×öÖкͷ´Ó¦ÊµÑéʱ£¬ÎÒ½«Ï¡ÑÎËáµÎÈëÇâÑõ»¯ÄÆÈÜÒºÖУ¬ÒâÍâ¿´µ½ÓÐÆøÅݲúÉú£®Ð¡Î°ÌáÐÑ£ºÊDz»ÊÇÄôíÁËÒ©Æ·£¿ÎÒ²éÑéºóÈ·ÈÏҩƷû´í£¬Ö»ÊÇÔÚÆ¿¿Ú·¢ÏÖÓа×É«·Ûĩ״ÎïÖÊ£®ÎÒÈÏΪÊÇÇâÑõ»¯ÄÆÈÜÒº±äÖÊÁË£®
£¨1£©NaOHÈÜÒº±äÖʵÄÔ­Òò______£®£¨Ð´»¯Ñ§·½³Ìʽ£©
£¨2£©ÀûÓÃÓëÉÏÊöʵÑ鲻ͬµÄÔ­Àí£¬ÎÒÓÖÉè¼ÆÁËÒ»¸öʵÑéÔÙ´ÎÈ·ÈϸÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ£®
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡ÉÙÁ¿ÑõÑõ»¯ÄÆÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó______
                
______  ¸ÃÇâÑõ»¯ÄÆÈÜÒº
ÒѱäÖÊ______
£¨3£©¸ÃÇâÑõ»¯ÄÆÈÜÒºÊDz¿·Ö±äÖÊ»¹ÊÇÈ«²¿±äÖÊ£¿
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
______ ______  ¸ÃÇâÑõ»¯ÄÆÈÜÒº______±äÖÊ
д³ö£¨3£©ÖÐËùÉæ¼°µ½µÄÒ»¸ö»¯Ñ§·½³Ìʽ______                               
£¨4£©ÈçºÎÓøñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¿¼òÊöʵÑé²½Öè______£®
£¨1£©ÇâÑõ»¯ÄÆÈÜÒºÎüÊÕ¿ÕÆøÖеĶþÑõ»¯Ì¼Éú³É̼ËáÄÆºÍË®£¬Òò´Ë£¬Óë¿ÕÆø½Ó´¥µÄÇâÑõ»¯ÄÆÈÜÒº»á±äÖÊ£»
»¯Ñ§·½³ÌʽΪ£ºCO2+2NaOH=Na2CO3+H2O£»
£¨2£©Ì¼ËáÑκÍËá·´Ó¦ÓÐÆøÌåÉú³É£¬ºÍº¬Ba2+£¨»òCa2+£©µÄÈÜÒº·´Ó¦»áÉú³É³Áµí£»
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡ÉÙÁ¿NaOHÈÜÒºÓëÊÔ¹ÜÖУ¬µÎ¼ÓÂÈ»¯¸ÆÈÜÒº ÈÜÒºÖвúÉú°×É«³Áµí ¸ÃÇâÑõ»¯ÄÆÒѱäÖÊ
Na2CO3+CaCl2=2NaCl+CaCO3¡ý
£¨3£©ÊÇ·ñÈ«²¿±äÖÊ£¬·ÖÎöÊÇ·ñÓÐÇâÑõ»¯ÄÆ´æÔÚ£»
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
¢ÙÈ¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓÈë×ãÁ¿µÄÂÈ»¯±µÈÜÒº
¢Ú¹ýÂË£¬ÔÚÂËÒºÖеμӼ¸µÎÎÞÉ«·Ó̪ÈÜÒº
¢ÙÉú³É°×É«³Áµí
¢Ú·Ó̪ÈÜÒº±äºì
ÇâÑõ»¯ÄƲ¿·Ö±äÖÊ
£¨3£©ÖÐËùÉæ¼°µ½µÄÒ»¸ö»¯Ñ§·½³Ìʽ£ºBaCl2+Na2CO3¨TBaCO3¡ý+2NaCl£»
£¨4£©ÓñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¬¾ÍÊdzýȥ̼ËáÄÆ£¬ÏòÈÜÒºÖмÓÊÊÁ¿µÄÂÈ»¯±µÈÜÒº£¬¹ýÂË£¬ËùµÃÂËҺΪÇâÑõ»¯ÄÆÈÜÒº£®
¹Ê´ð°¸Îª£º£¨1£©CO2+2NaOH=Na2CO3+H2O£»
£¨2£©
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡ÉÙÁ¿NaOHÈÜÒºÓëÊÔ¹ÜÖУ¬µÎ¼ÓÂÈ»¯¸ÆÈÜÒº ÈÜÒºÖвúÉú°×É«³Áµí ¸ÃÇâÑõ»¯ÄÆÒѱäÖÊ
Na2CO3+CaCl2=2NaCl+CaCO3¡ý
£¨3£©
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
¢ÙÈ¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓÈë×ãÁ¿µÄÂÈ»¯±µÈÜÒº
¢Ú¹ýÂË£¬ÔÚÂËÒºÖеμӼ¸µÎÎÞÉ«·Ó̪ÈÜÒº
¢ÙÉú³É°×É«³Áµí
¢Ú·Ó̪ÈÜÒº±äºì
ÇâÑõ»¯ÄƲ¿·Ö±äÖÊ
BaCl2+Na2CO3¨TBaCO3¡ý+2NaCl£»
£¨4£©ÏòÈÜÒºÖмÓÊÊÁ¿µÄÂÈ»¯±µÈÜÒº£¬¹ýÂË£¬ËùµÃÂËҺΪÇâÑõ»¯ÄÆÈÜÒº£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

27¡¢×öÖкͷ´Ó¦ÊµÑéʱ£¬ÎÒ½«Ï¡ÑÎËáµÎÈëÇâÑõ»¯ÄÆÈÜÒºÖУ¬ÒâÍâ¿´µ½ÓÐÆøÅݲúÉú£®Ð¡Î°ÌáÐÑ£ºÊDz»ÊÇÄôíÁËÒ©Æ·£¿ÎÒ²éÑéºóÈ·ÈÏҩƷû´í£¬Ö»ÊÇÔÚÆ¿¿Ú·¢ÏÖÓа×É«·Ûĩ״ÎïÖÊ£®ÎÒÈÏΪÊÇÇâÑõ»¯ÄÆÈÜÒº±äÖÊÁË£®
£¨1£©ÇâÑõ»¯ÄÆÈÜÒº±äÖʵÄÔ­ÒòÊÇ
Óë¿ÕÆøÖеÄCO2·´Ó¦£¬Éú³É°×É«³Áµí
£®
£¨2£©ÀûÓÃÓëÉÏÊöʵÑ鲻ͬµÄÔ­Àí£¬ÎÒÓÖÉè¼ÆÁËÒ»¸öʵÑéÔÙ´ÎÈ·ÈϸÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ£®
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡ÉÙÁ¿ÑõÑõ»¯ÄÆÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó_____________ ¸ÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ
£¨3£©¸ÃÇâÑõ»¯ÄÆÈÜÒºÊDz¿·Ö±äÖÊ»¹ÊÇÈ«²¿±äÖÊ£¿
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
1¡¢È¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬
2¡¢¼ÓÈë×ãÁ¿µÄ
ÈÜÒº£¬
3¡¢¹ýÂË£¬
4¡¢ÔÚÂËÒºÖеμÓ
ÈÜÒº
2¡¢²úÉú°×É«³Áµí
4¡¢
ÇâÑõ»¯ÄÆ

£¨²¿·Ö»òÈ«²¿£©±äÖÊ
д³ö£¨3£©ÖÐËùÉæ¼°µ½µÄÒ»¸ö»¯Ñ§·½³Ìʽ
BaCl2+Na2CO3¨TBaCO3¡ý+2NaCl
£®
£¨4£©ÈçºÎÓøñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¿¼òÊöʵÑé²½Ö裮

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ×öÖкͷ´Ó¦ÊµÑéʱ£¬ÎÒ½«Ï¡ÑÎËáµÎÈëÇâÑõ»¯ÄÆÈÜÒºÖУ¬ÒâÍâ¿´µ½ÆøÅݲúÉú£®Ð¡»ªÌáÐÑ£ºÊDz»ÊÇÄôíÁËÒ©Æ·£¿ÎÒ²éÑéºóÈ·ÈÏҩƷû´í£¬Ö»ÊÇÔÚÆ¿¿Ú·¢ÏÖÓа×É«·Ûĩ״ÎïÖÊ£®ÎÒÈÏΪÊÇÇâÑõ»¯ÄÆÈÜÒº±äÖÊÁË£®
£¨1£©ÇâÑõ»¯ÄÆÈÜÒº±äÖʵÄÔ­ÒòÊÇ
CO2+2NaOH¨TNa2CO3¡ý+H2O
CO2+2NaOH¨TNa2CO3¡ý+H2O
£®£¨Ð´³öÓйػ¯Ñ§·½³Ìʽ£©
£¨2£©ÀûÓÃÓëÉÏÊöʵÑ鲻ͬµÄÔ­Àí£¨²»ÓüÓÑÎËá·½·¨£©£¬ÎÒÓÖÉè¼ÆÁËÒ»¸öʵÑéÔÙ´ÎÈ·ÈϸÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ£®
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡ÉÙÁ¿ÇâÑõ»¯ÄÆÈÜÒºÓÚÊÔ¹ÜÖеμÓ
ÇâÑõ»¯¸ÆÈÜÒº
ÇâÑõ»¯¸ÆÈÜÒº
£®
Óа×É«³ÁµíÉú³É
Óа×É«³ÁµíÉú³É
¸ÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ£®
д³ö£¨2£©ÖеĻ¯Ñ§·´Ó¦·½³Ìʽ
Ca£¨OH£©2+Na2CO3¨TCaCO3¡ý+2NaOH
Ca£¨OH£©2+Na2CO3¨TCaCO3¡ý+2NaOH
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2008?ÇàÆÖÇø¶þÄ££©ÔÚ×öÖкͷ´Ó¦ÊµÑéʱ£¬ÎÒ½«Ï¡ÑÎËáµÎÈëÇâÑõ»¯ÄÆÈÜÒºÖУ¬ÒâÍâ¿´µ½ÆøÅݲúÉú£®Ð¡»ªÌáÐÑ£ºÊDz»ÊÇÄôíÁËÒ©Æ·£¿ÎÒ²éÑéºóÈ·ÈÏҩƷû´í£¬Ö»ÊÇÔÚÆ¿¿Ú·¢ÏÖÓа×É«·Ûĩ״ÎïÖÊ£®ÎÒÈÏΪÊÇÇâÑõ»¯ÄÆÈÜÒº±äÖÊÁË£®
£¨1£©ÇâÑõ»¯ÄÆÈÜÒº±äÖʵÄÔ­ÒòÊÇ
2NaOH+CO2¨TNa2CO3+H2O
2NaOH+CO2¨TNa2CO3+H2O
£®£¨Ð´³öÓйػ¯Ñ§·½³Ìʽ£©
£¨2£©ÀûÓÃÓëÉÏÊöʵÑ鲻ͬµÄÔ­Àí£¨²»ÓüÓÑÎËá·½·¨£©£¬ÎÒÓÖÉè¼ÆÁËÒ»¸öʵÑéÔÙ´ÎÈ·ÈϸÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ£®
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡ÉÙÁ¿ÇâÑõ»¯ÄÆÈÜÒºÓÚÊÔ¹ÜÖеμÓ
ÂÈ»¯¸ÆÈÜÒº
ÂÈ»¯¸ÆÈÜÒº
£®
ÈÜÒºÖвúÉú°×É«³Áµí
ÈÜÒºÖвúÉú°×É«³Áµí
¸ÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ£®
д³ö£¨2£©ÖеĻ¯Ñ§·´Ó¦·½³Ìʽ
Na2CO3+CaCl2=2NaCl+CaCO3¡ý
Na2CO3+CaCl2=2NaCl+CaCO3¡ý
£®
£¨3£©ÈçºÎÓøñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¿¼òÊö²Ù×÷²½Ö裺
ÏòÈÜÒºÖеμÓÇâÑõ»¯¸ÆÈÜÒºÖ±ÖÁ²»ÔÙ²úÉú³ÁµíΪֹ£¬¹ýÂË£¬ËùµÃÂËҺΪÇâÑõ»¯ÄÆÈÜÒº£®
ÏòÈÜÒºÖеμÓÇâÑõ»¯¸ÆÈÜÒºÖ±ÖÁ²»ÔÙ²úÉú³ÁµíΪֹ£¬¹ýÂË£¬ËùµÃÂËҺΪÇâÑõ»¯ÄÆÈÜÒº£®
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

×öÖкͷ´Ó¦ÊµÑéʱ£¬ÎÒ½«Ï¡ÑÎËáµÎÈëÇâÑõ»¯ÄÆÈÜÒºÖУ¬ÒâÍâ¿´µ½ÓÐÆøÅݲúÉú£®Ð¡Î°ÌáÐÑ£ºÊDz»ÊÇÄôíÁËÒ©Æ·£¿ÎÒ²éÑéºóÈ·ÈÏҩƷû´í£¬Ö»ÊÇÔÚÆ¿¿Ú·¢ÏÖÓа×É«·Ûĩ״ÎïÖÊ£®ÎÒÈÏΪÊÇÇâÑõ»¯ÄÆÈÜÒº±äÖÊÁË£®
£¨1£©ÇâÑõ»¯ÄÆÈÜÒº±äÖʵÄÔ­ÒòÊÇ
CO2+2NaOH=Na2CO3+H2O
CO2+2NaOH=Na2CO3+H2O
£®
£¨2£©ÀûÓÃÓëÉÏÊöʵÑ鲻ͬµÄÔ­Àí£¬ÎÒÓÖÉè¼ÆÁËÒ»¸öʵÑéÔÙ´ÎÈ·ÈϸÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ£®
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡ÉÙÁ¿ÇâÑõ»¯ÄÆÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó
ÂÈ»¯¸ÆÈÜÒº
ÂÈ»¯¸ÆÈÜÒº
ÈÜÒºÖвúÉú°×É«³Áµí
ÈÜÒºÖвúÉú°×É«³Áµí
¸ÃÇâÑõ»¯ÄÆÈÜÒºÒѱäÖÊ
£¨3£©¸ÃÇâÑõ»¯ÄÆÈÜÒºÊDz¿·Ö±äÖÊ»¹ÊÇÈ«²¿±äÖÊ£¿
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
1È¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬
2¼ÓÈë×ãÁ¿µÄ
ÂÈ»¯±µ
ÂÈ»¯±µ
ÈÜÒº£¬
3¹ýÂË£¬
4ÔÚÂËÒºÖеμÓ
ÎÞÉ«·Ó̪
ÎÞÉ«·Ó̪
ÈÜÒº

2²úÉú°×É«³Áµí

4
·Ó̪ÈÜÒº²»±äÉ«£¨»ò·Ó̪ÈÜÒº±äºì£©
·Ó̪ÈÜÒº²»±äÉ«£¨»ò·Ó̪ÈÜÒº±äºì£©

ÇâÑõ»¯ÄÆ
È«²¿£¨»ò²¿·Ö£©
È«²¿£¨»ò²¿·Ö£©

£¨Ì·Ö»òÈ«²¿£©±äÖÊ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

×öÖкͷ´Ó¦ÊµÑéʱ£¬ÎÒ½«Ï¡ÑÎËáµÎÈëÇâÑõ»¯ÄÆÈÜÒºÖУ¬ÒâÍâ¿´µ½ÓÐÆøÅݲúÉú£®Ð¡Î°ÌáÐÑ£¬ÊDz»ÊÇÄôíÁËÒ©Æ·£¿ÎÒ²éÑéºóÈ·ÈÏҩƷû´í£¬µ«ÔÚÆ¿¿Ú·¢ÏÖÓа×É«·Ûĩ״ÎïÖÊ£®ÎÒÈÏΪÊÇÇâÑõ»¯ÄÆÈÜÒº±äÖÊÁË£®£¨ÒÑÖª£ºNa2CO3ÈÜÒºµÄpH£¾7£©
£¨1£©ÀûÓÃÓëÉÏÊöʵÑ鲻ͬµÄÔ­Àí£¬ÎÒÓÖÉè¼ÆÁËÒ»¸öʵÑéÈ·ÈϸÃÇâÑõ»¯ÄÆÈÜÒºÒѾ­±äÖÊÁË£®
ʵÑé²½Öè ʵÑéÏÖÏó ʵÑé½áÂÛ
È¡ÉÙÁ¿ÑõÑõ»¯ÄÆÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎ¼Ó
Ï¡ÑÎËá
Ï¡ÑÎËá

ÓÐÆøÅݲúÉú
ÓÐÆøÅݲúÉú
¸ÃÇâÑõ»¯ÄÆÈÜÒº
ÒѱäÖÊ
£¨2£©ÎªÖ¤Ã÷¸ÃÇâÑõ»¯ÄÆÈÜÒºÊDz¿·Ö±äÖÊ£¬ÎÒÉè¼ÆÁËÒ»¸öʵÑ飮
    ÊµÑéÄ¿µÄ     ÊµÑé²½Öè     ÊµÑéÏÖÏó

Ö¤Ã÷¸ÃÈÜÒºÖк¬ÓÐ̼ËáÄÆ
Ö¤Ã÷¸ÃÈÜÒºÖк¬ÓÐ̼ËáÄÆ
È¡ÉÙÁ¿¹ÌÌåÈÜÓÚË®£¬ÏòÆäÖÐ
µÎ¼Ó×ãÁ¿µÄ
ÂÈ»¯±µ
ÂÈ»¯±µ
ÈÜÒº

²úÉú°×É«³Áµí
Ö¤Ã÷¸ÃÇâÑõ»¯ÄÆÈÜÒºÊÇ
²¿·Ö±äÖÊ
ÏòµÚÒ»²½ËùµÃÈÜÒºÖеμÓÎÞ
É«·Ó̪ÊÔÒº

·Ó̪±äºìÉ«
·Ó̪±äºìÉ«
£¨3£©ÈçºÎÓøñäÖʵÄÈÜÒºÀ´ÖÆÈ¡ÇâÑõ»¯ÄÆÈÜÒº£¿Óû¯Ñ§·½³Ìʽ±íʾ
Ba£¨OH£©2+Na2CO3=BaCO3¡ý+2NaOH
Ba£¨OH£©2+Na2CO3=BaCO3¡ý+2NaOH
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸