ÌâÄ¿ÄÚÈÝ

ÓÃ×÷ÑÀ¸àĦ²Á¼ÁµÄÇáÖÊ̼Ëá¸Æ£¬¹¤ÒµÉϳ£ÓÃʯ»ÒʯÀ´ÖƱ¸£¬Ö÷ÒªÉú²úÁ÷³ÌÈçÏ£º  
¢Ù ÔÚ¡°ìÑÉÕ¯¡±Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ____________¡£
¢Ú ÔÚ¡°³Áµí³Ø¡±ÖÐÉúʯ»ÒÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ____________¡£
¢Û Í¶Èëµ½¡°·´Ó¦³Ø¡±ÖеÄʯ»ÒÈéÊDz»¾ùÒ»¡¢²»Îȶ¨µÄ»ìºÏÎÊôÓÚ___________£¨Ñ¡ÌÈÜÒº¡¢Ðü×ÇÒº»òÈé×ÇÒº£©¡£
¢Ü ÈôÔÚ»¯Ñ§ÊµÑéÊÒÀï·ÖÀë¡°·´Ó¦³Ø¡±ÖеĻìºÏÎ¸Ã²Ù×÷µÄÃû³ÆÊÇ ____________ ¡£
¢Ý ±¾Á÷³ÌͼÖÐ___________£¨ÎïÖÊ£©¿ÉÒÔÌæ´ú¡°Ì¼ËáÄÆÈÜÒº¡±´ïµ½½µµÍÉú²ú³É±¾ºÍ½ÚÄܼõÅÅ¡£
¢ÙCaCO3CaO+CO2¡ü  
¢ÚH2O+CaO==Ca£¨OH£©2 
¢ÛÐü×ÇÒº  
¢Ü¹ýÂË     
¢Ý¶þÑõ»¯Ì¼£¨CO2£©
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2013?½ðÉ½ÇøÒ»Ä££©¶ÔÑÀ¸àµÄ̽¾¿ÒªÓõ½Ðí¶à»¯Ñ§ÖªÊ¶£®
£¨1£©Ï±íÁгöÁËÈýÖÖÑÀ¸àÖеÄÄ¥²Á¼Á£®ÇëÔÚ±íÖÐÌîдÈýÖÖÄ¥²Á¼ÁËùÊôµÄÎïÖÊÀà±ð£®
¶ùͯÑÀ¸à ·À³ôÑÀ¸à ͸Ã÷ÑÀ¸à
Ħ²Á¼Á ÇâÑõ»¯ÂÁ ̼Ëá¸Æ ¶þÑõ»¯¹è
ÎïÖʵÄÀà±ð
£¨Ö¸Ëá¡¢¼î¡¢ÑΡ¢Ñõ»¯Î
¼î
¼î
ÑÎ
Ñõ»¯Îï
Ñõ»¯Îï
£¨2£©¸ù¾ÝÄãµÄÍÆ²â£¬ÑÀ¸àÄ¥²Á¼ÁµÄÈܽâÐÔÊÇ
ÄÑÈÜ
ÄÑÈÜ
£¨Ìî¡°Ò×ÈÜ¡±»ò¡°ÄÑÈÜ¡±£©£®
£¨3£©ÓÃ×÷ÑÀ¸àĦ²Á¼ÁµÄÇáÖÊ̼Ëá¸Æ£¬¹¤ÒµÉϳ£ÓÃʯ»ÒʯÀ´ÖƱ¸£¬Ö÷ÒªÉú²úÁ÷³ÌÈçÏ£º

¢ÙÔÚ¡°ìÑÉÕ¯¡±Öз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
CaCO3
 ¸ßΠ
.
 
CaO+CO2¡ü
CaCO3
 ¸ßΠ
.
 
CaO+CO2¡ü
£¬¸Ã·´Ó¦ÀàÐÍÊôÓÚ
·Ö½â
·Ö½â
·´Ó¦£®
¢ÚÔÚ¡°³Áµí³Ø¡±ÖÐÉúʯ»ÒÓëË®·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
CaO+H2O=Ca£¨OH£©2
CaO+H2O=Ca£¨OH£©2
£®
¢ÛͶÈëµ½¡°·´Ó¦³Ø¡±ÖеÄʯ»ÒÈéÊDz»¾ùÒ»¡¢²»Îȶ¨µÄ»ìºÏÎÊôÓÚ
Ðü×ÇÒº
Ðü×ÇÒº
£¨Ñ¡ÌÈÜÒº¡¢Ðü×ÇÒº»òÈé×ÇÒº£©£®
¢Ü·´Ó¦³ØÖз¢ÉúµÄ»¯Ñ§·´Ó¦ÊÇCa£¨OH£©2+Na2CO3¨TCaCO3¡ý+2NaOH£¬ÈôÔÚ»¯Ñ§ÊµÑéÊÒÀï·ÖÀë³ö¡°·´Ó¦³Ø¡±ÖеijÁµíÎ¸Ã²Ù×÷µÄÃû³ÆÊÇ
¹ýÂË
¹ýÂË
£®
¢Ý±¾Á÷³ÌͼÖÐ
¶þÑõ»¯Ì¼
¶þÑõ»¯Ì¼
£¨ÌîÎïÖÊÃû³Æ£©¿ÉÒÔÌæ´ú¡°Ì¼ËáÄÆÈÜÒº¡±´ïµ½½µµÍÉú²ú³É±¾ºÍ½ÚÄܼõÅÅ£®
£¨2013?½­ÄþÇøÒ»Ä££©¶ÔÑÀ¸àµÄ̽¾¿ÒªÓõ½Ðí¶à»¯Ñ§ÖªÊ¶£®
£¨1£©ÓÃ×÷ÑÀ¸àĦ²Á¼ÁµÄÇáÖÊ̼Ëá¸Æ£¬¹¤ÒµÉϳ£ÓÃʯ»ÒʯÀ´ÖƱ¸£¬Ä³ÊµÑéС×éͬѧÉè¼ÆÁË2ÖÖת»¯Á÷³Ì£¬ÈçÏÂͼ1Ëùʾ£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
a£®Ð¡ÍõͬѧÖ÷ÕÅÓÃÁ÷³Ì¢Ù¡¢¢Ú¡¢¢ÜºÍ²Ù×÷¢òµÄÉè¼Æ£¬ÈÏΪÆä¹¤ÒÕ¼òµ¥£®
Çëд³ö·´Ó¦¢ÙºÍ¢ÜµÄ»¯Ñ§·½³Ìʽ£º¢Ù
CaCO3
 ¸ßΠ
.
 
CaO+CO2¡ü
CaCO3
 ¸ßΠ
.
 
CaO+CO2¡ü
£»¢Ü
Ca£¨OH£©2+Na2CO3¨TCaCO3¡ý+2NaOH
Ca£¨OH£©2+Na2CO3¨TCaCO3¡ý+2NaOH
£»
b£®²Ù×÷¢ò°üÀ¨½Á°è¡¢
¹ýÂË
¹ýÂË
¡¢Ï´µÓ¡¢¸ÉÔïµÈ¹¤Ðò£®
c£®Ð¡ÀîͬѧÈÏΪÁ÷³Ì¢Ù¡¢¢Ú¡¢¢ÛºÍ²Ù×÷I±ÈСÍõµÄÖ÷ÕŸüºÃ£¬ÆäÀíÓÉÊÇ£º
ÀûÓòúÉúµÄ¶þÑõ»¯Ì¼±ÈÁí¼Ó̼ËáÄÆÈÜÒº¸ü¾­¼Ã£¬Éú³É³É±¾¸üµÍ£¬Èô²úÉúµÄ¶þÑõ»¯Ì¼²»»ØÊÕÖ±½ÓÅÅ·Å£¬²»ÀûÓÚ¡°½ÚÄܼõÅÅ¡±
ÀûÓòúÉúµÄ¶þÑõ»¯Ì¼±ÈÁí¼Ó̼ËáÄÆÈÜÒº¸ü¾­¼Ã£¬Éú³É³É±¾¸üµÍ£¬Èô²úÉúµÄ¶þÑõ»¯Ì¼²»»ØÊÕÖ±½ÓÅÅ·Å£¬²»ÀûÓÚ¡°½ÚÄܼõÅÅ¡±
£»
²Ù×÷I°üÀ¨½Á°èºÍ΢Èȵȹ¤Ðò£®
£¨2£©¸ÃʵÑéС×éΪÁ˲ⶨijʯ»ÒʯÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊý£¬È¡25gʯ»ÒʯÑùÆ··ÅÔÚÉÕ±­ÖУ¬È»ºóÏòÆäÖÐÖðµÎ¼ÓÈëÒ»¶¨Á¿Ä³ÖÊÁ¿·ÖÊýµÄÏ¡ÑÎËᣬʹ֮ÓëÑùÆ·³ä·Ö·´Ó¦£¨ÔÓÖʲ»²Î¼Ó·´Ó¦£©£®Ëæ×Å·´Ó¦½øÐУ¬¼ÓÈëÏ¡ÑÎËáµÄÖÊÁ¿Óë·´Ó¦µÃµ½ÆøÌåµÄÖÊÁ¿±ä»¯¹ØÏµÈçͼ2Ëùʾ£®ÇëÍê³ÉÏÂÁмÆËãÄÚÈÝ£º
¢ÙÑùÆ·ÖÐ̼Ëá¸ÆµÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿
¢ÚËùÓÃÏ¡ÑÎËáÈÜÖʵÄÖÊÁ¿·ÖÊýÊǶàÉÙ£¿
£¨Ð´³ö¼ÆËã¹ý³Ì£©
½â£ºÉèÑùÆ·ÖÐCaCO3µÄÖÊÁ¿Îªx£¬²Î¼Ó·´Ó¦µÄÏ¡ÑÎËáÖÐHClµÄÖÊÁ¿Îªy
CaCO3+2HCl=CaCl2+CO2¡ü+H2O
1007344
xy8.8g
¸ù¾Ý
100
44
=
X
8.8g
½âµÃx=20 ¸ù¾Ý
73
44
=
Y
8.8g
½âµÃy=14.6g
ÑùÆ·ÖÐCaCO3µÄÖÊÁ¿·ÖÊýΪ£º
20g
25g
¡Á100%=80%
Ï¡ÑÎËáµÄÈÜÖÊÖÊÁ¿·ÖÊýΪ£º
14.6g
146g
¡Á100%=10%
½â£ºÉèÑùÆ·ÖÐCaCO3µÄÖÊÁ¿Îªx£¬²Î¼Ó·´Ó¦µÄÏ¡ÑÎËáÖÐHClµÄÖÊÁ¿Îªy
CaCO3+2HCl=CaCl2+CO2¡ü+H2O
1007344
xy8.8g
¸ù¾Ý
100
44
=
X
8.8g
½âµÃx=20 ¸ù¾Ý
73
44
=
Y
8.8g
½âµÃy=14.6g
ÑùÆ·ÖÐCaCO3µÄÖÊÁ¿·ÖÊýΪ£º
20g
25g
¡Á100%=80%
Ï¡ÑÎËáµÄÈÜÖÊÖÊÁ¿·ÖÊýΪ£º
14.6g
146g
¡Á100%=10%
£®

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

¾«Ó¢¼Ò½ÌÍø