ÌâÄ¿ÄÚÈÝ
£¨10·Ö£©»¯Ñ§¿Îºó£¬»¯Ñ§ÐËȤС×éµÄͬѧÔÚÕûÀíʵÑé×Àʱ£¬·¢ÏÖÓÐһƿÇâÑõ»¯ÄÆÈÜҺûÓÐÈûÏðƤÈû£¬Õ÷µÃÀÏʦͬÒâºó£¬¿ªÕ¹ÁËÒÔÏÂ̽¾¿£º
[Ìá³öÎÊÌâ1] ¸ÃÇâÑõ»¯ÄÆÈÜÒºÊÇ·ñ±äÖÊÁËÄØ£¿
[ʵÑé̽¾¿1]
|
ʵÑé²Ù×÷ |
ʵÑéÏÖÏó |
ʵÑé½áÂÛ |
|
È¡ÉÙÁ¿¸ÃÈÜÒºÓÚÊÔ¹ÜÖУ¬ÏòÈÜÒºÖеμÓÏ¡ÑÎËᣬ²¢²»¶ÏÕñµ´¡£ |
|
ÇâÑõ»¯ÄÆÈÜÒºÒ»¶¨±äÖÊÁË¡£ |
[Ìá³öÎÊÌâ2] ¸ÃÇâÑõ»¯ÄÆÈÜÒºÊÇÈ«²¿±äÖÊ»¹ÊDz¿·Ö±äÖÊÄØ£¿
[²ÂÏëÓë¼ÙÉè]
²ÂÏë1£ºÇâÑõ»¯ÄÆÈÜÒº²¿·Ö±äÖÊ¡£ ²ÂÏë2£ºÇâÑõ»¯ÄÆÈÜҺȫ²¿±äÖÊ¡£
[²éÔÄ×ÊÁÏ] ÂÈ»¯±µÈÜÒº³ÊÖÐÐÔ¡£
[ʵÑé̽¾¿2]
|
ʵÑé²½Öè |
ʵÑéÏÖÏó |
ʵÑé½áÂÛ |
|
£¨1£©È¡ÉÙÁ¿¸ÃÈÜÒºÓÚÊÔ¹ÜÖУ¬ÏòÈÜÒºÖеμӹýÁ¿µÄÂÈ»¯±µÈÜÒº£¬²¢²»¶ÏÕñµ´¡£ |
ÓÐ Éú³É¡£ |
˵Ã÷ÔÈÜÒºÖÐÒ»¶¨ÓÐ̼ËáÄÆ¡£ |
|
£¨2£©È¡²½Ö裨1£©ÊÔ¹ÜÖеÄÉÙÁ¿ÉϲãÇåÒº£¬µÎ¼Ó·Ó̪ÈÜÒº¡£ |
ÈÜÒº±äºìÉ«¡£ |
˵Ã÷ÔÈÜÒºÖÐÒ»¶¨ÓÐ ¡£ |
[ʵÑé½áÂÛ] ¸ÃÇâÑõ»¯ÄÆÈÜÒº £¨Ìî¡°²¿·Ö¡±»ò¡°È«²¿¡±£©±äÖÊ¡£
[·´Ë¼ÓëÆÀ¼Û]
ÔÚÉÏÊö[ʵÑé̽¾¿2]ÖУ¬Ð¡Ã÷Ìá³ö¿ÉÓÃÇâÑõ»¯¸ÆÈÜÒº´úÌæÂÈ»¯±µÈÜÒº£¬ÄãÈÏΪ¸Ã·½°¸ £¨Ìî¡°¿ÉÐС±»ò¡°²»¿ÉÐС±£©£¬ÀíÓÉÊÇ
[Àí½âÓëÓ¦ÓÃ] £¨1£© ÇâÑõ»¯ÄÆÈÜÒºÈÝÒ×±äÖÊ£¬±ØÐëÃÜ·â±£´æ¡£ÊµÑéÊÒ±ØÐëÃÜ·â±£´æµÄÒ©Æ·»¹Óкܶ࣬ÊÔÁí¾ÙÒ»Àý£º ¡£
£¨2£©È¡ÉÏÊö²¿·Ö±äÖʵÄÇâÑõ»¯ÄÆÈÜÒº100g£¬ÏòÆäÖмÓÈë×ãÁ¿µÄÇâÑõ»¯±µÈÜÒº£¬ÍêÈ«·´Ó¦ºóµÃµ½1.97£ç°×É«³Áµí£¬ÇóÉÏÊöÇâÑõ»¯ÄÆÈÜÒºÖÐ̼ËáÄÆµÄÖÊÁ¿·ÖÊý¡££¨Ð´³ö¼ÆËã¹ý³Ì£©
[ʵÑé̽¾¿1]£¨Ã¿¿Õ1·Ö£©
|
ʵÑé²Ù×÷ |
ʵÑéÏÖÏó |
ʵÑé½áÂÛ |
|
È¡ÉÙÁ¿¸ÃÈÜÒºÓÚÊÔ¹ÜÖУ¬ÏòÈÜÒºÖеμÓÏ¡ÑÎËᣬ²¢²»¶ÏÕñµ´¡£ |
ÓÐÆøÅÝ |
ÇâÑõ»¯ÄÆÈÜÒºÒ»¶¨±äÖÊÁË¡£ |
[ʵÑé̽¾¿2]
|
ʵÑé²½Öè |
ʵÑéÏÖÏó |
ʵÑé½áÂÛ |
|
£¨1£©È¡ÉÙÁ¿¸ÃÈÜÒºÓÚÊÔ¹ÜÖУ¬ÏòÈÜÒºÖеμӹýÁ¿µÄÂÈ»¯±µÈÜÒº£¬²¢²»¶ÏÕñµ´¡£ |
ÓÐ °×É«³Áµí Éú³É¡£ |
˵Ã÷ÔÈÜÒºÖÐÒ»¶¨ÓÐ̼ËáÄÆ¡£ |
|
£¨2£©È¡²½Ö裨1£©ÊÔ¹ÜÖеÄÉÙÁ¿ÉϲãÇåÒº£¬µÎ¼Ó·Ó̪ÈÜÒº¡£ |
ÈÜÒº±äºìÉ«¡£ |
˵Ã÷ÔÈÜÒºÖÐÒ»¶¨ÓÐ ÇâÑõ»¯ÄÆ ¡£ |
[ʵÑé½áÂÛ] ¸ÃÇâÑõ»¯ÄÆÈÜÒº ²¿·Ö £¨Ìî¡°²¿·Ö¡±»ò¡°È«²¿¡±£©±äÖÊ¡£
¸Ã·½°¸²»¿ÉÐÐ £¨Ìî¡°¿ÉÐС±»ò¡°²»¿ÉÐС±£©£¬ÀíÓÉÊÇÇâÑõ»¯¸ÆÓë̼ËáÄÆ·´Ó¦»áÉú³ÉÇâÑõ»¯ÄÆ £¨´ð°¸ºÏÀí¼´¿É£©
[Àí½âÓëÓ¦ÓÃ] £¨1£© ÇâÑõ»¯ÄÆÈÜÒºÈÝÒ×±äÖÊ£¬±ØÐëÃÜ·â±£´æ¡£ÊµÑéÊÒ±ØÐëÃÜ·â±£´æµÄÒ©Æ·»¹Óкܶ࣬ÊÔÁí¾ÙÒ»Àý£º ŨÁòËᣨ´ð°¸ºÏÀí¼´¿É£© ¡£
£¨2£©1.06%
¡¾½âÎö¡¿
ÊÔÌâ·ÖÎö£ºÔÚʵÑé̽¾¿Ò»ÖУ¬ÑùÆ·ÊÇ·ñ±äÖÊ£¬¾ÍÊÇ̽¾¿ÆäÖÐÓÐÎÞ̼ËáÄÆ£¬¼ÓÈëÑÎËáÓÐÆøÅÝð³ö˵Ã÷ÆäÖÐÒ»¶¨º¬ÓÐ̼ËáÄÆ£¬ËµÃ÷ÒѾ±äÖÊÁË£»Ì½¾¿¶þÖ÷ÒªÊÇ̽¾¿ÑùÆ·ÊÇÍêÈ«±äÖʾÍÊÇ̽¾¿ÑùÆ·ÍêÈ«ÊÇ̼ËáÄÆ»¹ÊÇ̼ËáÄÆºÍÇâÑõ»¯ÄƵĻìºÏÎ¼ÓÈë¹ýÁ¿ÂÈ»¯±µ£¬ÂÈ»¯±µºÍÉú³ÉÎïÈ«²¿ÎªÖÐÐÔ£¬²âÆäËá¼î¶ÈÈÔΪ¼îÐÔ£¬ËµÃ÷ÓÐÇâÑõ»¯ÄÆÊ£Ó࣬ÔÑùƷΪ²¿·Ö±äÖÊ£¬Èç¹ûΪÖÐÐÔÔòÑùÆ·È«²¿ÎªÌ¼ËáÄÆ£¬ÑùÆ·ÍêÈ«±äÖÊ¡£²»¿ÉÒÔÓÃÇâÑõ»¯¸Æ´úÌæÂÈ»¯±µ£¬ÒòΪÇâÑõ»¯¸ÆºÍ̼ËáÄÆ·´Ó¦Éú³É̼Ëá¸ÆºÍÇâÑõ»¯ÄÆ£¬ÎÞ·¨Í¨¹ý²â¶¨·´Ó¦ºóÈÜÒºpHÖµÀ´²â¶¨ÑùÆ·ÊÇ·ñÍêÈ«±äÖÊ¡£Àí½âÓëÓ¦ÓÃÖУ¬ÊµÑéÊÒÃÜ·â±£´æµÄÒ©Æ·ºÜ¶à£¬±ÈÈçŨÑÎËᣬŨÁòËáµÈ¡£
É裺ÑùÆ·ÖÐ̼ËáÄÆµÄÖÊÁ¿ÊÇX
Na2CO3 + Ba(OH)2= BaCO3 + 2NaOH
106 197
X 1.97g
![]()
Ôò̼ËáÄÆµÄÖÊÁ¿·ÖÊýΪ£º![]()
ÉÏÊöÇâÑõ»¯ÄÆÈÜÒºÖÐ̼ËáÄÆµÄÖÊÁ¿·ÖÊýΪ1.06%
¿¼µã£ºÇâÑõ»¯ÄƵÄÐÔÖÊ
µãÆÀ£ºÕâµÀÌâÄ¿½ÏΪ×ۺϣ¬ÓйØÓÚʵÑé̽¾¿µÄ¿¼²ìÒ²ÓйØÓÚ¼ÆËãÌâµÄ¿¼²ì£¬¿¼²ì½ÏΪ³£¹æ£¬×¢Òâ×ÐϸÉóÌ⣬עÒâÌâÄ¿¸÷СÎÊǰºóÖ®¼äµÄÁªÏµ¡£