ÑàÑàͬѧÔÚʵÑéÊÒ·¢ÏÖһƿ³¨¿Ú·ÅÖõÄNaOHÈÜÒº£¬²»Öª¸ÃÈÜÒºÊÇ·ñ±äÖÊ£¬ÓÚÊÇËý½øÐÐÈçÏÂʵÑ飬ÇëÄã²ÎÓë²¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÑàÑà²Â²â¸ÃÈÜÒºÒѱäÖÊ£¬ÆäÒÀ¾ÝÊÇ£ºNaOHÈÜÒºÄÜÎüÊÕ¿ÕÆøÖеÄÉú³ÉNa2CO3£®
£¨2£©È¡ÉÙÁ¿Ô­ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈë±¥ºÍʯ»ÒË®£¬·¢ÏÖʯ»ÒË®£¬ËµÃ÷Ô­ÈÜÒºÒѾ­±äÖÊ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£®
£¨3£©ÎªÁ˽øÒ»²½Ì½¾¿ÇâÑõ»¯ÄÆÈÜÒºÊÇ·ñÍêÈ«±äÖÊ£¬Çë°ïÖúÍê³ÉÏÂÁÐʵÑ鱨¸æ£º
ʵÑé²½ÖèʵÑéÏÖÏóʵÑé½áÂÛ
¢ÙÈ¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓÈë×ãÁ¿ÈÜÒºÈÜÒºÖ»Óв¿·Ö±äÖÊ
¢ÚÈ¡ÉÙÁ¿³ä·Ö·´Ó¦ºóµÄÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈëÎÞÉ«·Ó̪ÊÔÒº£®
a¡¢£»
b¡¢£»
c¡¢£®
¡¾´ð°¸¡¿·ÖÎö£º£¨1£©NaOHÈÜÒºÒ×ÎüÊÕ¿ÕÆøÖеÄCO2Éú³ÉNa2CO3£»
£¨2£©ÈôÈÜÒºÒѱäÖÊÔòÈÜÒºÖдæÔÚNa2CO3£¬Na2CO3+Ca£¨OH£©2=CaCO3¡ý+2NaOH£¬¿É¼ÓÈë±¥ºÍʯ»ÒË®¿´ÈÜÒºÊÇ·ñ±ä»ë×ÇÀ´ÅжÏÈÜÒºÊÇ·ñ±äÖÊ£»
£¨3£©ÒòΪNa2CO3+CaCl2=CaCO3¡ý+2NaCl£¬¿É¼ÓÈëCaCl2ÈÜÒº¿´ÈÜÒºÊÇ·ñ±ä»ë×ÇÀ´ÅжÏÈÜÒºÊÇ·ñ±äÖÊ£¬·Ó̪ÊÔÒºÓö¼î±äºìÉ«Òò´ËÔÚ³ý¾¡Ì¼Ëá¸ùµÄÇé¿öÏ¿ɵμӷÓ̪ÊÔÒºÀ´ÅжÏÈÜÒºÊÇ·ñÍêÈ«±äÖÊ£®
½â´ð£º½â£º£¨1£©ÒòΪ2NaOH+CO2¨TNa2CO3+H2O£¬¹Ê´ð°¸Îª£º¶þÑõ»¯Ì¼
£¨2£©Na2CO3+Ca£¨OH£©2=CaCO3¡ý+2NaOH£¬CaCO3Ϊ°×É«³Áµí£®¹Ê´ð°¸Îª£º±ä»ë×Ç£»Na2CO3+Ca£¨OH£©2=CaCO3¡ý+2NaOH
£¨3£©¢ÙNa2CO3+CaCl2=CaCO3¡ý+2NaCl ¹Ê´ð°¸Îª£ºCaCl2£»Éú³É°×É«³Áµí£®
     ¢Ú·Ó̪ÊÔÒºÓö¼î±äºìÉ«£¬Èç¹ûÇâÑõ»¯ÄÆÍêÈ«±äÖÊÔò²»ÏÔºìÉ«£¬½áÂÛÊÇÈÜÒºÖ»Óв¿·Ö±äÖÊ£®¹Ê´ð°¸Îª£ºÈÜÒº±äºìÉ«
µãÆÀ£º±¾Ì⿼²éÁËѧÉú¡°NaOHÈÜÒºÎüÊÕ¿ÕÆøÖеÄCO2Éú³ÉNa2CO3£»Na2CO3ÈÜÒºÖмÓÈë¸ÆÑÎÈÜÒºÉú³ÉCaCO3°×É«³ÁµíºÍ·Ó̪ÊÔÒº±äÉ«¡±µÄһЩ֪ʶ£¬ÌâĿͨ¹ýÒ©Æ·ÊÇ·ñ±äÖʵÄ̽¾¿ÅàÑøÁËѧÉúµÄʵÑé̽¾¿ÄÜÁ¦£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÑàÑàͬѧÔÚʵÑéÊÒ·¢ÏÖһƿ³¨¿Ú·ÅÖõÄNaOHÈÜÒº£¬²»Öª¸ÃÈÜÒºÊÇ·ñ±äÖÊ£¬ÓÚÊÇËý½øÐÐÈçÏÂʵÑ飬ÇëÄã²ÎÓë²¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÑàÑà²Â²â¸ÃÈÜÒºÒѱäÖÊ£¬ÆäÒÀ¾ÝÊÇ£ºNaOHÈÜÒºÄÜÎüÊÕ¿ÕÆøÖеÄ
¶þÑõ»¯Ì¼
¶þÑõ»¯Ì¼
Éú³ÉNa2CO3£®
£¨2£©È¡ÉÙÁ¿Ô­ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈë±¥ºÍʯ»ÒË®£¬·¢ÏÖʯ»ÒË®
±ä»ë×Ç
±ä»ë×Ç
£¬ËµÃ÷Ô­ÈÜÒºÒѾ­±äÖÊ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Na2CO3+Ca£¨OH£©2¨TCaCO3¡ý+2NaOH
Na2CO3+Ca£¨OH£©2¨TCaCO3¡ý+2NaOH
£®
£¨3£©ÎªÁ˽øÒ»²½Ì½¾¿ÇâÑõ»¯ÄÆÈÜÒºÊÇ·ñÍêÈ«±äÖÊ£¬Çë°ïÖúÍê³ÉÏÂÁÐʵÑ鱨¸æ£º
ʵÑé²½ÖèʵÑéÏÖÏóʵÑé½áÂÛ
¢ÙÈ¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬
¼ÓÈë×ãÁ¿
ÈÜÒº£®
ÈÜÒºÖ»Óв¿·Ö±äÖÊ
¢ÚÈ¡ÉÙÁ¿³ä·Ö·´Ó¦ºóµÄÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈëÎÞÉ«·Ó̪ÊÔÒº£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

25¡¢ÑàÑàͬѧÔÚʵÑéÊÒ·¢ÏÖһƿ³¨¿Ú·ÅÖõÄNaOHÈÜÒº£¬²»Öª¸ÃÈÜÒºÊÇ·ñ±äÖÊ£¬ÓÚÊÇËý½øÐÐÈçÏÂʵÑ飬ÇëÄã²ÎÓë²¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÑàÑà²Â²â¸ÃÈÜÒºÒѱäÖÊ£¬ÆäÒÀ¾ÝÊÇ£ºNaOHÈÜÒºÄÜÎüÊÕ¿ÕÆøÖеÄ
¶þÑõ»¯Ì¼£¨»òCO2£©
Éú³ÉNa2CO3£®
£¨2£©È¡ÉÙÁ¿Ô­ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈë±¥ºÍʯ»ÒË®£¬·¢ÏÖʯ»ÒË®
±ä»ë×Ç
£¬ËµÃ÷Ô­ÈÜÒºÒѾ­±äÖÊ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
Na2CO3+Ca£¨OH£©2=CaCO3¡ý+2NaOH
£®
£¨3£©ÎªÁ˽øÒ»²½Ì½¾¿ÇâÑõ»¯ÄÆÈÜÒºÊÇ·ñÍêÈ«±äÖÊ£¬Çë°ïÖúÍê³ÉÏÂÁÐʵÑ鱨¸æ£º

a¡¢
CaCl2£¨ºÏÀí´ð°¸¾ù¿É£©
 
b¡¢
Éú³É°×É«³Áµí

c¡¢
ÈÜÒº±äºìÉ«

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

ÑàÑàͬѧÔÚʵÑéÊÒ·¢ÏÖһƿ³¨¿Ú·ÅÖõÄNaOHÈÜÒº£¬²»Öª¸ÃÈÜÒºÊÇ·ñ±äÖÊ£¬ÓÚÊÇËý½øÐÐÈçÏÂʵÑ飬ÇëÄã²ÎÓë²¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÑàÑà²Â²â¸ÃÈÜÒºÒѱäÖÊ£¬ÆäÒÀ¾ÝÊÇ£ºNaOHÈÜÒºÄÜÎüÊÕ¿ÕÆøÖеÄ________Éú³ÉNa2CO3£®
£¨2£©È¡ÉÙÁ¿Ô­ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈë±¥ºÍʯ»ÒË®£¬·¢ÏÖʯ»ÒË®________£¬ËµÃ÷Ô­ÈÜÒºÒѾ­±äÖÊ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________£®
£¨3£©ÎªÁ˽øÒ»²½Ì½¾¿ÇâÑõ»¯ÄÆÈÜÒºÊÇ·ñÍêÈ«±äÖÊ£¬Çë°ïÖúÍê³ÉÏÂÁÐʵÑ鱨¸æ£º
ʵÑé²½ÖèʵÑéÏÖÏóʵÑé½áÂÛ
¢ÙÈ¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬¼ÓÈë×ãÁ¿________ÈÜÒº________ÈÜÒºÖ»Óв¿·Ö±äÖÊ
¢ÚÈ¡ÉÙÁ¿³ä·Ö·´Ó¦ºóµÄÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈëÎÞÉ«·Ó̪ÊÔÒº£®________
a¡¢________£»
b¡¢________£»
c¡¢________£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º2007Äê¹ã¶«Ê¡·ðɽÊÐÈýˮʵÑéÖÐѧ¸ßÒ»ÈëѧÃþµ×²âÊÔ»¯Ñ§ÊÔ¾í£¨¶þ£©£¨½âÎö°æ£© ÌâÐÍ£º½â´ðÌâ

£¨2006?¸£ÖÝ£©ÑàÑàͬѧÔÚʵÑéÊÒ·¢ÏÖһƿ³¨¿Ú·ÅÖõÄNaOHÈÜÒº£¬²»Öª¸ÃÈÜÒºÊÇ·ñ±äÖÊ£¬ÓÚÊÇËý½øÐÐÈçÏÂʵÑ飬ÇëÄã²ÎÓë²¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÑàÑà²Â²â¸ÃÈÜÒºÒѱäÖÊ£¬ÆäÒÀ¾ÝÊÇ£ºNaOHÈÜÒºÄÜÎüÊÕ¿ÕÆøÖеÄ______Éú³ÉNa2CO3£®
£¨2£©È¡ÉÙÁ¿Ô­ÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈë±¥ºÍʯ»ÒË®£¬·¢ÏÖʯ»ÒË®______£¬ËµÃ÷Ô­ÈÜÒºÒѾ­±äÖÊ£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______£®
£¨3£©ÎªÁ˽øÒ»²½Ì½¾¿ÇâÑõ»¯ÄÆÈÜÒºÊÇ·ñÍêÈ«±äÖÊ£¬Çë°ïÖúÍê³ÉÏÂÁÐʵÑ鱨¸æ£º
ʵÑé²½ÖèʵÑéÏÖÏóʵÑé½áÂÛ
¢ÙÈ¡ÉÙÁ¿ÑùÆ·ÓÚÊÔ¹ÜÖУ¬
¼ÓÈë×ãÁ¿______ÈÜÒº£®
______ÈÜÒºÖ»Óв¿·Ö±äÖÊ
¢ÚÈ¡ÉÙÁ¿³ä·Ö·´Ó¦ºóµÄÈÜÒºÓÚÊÔ¹ÜÖУ¬µÎÈëÎÞÉ«·Ó̪ÊÔÒº£®______

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸