ÌâÄ¿ÄÚÈÝ

15£®Ä³Æ·ÅÆÑÀ¸à²àÃæµÄ±êÇ©Èçͼ£®
£¨1£©µ¥·úÁ×ËáÄÆµÄ»¯Ñ§Ê½ÊÇNa2PO3F£¬ÆäÖÐPÔªËØµÄ»¯ºÏ¼ÛΪ+5£¬ÔòFÔªËØµÄ»¯ºÏ¼ÛΪ-1£»¸ÃÎïÖÊÖÐNa¡¢OÔªËØµÄÖÊÁ¿±ÈΪ23£º24£®
£¨2£©ÑÀ¸àË®ÈÜÒºÄÜʹÎÞÉ«·Ó̪±äºì£¬ÔòÆä³Ê¼îÐÔ£¨Ìî¡°ËáÐÔ¡±¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©£®
£¨3£©¼ìÑéĦ²Á¼ÁÖеÄÒõÀë×ÓCO32-£¬ÏȺó·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCaCO3+2HCl=CaCl2+H2O+CO2¡ü£¬¸Ã·´Ó¦µÄ»ù±¾ÀàÐÍΪ¸´·Ö½â·´Ó¦£»CO2+Ca£¨OH£©2=CaCO3¡ý+H2O£®
£¨4£©ÊÔÑé±íÃ÷ÑÀ¸àÖе¥·úÁ×ËáÄÆµÄÖÊÁ¿·ÖÊý´ïµ½0.76%¡«0.80%ʱ£¬·ÀÈ£³ÝµÄЧ¹û½ÏºÃ£¬Çëͨ¹ý¼ÆËãÅжϸÃÑÀ¸à¾ßÓУ¨Ìî¡°¾ßÓС±»ò¡°²»¾ßÓС±£©½ÏºÃµÄ·ÀÈ£³ÝЧ¹û£®

·ÖÎö £¨1£©¸ù¾Ý»¯ºÏ¼ÛÔ­Ôò£¬ÓÉ»¯Ñ§Ê½Çó³öÔªËØµÄ»¯ºÏ¼Û£¬¸ù¾Ý»¯Ñ§Ê½Çó³öÔªËØµÄÖÊÁ¿±È£»
£¨2£©¸ù¾Ýָʾ¼ÁµÄ±äÉ«·ÖÎöÈÜÒºµÄËá¼îÐÔ£»
£¨3£©¸ù¾Ý̼Ëá¸ùÀë×ӵļìÑé·½·¨¼°·´Ó¦µÄÌØµã·ÖÎö»Ø´ð£»
£¨4£©¸ù¾ÝÑÀ¸àÖе¥·úÁ×ËáÄÆµÄÖÊÁ¿·ÖÊý·ÖÎöÅжϣ®

½â´ð ½â£º£¨1£©µ¥·úÁ×ËáÄÆµÄ»¯Ñ§Ê½ÊÇNa2PO3F£¬ÓÉÓÚÄÆµÄ»¯ºÏ¼ÛΪ+1£¬ÑõµÄ»¯ºÏ¼ÛΪ-2£¬PÔªËØµÄ»¯ºÏ¼ÛΪ+5£¬ÉèFÔªËØµÄ»¯ºÏ¼ÛΪx£®£¨+1£©¡Á2+£¨+5£©+£¨-2 £©¡Á3+x=0£¬½âµÃx=-1£¬¸ÃÎïÖÊÖÐNa¡¢OÔªËØµÄÖÊÁ¿±ÈΪ£º£¨23¡Á2£©£º£¨16¡Á3£©=23£º24£®
£¨2£©ÑÀ¸àË®ÈÜÒºÄÜʹÎÞÉ«·Ó̪±äºì£¬ÔòÆä³Ê¼îÐÔ£®
£¨3£©¼ìÑéĦ²Á¼ÁÖеÄÒõÀë×ÓCO32-£¬³£ÓüÓÈëÏ¡ÑÎËᣬÔÙ½«Éú³ÉµÄÆøÌåͨÈë³ÎÇåµÄʯ»ÒË®ÖÐÀ´¼ìÑ飬ÏȺó·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ CaCO3+2HCl=CaCl2+H2O+CO2¡ü£¬¸Ã·´Ó¦µÄ»ù±¾ÀàÐÍΪ¸´·Ö½â·´Ó¦£»CO2+Ca£¨OH£©2=CaCO3¡ý+H2O£®
£¨4£©µ¥·úÁ×ËáÄÆÖÐFÔªËØµÄÖÊÁ¿·ÖÊýΪ£º$\frac{19}{23¡Á2+31+16¡Á3+19}$¡Á100%=$\frac{19}{144}$¡Á100%¡Ö13.19%£»º¬114mg=0.114g·úµÄµ¥·úÁ×ËáÄÆµÄÖÊÁ¿Îª0.114¡Â13.19%¡Ö0.864g£¬
ÑÀ¸àÖе¥·úÁ×ËáÄÆµÄÖÊÁ¿·ÖÊýΪ$\frac{0.864g}{110g}$¡Á100%¡Ö0.79%£¬
¼ÆËã½á¹ûΪ0.79%ÔÚ0.76%ºÍ0.80%Ö®¼ä£¬¾ßÓнϺõķÀÈ£³ÝЧ¹û£®
¹Ê´ðΪ£º£¨1£©-1£¬23£º24£»£¨2£©¼îÐÔ£»£¨3£©CaCO3+2HCl=CaCl2+H2O+CO2¡ü£¬¸´·Ö½â£¬CO2+Ca£¨OH£©2=CaCO3¡ý+H2O£»£¨4£©¾ßÓУ®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁË»¯Ñ§ÓÃÓïµÄÊéд¡¢ÒÔ¼°¸ù¾Ý»¯Ñ§Ê½µÄ¼ÆËãµÈ£¬ÊôÓڿα¾µÄ»ù´¡ÖªÊ¶£¬Ó¦¼Óǿѧϰ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø