ÖÀí»¯ÐÔÖʼûÏÂ±í£º

ÈÛ µã

·Ð µã

ÈÈ ÎÈ ¶¨ ÐÔ

Óë ¼î ·´ Ó¦

101¡æ¡«102¡æ

150¡æ¡«160¡æ

Éý»ª

100.1¡æ·Ö½â³öË®£¬175¡æ·Ö½â³ÉCO2¡¢CO¡¢H2O

Óë Ca(OH)2·´Ó¦²úÉú°×É«³Áµí(CaC2O4)

£¨1£©Í¼1ÊǼÓÈÈ×°Öá£×îÊÊÒ˵ļÓÈÈ·Ö½â²ÝËá¾§Ìå×°ÖÃÊÇ ¡£

£¨2£©Í¼ 2 ÊÇÑéÖ¤ÈÈ·Ö½â²úÎïÖк¬ CO ¡¢CO2µÄ×°Öá£

¢Ù×°Öà A µÄÖ÷Òª×÷ÓÃÊÇ_______________ _ _____¡£

¢ÚÆøÄÒµÄ×÷ÓÃÊÇ____________________¡£

¢ÛÖ¤Ã÷´æÔÚ CO2µÄÏÖÏóÊÇ_ £¬BÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ ¡£

¢ÜÖ¤Ã÷´æÔÚ CO µÄÏÖÏóÊÇ_______ _______¡£

£¨3£©Îª²â¶¨ÑùÆ·ÖвÝËá¾§ÌåµÄÖÊÁ¿·ÖÊý£¬Éè¼ÆÈçÏ·½°¸£º³ÆÈ¡Ò»¶¨Á¿ÑùÆ·£¬ÓÃÉÏÊö×°ÖýøÐÐʵÑ飬³ÆÁ¿×°ÖÃD·´Ó¦Ç°ºóµÄÖÊÁ¿²î¡£Óɴ˼ÆËã³öµÄʵÑé½á¹û±Èʵ¼ÊֵƫµÍ£¬ÅųýÒÇÆ÷ºÍ²Ù×÷ÒòËØ£¬Æä¿ÉÄÜÔ­Òò_______£¨Ð´Ò»Ìõ¼´¿É£©¡£

£¨4£©³ÆÈ¡17.5g²ÝËá¾§ÌåÑùÆ·ÅäÖÆ50.00gÈÜÒº£¬¼ÓÊÊÁ¿µÄÏ¡ÁòËᣬȻºóµÎ¼ÓKMnO4 ÈÜÒº£¨º¬KMnO47.9¿Ë£© Ç¡ºÃ·´Ó¦ÍêÈ«¡£

£¨ÒÑÖª£º2KMnO4+5H2C2O4+3H2SO4=K2SO4+2MnSO4+10CO2¡ü+8H2O£©

Çë¼ÆËãÑùÆ·ÖвÝËá¾§Ì壨H2C2O4¡¤2H2O £©µÄÖÊÁ¿·ÖÊý¡££¨Ð´³ö¼ÆËã¹ý³Ì£©

[ÓйØÎïÖʵÄÏà¶Ô·Ö×ÓÖÊÁ¿£ºMr£¨H2C2O4£©=90£¬Mr£¨H2C2O42H2O£©=126£¬Mr£¨KMnO4£©=158 ]

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø