ÏÖʵÑéÊÒÓÐһƿ¾ÃÖõÄÇâÑõ»¯ÄƹÌÌ壬СºìͬѧÏë̽¾¿Ò»ÏÂ
¸Ã¹ÌÌåµÄ³É·Ö¡£ËýÊ×ÏȲéÔÄÁË×ÊÁÏ£¬µÃÖª£º
1.CaCl2+Na2CO3=2NaCl+CaCO3¡ý 2.CaCl2ÈÜÒº³ÊÖÐÐÔ¡£
È»ºóСºì×öÁËÈçÏÂʵÑ飺
£¨1£©È¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓË®Èܽ⣬ȻºóµÎ¼Ó·Ó̪ÊÔÒº£¬ÈÜÒº
±äºì£¬ÔòÑùÆ·ÖÐÒ»¶¨º¬ÓÐÇâÑõ»¯ÄÆ£¬¸Ã½áÂÛ²»ÕýÈ·µÄÔÒòÊÇ
_______________________________________¡£
£¨2£©È¡ÉÙÁ¿ÑùÆ·£¬¼ÓÈëÏ¡ÑÎËᣬÓÐÆøÌåÉú³É£¬ÔòÑùÆ·ÖÐÒ»¶¨º¬ÓÐ____________¡£
£¨3£©È¡ÉÙÁ¿ÑùÆ·¼ÓË®Èܽâºó£¬¼Ó¹ýÁ¿ÂÈ»¯¸ÆÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬¹ýÂË£¬ÏòÂËÒºÖмÓÈë·Ó̪ÊÔÒº£¬±äºì£¬¸Ã²½ÊµÑéµÄÄ¿µÄÊÇ_________________________
__¡£
ÈôÏòÂËÒºÖÐͨÈëÉÙÁ¿¶þÑõ»¯Ì¼£¬²úÉúµÄÏÖÏóÊÇ____________£¬Çëд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________________________________________¡£
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê±±¾©ÊÐÑÓÇìÏØÖп¼Ò»Ä£»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌ½¾¿Ìâ
£¨5·Ö£©ÏÖʵÑéÊÒÓÐһƿ¾ÃÖõÄÇâÑõ»¯ÄƹÌÌ壬СºìͬѧÏë̽¾¿Ò»ÏÂ
¸Ã¹ÌÌåµÄ³É·Ö¡£
![]()
ËýÊ×ÏȲéÔÄÁË×ÊÁÏ£¬µÃÖª£º
1.CaCl2+Na2CO3=2NaCl+CaCO3¡ý 2.CaCl2ÈÜÒº³ÊÖÐÐÔ¡£
È»ºóСºì×öÁËÈçÏÂʵÑ飺
£¨1£©È¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓË®Èܽ⣬ȻºóµÎ¼Ó·Ó̪ÊÔÒº£¬ÈÜÒº±äºì£¬ÔòÑùÆ·ÖÐÒ»¶¨º¬ÓÐÇâÑõ»¯ÄÆ£¬¸Ã½áÂÛ²»ÕýÈ·µÄÔÒòÊÇ______ __________¡£
£¨2£©È¡ÉÙÁ¿ÑùÆ·£¬¼ÓÈëÏ¡ÑÎËᣬÓÐÆøÌåÉú³É£¬ÔòÑùÆ·ÖÐÒ»¶¨º¬ÓÐ____________¡£
£¨3£©È¡ÉÙÁ¿ÑùÆ·¼ÓË®Èܽâºó£¬¼Ó¹ýÁ¿ÂÈ»¯¸ÆÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬¹ýÂË£¬ÏòÂËÒºÖмÓÈë·Ó̪ÊÔÒº£¬±äºì£¬¸Ã²½ÊµÑéµÄÄ¿µÄÊÇ___________________________¡£
ÈôÏòÂËÒºÖÐͨÈëÉÙÁ¿¶þÑõ»¯Ì¼£¬²úÉúµÄÏÖÏóÊÇ____________£¬Çëд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÖʵÑéÊÒÓÐһƿ¾ÃÖõÄÇâÑõ»¯ÄƹÌÌ壬СºìͬѧÏë̽¾¿Ò»ÏÂ
¸Ã¹ÌÌåµÄ³É·Ö¡£ËýÊ×ÏȲéÔÄÁË×ÊÁÏ£¬µÃÖª£º
1.CaCl2+Na2CO3=2NaCl+CaCO3¡ý 2.CaCl2ÈÜÒº³ÊÖÐÐÔ¡£
È»ºóСºì×öÁËÈçÏÂʵÑ飺
£¨1£©È¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓË®Èܽ⣬ȻºóµÎ¼Ó·Ó̪ÊÔÒº£¬ÈÜÒº±äºì£¬ÔòÑùÆ·ÖÐÒ»¶¨º¬ÓÐÇâÑõ»¯ÄÆ£¬¸Ã½áÂÛ²»ÕýÈ·µÄÔÒòÊÇ_______________________________________¡£
£¨2£©È¡ÉÙÁ¿ÑùÆ·£¬¼ÓÈëÏ¡ÑÎËᣬÓÐÆøÌåÉú³É£¬ÔòÑùÆ·ÖÐÒ»¶¨º¬ÓÐ____________¡£
£¨3£©È¡ÉÙÁ¿ÑùÆ·¼ÓË®Èܽâºó£¬¼Ó¹ýÁ¿ÂÈ»¯¸ÆÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬¹ýÂË£¬ÏòÂËÒºÖмÓÈë·Ó̪ÊÔÒº£¬±äºì£¬¸Ã²½ÊµÑéµÄÄ¿µÄÊÇ___________________________¡£
ÈôÏòÂËÒºÖÐͨÈëÉÙÁ¿¶þÑõ»¯Ì¼£¬²úÉúµÄÏÖÏóÊÇ____________£¬Çëд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________________________________________¡£
![]()
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÖʵÑéÊÒÓÐһƿ¾ÃÖõÄÇâÑõ»¯ÄƹÌÌ壬СºìͬѧÏë̽¾¿Ò»ÏÂ
¸Ã¹ÌÌåµÄ³É·Ö¡£
![]()
ËýÊ×ÏȲéÔÄÁË×ÊÁÏ£¬µÃÖª£º
1.CaCl2+Na2CO3=2NaCl+CaCO3¡ý 2.CaCl2ÈÜÒº³ÊÖÐÐÔ¡£
È»ºóСºì×öÁËÈçÏÂʵÑ飺
£¨1£©È¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓË®Èܽ⣬ȻºóµÎ¼Ó·Ó̪ÊÔÒº£¬ÈÜÒº±äºì£¬ÔòÑùÆ·ÖÐÒ»¶¨º¬ÓÐÇâÑõ»¯ÄÆ£¬¸Ã½áÂÛ²»ÕýÈ·µÄÔÒòÊÇ______ __________¡£
£¨2£©È¡ÉÙÁ¿ÑùÆ·£¬¼ÓÈëÏ¡ÑÎËᣬÓÐÆøÌåÉú³É£¬ÔòÑùÆ·ÖÐÒ»¶¨º¬ÓÐ____________¡£
£¨3£©È¡ÉÙÁ¿ÑùÆ·¼ÓË®Èܽâºó£¬¼Ó¹ýÁ¿ÂÈ»¯¸ÆÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬¹ýÂË£¬ÏòÂËÒºÖмÓÈë·Ó̪ÊÔÒº£¬±äºì£¬¸Ã²½ÊµÑéµÄÄ¿µÄÊÇ___________________________¡£
ÈôÏòÂËÒºÖÐͨÈëÉÙÁ¿¶þÑõ»¯Ì¼£¬²úÉúµÄÏÖÏóÊÇ____________£¬Çëд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com