ÌâÄ¿ÄÚÈÝ

ÈËÌåѪҺÀïCa2+Àë×ÓµÄŨ¶ÈÒ»°ã²ÉÓÃg/cm3À´±íʾ¡£³éȡһ¶¨Ìå»ýµÄѪÑù£¬¼ÓÊÊÁ¿µÄ²ÝËáï§£Û(NH4)2C2O4£ÝÈÜÒº£¬¿ÉÎö³ö²ÝËá¸Æ(CaC2O4)³Áµí£¬½«´Ë²ÝËá¸Æ³ÁµíÏ´µÓºóÈÜÓÚÇ¿Ëá¿ÉµÃÒ»ÖÖÈõËá²ÝËá(H2C2O4)£¬ÔÙÓÃKMnO4ÈÜÒºµÎ¶¨¼´¿É²â¶¨ÑªÒºÑùÆ·ÖÐCa2+µÄŨ¶È¡£Ä³Ñо¿ÐÔѧϰС×éÉè¼ÆÈçÏÂʵÑé²½Öè²â¶¨ÑªÒºÑùÆ·ÖÐCa2+µÄŨ¶È£º

¡¾ÅäÖÆKMnO4±ê×¼ÈÜÒº¡¿ÏÂͼÊÇÅäÖÆ50mLKMnO4±ê×¼ÈÜÒºµÄ¹ý³ÌʾÒâͼ¡£

£¨1£©ÉÏÊö¹ý³ÌÖÐÓÐÁ½´¦´íÎó£¬ÇëÄã¹Û²ìͼʾÅÐ¶ÏÆäÖв»ÕýÈ·µÄ²Ù×÷ÊÇ£¨ÌîÐòºÅ£©____________£»

£¨2£©Èç¹û°´ÕÕͼʾµÄ²Ù×÷ÅäÖÆÈÜÒº£¬ËùµÃµÄʵÑé½á¹û½«____________£¨ÌîÆ«´ó»òƫС£©¡£

¡¾²â¶¨ÑªÒºÑùÆ·ÖÐCa2+µÄŨ¶È¡¿³éȡѪÑù20.00mL£¬¾­¹ýÉÏÊö´¦ÀíºóµÃµ½²ÝËᣬÔÙÓÃËữµÄ0.020mol/L KMnO4ÈÜÒºµÎ¶¨£¬Ê¹²ÝËáת»¯³ÉCO2Òݳö£¬Õâʱ¹²ÏûºÄ12.00mL KMnO4ÈÜÒº¡£

£¨3£©µÎ¶¨Ê±¸ßÃÌËá¼ØÐèÒªËữ£¬´ÓÀ¨ºÅÖÐÑ¡ÔñËữ¸ßÃÌËá¼ØÈÜÒºËùÓõÄËᣨÁòËá¡¢ÑÎËá¡¢ÏõËᣩ £¬È·¶¨·´Ó¦´ïµ½ÖÕµãµÄÏÖÏó_____________________________¡£

£¨4£©Ð´³ö²ÝËá¸úËáÐÔKMnO4ÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º £»·´Ó¦ÖÐÈô×ªÒÆ0.2mol µç×ÓÉú³É±ê×¼×´¿öÏÂCO2ÆøÌåµÄÌå»ýΪ L¡£

£¨5£©¸ù¾ÝËù¸øÊý¾Ý¼ÆËãѪҺÑùÆ·ÖÐCa2+Àë×ÓµÄŨ¶ÈΪ________mg/cm3¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø