Ë®µÄÓ²¶È±íʾ·½·¨ÊÇ£º½«Ë®ÖеÄCa2+¡¢Mg2+ÖÊÁ¿ÕÛËã³ÉCaOµÄÖÊÁ¿£®Í¨³£°Ñ1LË®Öк¬ÓÐ10mg CaO³ÆΪ1¶È£¬1LË®Öк¬ÓÐ20mg CaO ¼´Îª2¶È£¬ÒÔ´ËÀàÍÆ£®8¶ÈÒÔÉÏΪӲˮ£¬8¶ÈÒÔÏÂΪÈíË®£®ÎÒ¹ú¹æ¶¨ÒûÓÃË®µÄÓ²¶È²»Äܳ¬¹ý25¶È£®
£¨1£©ÈÕ³£Éú»îÖУ¬¿ÉÓÃ______¼ìÑéijˮÑùÊÇӲˮ»¹ÊÇÈíË®£®
£¨2£©Ca£¨HCO3£©2¼ÓÈÈʱ»á·Ö½â²úÉúÒ»ÖÖ°×É«³Áµí£¨Ë®¹¸µÄÖ÷Òª³É·Ö£©ºÍÁ½ÖÖ³£¼ûµÄÑõ»¯ÎÇëд³ö·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ______ CaCO3+H2O+CO2¡ü
¡¾´ð°¸¡¿·ÖÎö£º£¨1£©¸ù¾ÝӲˮºÍÈíË®µÄÇø±ð¿¼ÂÇ£»
£¨2£©¸ù¾Ý·½³ÌʽµÄд·¨ºÍӲˮÈí»¯µÄ·½·¨¿¼ÂÇ£»
£¨3£©¸ù¾Ý·½³ÌʽµÄд·¨¿¼ÂÇ£»
£¨4£©¸ù¾ÝÓ²¶ÈµÄ¼ÆËã·½·¨¿¼ÂÇ£»
£¨5£©¸ù¾Ýµç½âË®µÄ·½³Ìʽ¼ÆËã¼´¿É£®
½â´ð£º½â£º£¨1£©Ó÷ÊÔíË®¿ÉÒÔ¼ø±ðӲˮºÍÈíË®£¬ÅÝÄ­¶àµÄÊÇÈíË®£¬ÅÝÄ­ÉÙµÄÊÇӲˮ£»
£¨2£©·´Ó¦ÎïÊÇ̼ËáÇâ¸Æ£¬Éú³ÉÎïÊÇ̼Ëá¸ÆºÍË®¡¢¶þÑõ»¯Ì¼£»Ó²Ë®Èí»¯µÄ·½·¨ÊÇÕôÁó»òÖó·Ð£»
£¨3£©·´Ó¦ÎïÊÇ̼Ëá¸ÆºÍÑÎËᣬÉú³ÉÎïÊÇÂÈ»¯¸Æ¡¢Ë®¡¢¶þÑõ»¯Ì¼£¬Óù۲취Åäƽ£»
£¨4£©¸ù¾Ý¸ÆÏà¶ÔÔ­×ÓÖÊÁ¿Îª40£¬Ñõ»¯¸ÆµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª56£¬ÁеÈʽ£º½âµÃX=0.28g=280mg£¬Ó²¶È£º×2=28£»
£¨5£©Éèµç½â5.4¿ËÕôÁóË®£¬¿ÉµÃµ½ÇâÆøÖÊÁ¿ÎªX£¬ÑõÆøÖÊÁ¿ÎªY£º
2H2O2H2¡ü+O2¡ü
36          4   32
5.4g        X   Y
¸ù¾Ý£º½âµÃX=0.6g£¬¸ù¾Ý£º½âµÃY=4.8g£®
¹Ê´ð°¸Îª£º
£¨1£©·ÊÔíË®£»
£¨2£©Ca£¨HCO3£©2CaCO3+H2O+CO2¡ü£»Öó·Ð£»
£¨3£©CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü£»
£¨4£©280£»28£»
£¨5£©0.6g£»4.8g£®
µãÆÀ£º½â´ð±¾ÌâµÄ¹Ø¼üÊÇÖªµÀӲˮÓëÈíË®µÄÇø±ðºÍӲˮÈí»¯µÄ·½·¨£¬ÖªµÀ·½³ÌʽµÄд·¨×¢ÒâÊÂÏ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ë®µÄÓ²¶È±íʾ·½·¨ÊÇ£º½«Ë®ÖеÄCa2+¡¢Mg2+ÖÊÁ¿ÕÛËã³ÉCaOµÄÖÊÁ¿£®Í¨³£°Ñ1LË®Öк¬ÓÐ10mg CaO³ÆΪ1¶È£¬1LË®Öк¬ÓÐ20mg CaO¼´Îª2¶È£¬ÒÔ´ËÀàÍÆ£®8¶ÈÒÔÉÏΪӲˮ£¬8¶ÈÒÔÏÂΪÈíË®£®ÎÒ¹ú¹æ¶¨ÒûÓÃË®µÄÓ²¶È²»Äܳ¬¹ý25¶È£®
£¨1£©ÈÕ³£Éú»îÖУ¬¿ÉÓÃ
 
¼ìÑéijˮÑùÊÇӲˮ»¹ÊÇÈíË®£®
£¨2£©Ca£¨HCO3£©2¼ÓÈÈʱ»á·Ö½â²úÉúÒ»ÖÖ°×É«³Áµí£¨Ë®¹¸µÄÖ÷Òª³É·Ö£©ºÍÁ½ÖÖ³£¼ûµÄÑõ»¯ÎÇëд³ö·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£®ºÓË®¡¢¾®Ë®ÖÐÈÜÓÐÒ»¶¨Á¿µÄCa£¨HCO3£©2£¬ÒûÓÃÇ°¿É²ÉÈ¡
 
µÄ·½·¨À´½µµÍË®µÄÓ²¶È£®
£¨3£©¼ÒÓÃÈÈˮƿһµ©²úÉúË®¹¸£¬Æä±£ÎÂÐÔÄܻήµÍ£®ÓÃÏ¡ÑÎËá¿É³ýÈ¥ÉÏÊöË®¹¸£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨4£©È¡Ä³Ë®Ñù1L£¬¾­ÊµÑé²â¶¨£¬ÆäÖк¬Ca2+0.2g£¬´ËË®ÑùµÄÓ²¶ÈԼΪ
 
£¬ÊÇ·ñ·ûºÏÒûÓÃË®±ê×¼£¿
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ë®µÄÓ²¶È±íʾ·½·¨ÊÇ£º½«Ë®ÖеÄCa2+¡¢Mg2+ÖÊÁ¿ÕÛËã³ÉCaOµÄÖÊÁ¿£®Í¨³£°Ñ1LË®Öк¬ÓÐ10mg CaO³ÆΪ1¶È£¬1LË®Öк¬ÓÐ20mg CaO¼´Îª2¶È£¬ÒÔ´ËÀàÍÆ£®8¶ÈÒÔÉÏΪӲˮ£¬8¶ÈÒÔÏÂΪÈíË®£®ÎÒ¹ú¹æ¶¨ÒûÓÃË®µÄÓ²¶È²»Äܳ¬¹ý25¶È£®
£¨1£©ÈÕ³£Éú»îÖУ¬¿ÉÓÃ
·ÊÔíË®
·ÊÔíË®
¼ìÑéijˮÑùÊÇӲˮ»¹ÊÇÈíË®£®
£¨2£©Ca£¨HCO3£©2¼ÓÈÈʱ»á·Ö½â²úÉúÒ»ÖÖ°×É«³Áµí£¨Ë®¹¸µÄÖ÷Òª³É·Ö£©ºÍÁ½ÖÖ³£¼ûµÄÑõ»¯ÎÇëд³ö·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
Ca£¨HCO3£©2
  ¡÷  
.
 
CaCO3¡ý+H2O+CO2¡ü
Ca£¨HCO3£©2
  ¡÷  
.
 
CaCO3¡ý+H2O+CO2¡ü
£®ºÓË®¡¢¾®Ë®ÖÐÈÜÓÐÒ»¶¨Á¿µÄCa£¨HCO3£©2£¬ÒûÓÃÇ°¿É²ÉÈ¡
Öó·Ð
Öó·Ð
µÄ·½·¨À´½µµÍË®µÄÓ²¶È£®
£¨3£©¼ÒÓÃÈÈˮƿһµ©²úÉúË®¹¸£¬Æä±£ÎÂÐÔÄܻήµÍ£®ÓÃ
´×Ëá»òÏ¡ÑÎËá
´×Ëá»òÏ¡ÑÎËá
¿É³ýÈ¥ÉÏÊöË®¹¸£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ë®µÄÓ²¶È±íʾ·½·¨ÊÇ£º½«Ë®ÖеÄCa2+¡¢Mg2+ÖÊÁ¿ÕÛËã³ÉCaOµÄÖÊÁ¿£®Í¨³£°Ñ1LË®Öк¬ÓÐ10mg CaO³ÆΪ1¶È£¬1LË®Öк¬ÓÐ20mg CaO ¼´Îª2¶È£¬ÒÔ´ËÀàÍÆ£®8¶ÈÒÔÉÏΪӲˮ£¬8¶ÈÒÔÏÂΪÈíË®£®ÎÒ¹ú¹æ¶¨ÒûÓÃË®µÄÓ²¶È²»Äܳ¬¹ý25¶È£®
£¨1£©ÈÕ³£Éú»îÖУ¬¿ÉÓÃ
·ÊÔíË®
·ÊÔíË®
¼ìÑéijˮÑùÊÇӲˮ»¹ÊÇÈíË®£®
£¨2£©Ca£¨HCO3£©2¼ÓÈÈʱ»á·Ö½â²úÉúÒ»ÖÖ°×É«³Áµí£¨Ë®¹¸µÄÖ÷Òª³É·Ö£©ºÍÁ½ÖÖ³£¼ûµÄÑõ»¯ÎÇëд³ö·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
Ca£¨HCO3£©2
  ¡÷  
.
 
CaCO3+H2O+CO2¡ü
Ca£¨HCO3£©2
  ¡÷  
.
 
CaCO3+H2O+CO2¡ü
£®ºÓË®¡¢¾®Ë®ÖÐÈÜÓÐÒ»¶¨Á¿µÄCa£¨HCO3£©2£¬ÒûÓÃÇ°¿É²ÉÈ¡
Öó·Ð
Öó·Ð
µÄ·½·¨À´½µµÍË®µÄÓ²¶È£®
£¨3£©¼ÒÓÃÈÈˮƿһµ©²úÉúË®¹¸£¬Æä±£ÎÂÐÔÄܻήµÍ£®ÓÃÏ¡ÑÎËá¿É³ýÈ¥ÉÏÊöË®¹¸£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
£®
£¨4£©È¡Ä³Ë®Ñù1L£¬¾­ÊµÑé²â¶¨£¬ÆäÖк¬Ca2+0.2g£¬ÕÛËã³ÉCaOµÄÖÊÁ¿Îª
280
280
mg£¬´ËË®ÑùµÄÓ²¶ÈԼΪ
28
28
£®
£¨5£©¹¤ÒµÉÏÓÃÕôÁóµÄ·½·¨½µµÍË®µÄÓ²¶È´Ó¶øµÃµ½´¿Ë®£®Èç¹ûÍêÈ«µç½â5.4¿ËÕôÁóË®£¬¿ÉµÃµ½ÇâÆø¡¢ÑõÆø¸÷¶àÉÙ¿Ë£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

»¯Ñ§ÓëÈËÀàÉú»îϢϢÏà¹Ø£¬ËüÄÜÉø͸µ½ÈËÀàÉú»îµÄÒ¡¢Ê³¡¢×¡¡¢Ðеȸ÷¸ö·½Ã森
Ë®µÄÓ²¶È±íʾ·½·¨ÊÇ£º½«Ë®ÖеÄCa2+¡¢Mg2+ÖÊÁ¿ÕÛËã³ÉCaOµÄÖÊÁ¿£®Í¨³£°Ñ1LË®Öк¬ÓÐ10mgCaO³ÆΪ1¶È£¬1LË®Öк¬ÓÐ20mg CaO¼´Îª2¶È£¬ÒÔ´ËÀàÍÆ£®8¶ÈÒÔÉÏΪӲˮ£¬8¶ÈÒÔÏÂΪÈíË®£®ÎÒ¹ú¹æ¶¨ÒûÓÃË®µÄÓ²¶È²»Äܳ¬¹ý25¶È£®
£¨1£©ÈÕ³£Éú»îÖУ¬¿ÉÓÃ
·ÊÔíË®
·ÊÔíË®
¼ìÑéijˮÑùÊÇӲˮ»¹ÊÇÈíË®£®
£¨2£©Ca£¨HCO3£©2¼ÓÈÈʱ»á·Ö½â²úÉúÒ»ÖÖ°×É«³Áµí£¨Ë®¹¸µÄÖ÷Òª³É·Ö£©ºÍÁ½ÖÖ³£¼ûµÄÑõ»¯ÎÇëд³ö·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
Ca£¨HCO3£©2
  ¡÷  
.
 
CaCO3¡ý+H2O+CO2¡ü
Ca£¨HCO3£©2
  ¡÷  
.
 
CaCO3¡ý+H2O+CO2¡ü
£®ºÓË®¡¢¾®Ë®ÖÐÈÜÓÐÒ»¶¨Á¿µÄCa£¨HCO3£©2£¬ÒûÓÃÇ°¿É²ÉÈ¡
Öó·Ð
Öó·Ð
µÄ·½·¨À´½µµÍË®µÄÓ²¶È£®
£¨3£©¼ÒÓÃÈÈˮƿһµ©²úÉúË®¹¸£¬Æä±£ÎÂÐÔÄܻήµÍ£®ÓÃ
´×Ëá»òÏ¡ÑÎËá
´×Ëá»òÏ¡ÑÎËá
¿É³ýÈ¥ÉÏÊöË®¹¸£®
£¨4£©Ð¡ÑôÈ¡ÏÂÁÐÉú»îÖеÄÎïÖÊ£¬²âµÃÆäpHÈçϱíËùʾ£º
ÎïÖÊ ·ÊÔíË® ÓêË® ÌÇË® ÄûÃÊÖ­ Ï´µÓ¼Á
pH 10.2 5.9 7.0 2.5 12.2
ÓÉ´ËÅжϣº
¢ÙÕâЩÎïÖÊÖÐËáÐÔ×îÇ¿µÄÊÇ
ÄûÃÊÖ­
ÄûÃÊÖ­
£»
¢ÚÕâЩÎïÖÊÄÜʹÎÞÉ«·Ó̪ÊÔÒº±äºìµÄÊÇ
·ÊÔíË®»òÏ´µÓ¼Á
·ÊÔíË®»òÏ´µÓ¼Á
£¨ÌîÒ»ÖÖÎïÖʼ´¿É£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ë®µÄÓ²¶È±íʾ·½·¨ÊÇ£º½«Ë®ÖеÄCa2+¡¢Mg2+ÖÊÁ¿ÕÛËã³ÉCaOµÄÖÊÁ¿£®Í¨³£°Ñ1LË®Öк¬ÓÐ10mg CaO³ÆΪ1¶È£¬1LË®Öк¬ÓÐ20mg CaO¼´Îª2¶È£¬ÒÔ´ËÀàÍÆ£®8¶ÈÒÔÉÏΪӲˮ£¬8¶ÈÒÔÏÂΪÈíË®£®ÎÒ¹ú¹æ¶¨ÒûÓÃË®µÄÓ²¶È²»Äܳ¬¹ý25¶È£®
£¨1£©ÈÕ³£Éú»îÖУ¬¿ÉÓÃ
·ÊÔíË®
·ÊÔíË®
¼ìÑéijˮÑùÊÇӲˮ»¹ÊÇÈíË®£®
£¨2£©Ca£¨HCO3£©2¼ÓÈÈʱ»á·Ö½â²úÉúÒ»ÖÖ°×É«³Áµí£¨Ë®¹¸µÄÖ÷Òª³É·Ö£©ºÍÁ½ÖÖ³£¼ûµÄÑõ»¯ÎÇëд³ö·¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
2Ca£¨HCO3£©2
  ¡÷  
.
 
CaCO3¡ý+H2O+CO2¡ü
2Ca£¨HCO3£©2
  ¡÷  
.
 
CaCO3¡ý+H2O+CO2¡ü
£®ºÓË®¡¢¾®Ë®ÖÐÈÜÓÐÒ»¶¨Á¿µÄCa£¨HCO3£©2£¬ÒûÓÃÇ°¿É²ÉÈ¡
¼ÓÈÈÖó·Ð
¼ÓÈÈÖó·Ð
µÄ·½·¨À´½µµÍË®µÄÓ²¶È£®
£¨3£©¼ÒÓÃÈÈˮƿһµ©²úÉúË®¹¸£¬Æä±£ÎÂÐÔÄܻήµÍ£®ÓÃÏ¡ÑÎËá¿É³ýÈ¥ÉÏÊöË®¹¸£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
CaCO3+2HCl¨TCaCl2+H2O+CO2¡ü
£®
£¨4£©Ë×»°Ëµ¡°Ë®µÎʯ´©¡±£¬ËµµÄÊÇʯ»ÒʯµÄÖ÷Òª³É·ÖÔÚÓêË®ºÍ¶þÑõ»¯Ì¼µÄ¹²Í¬×÷ÓÃÏÂÉú³ÉÁË¿ÉÈܵÄCa£¨HCO3£©2£¬ÇëÄãд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
CaCO3+H2O+CO2¨TCa£¨HCO3£©2
CaCO3+H2O+CO2¨TCa£¨HCO3£©2
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸