Сºì²Î¼Ó»¯Ñ§¿ÎÍâС×éµÄ»î¶¯£¬»î¶¯µÄÄÚÈÝÊÇÔÚʵÑéÊÒÁ·Ï°ÅäÖÆÅ©Ò©²¨¶û¶àÒº¡£ÆäÅäÖƹý³ÌÊÇ£º³ÆÈ¡1 gµ¨·¯·ÅÈëAÈÝÆ÷ÖУ¬ÔÙ¼Ó90 mLË®½Á°è£¬ÍêÈ«ÈܽâÖƳÉÁòËáÍ­ÈÜÒº£»³ÆÈ¡1 gÉúʯ»Ò·ÅÈëBÈÝÆ÷ÖУ¬ÏȼÓÉÙÁ¿Ë®½Á°è£¬Ê¹Éúʯ»Ò±ä³ÉÊìʯ»Ò£¬ÔÙ¼ÓÈë10 mLË®½Á°è£¬ÖƳÉʯ»ÒÈé¡£½«ÁòËáÍ­ÈÜÒºÂýÂýµØµ¹Èëʯ»ÒÈéÖУ¬Í¬Ê±²»¶Ï½Á°è£¬¼´³É²¨¶û¶àÒº¡£ÇëÍê³É£º
(1) д³öÅäÖƹý³ÌÖÐÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ù________________________________________________________________,
¢Ú________________________________________________________________ ¡£
(2) СºìÓõÄA ÈÝÆ÷ÄÜ·ñÑ¡ÓÃÌúÖÆÈÝÆ÷£¿___________ £¬ÀíÓÉÊÇ________________________¡£
(3) ÅäÖƹý³ÌÖУ¬Ð¡ºìÐèÓõ½ÄÄЩÖ÷Òª»¯Ñ§ÒÇÆ÷£¿Çëд³öÒÇÆ÷µÄÃû³Æ£º___________ ¡¢___________ ¡¢___________ ¡¢___________ ¡£
(1)CaO+H2O====Ca(OH)2£»CuSO4+Ca(OH)2====Cu(OH)2¡ý+CaSO4 £»
(2)²»ÄÜ£»ÒòΪÌú»áÓëÁòËáÍ­ÈÜÒº·¢Éú·´Ó¦£¬µ¼ÖÂÌúÖÆÈÝÆ÷¸¯Ê´ºÍũҩʧЧ£»
(3)ÍÐÅÌÌìƽ£»Á¿Í²£»ÉÕ±­£»²£Á§°ô
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

11¡¢Ð¡ºì²Î¼Ó»¯Ñ§¿ÎÍâС×éµÄ»î¶¯£¬»î¶¯µÄÄÚÈÝÊÇÔÚʵÑéÊÒÁ·Ï°ÅäÖÆÅ©Ò©²¨¶û¶àÒº£®ÆäÅäÖƹý³ÌÊÇ£º³ÆÈ¡1gµ¨·¯·ÅÈëAÈÝÆ÷ÖУ¬ÔÙ¼Ó90mLË®½Á°è£¬ÍêÈ«ÈܽâÖƳÉÁòËáÍ­ÈÜÒº£»³ÆÈ¡1gÉúʯ»Ò·ÅÈëBÈÝÆ÷ÖУ¬ÏȼÓÉÙÁ¿Ë®½Á°è£¬Ê¹Éúʯ»Ò±ä³ÉÊìʯ»Ò£¬ÔÙ¼ÓÈë10mLË®½Á°è£¬ÖƳÉʯ»ÒÈ飮½«ÁòËáÍ­ÈÜÒºÂýÂýµØµ¹Èëʯ»ÒÈéÖУ¬Í¬Ê±²»¶Ï½Á°è£¬¼´³É²¨¶û¶àÒº£®ÇëÍê³É£º
£¨1£©Ð´³öÅäÖƹý³ÌÖÐÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ù
CaO+H2O¨TCa£¨OH£©2
£¬
¢Ú
CuSO4+Ca£¨OH£©2¨TCu£¨OH£©2¡ý+CaSO4
£®
£¨2£©Ð¡ºìÓõÄAÈÝÆ÷ÄÜ·ñÑ¡ÓÃÌúÖÆÈÝÆ÷£¿
²»ÄÜ
£¬ÀíÓÉÊÇ
Ìú»áÓëÁòËáÍ­ÈÜÒº·¢Éú·´Ó¦£¬µ¼ÖÂÌúÖÆÈÝÆ÷¸¯Ê´ºÍũҩʧЧ
£®
£¨3£©ÅäÖƹý³ÌÖУ¬Ð¡ºìÐèÓõ½ÄÄЩÖ÷Òª»¯Ñ§ÒÇÆ÷£¿Çëд³öÒÇÆ÷µÄÃû³Æ£º
ÍÐÅÌÌìƽ
¡¢
Á¿Í²
¡¢
ÉÕ±­
¡¢
²£Á§°ô
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

23¡¢Ð¡ºì²Î¼Ó»¯Ñ§¿ÎÍâС×éµÄ»î¶¯£¬»î¶¯µÄÄÚÈÝÊÇÔÚʵÑéÊÒÁ·Ï°ÅäÖƲ¨¶û¶àҺũҩ£®ÆäÅäÖƹý³ÌÊÇ£º³ÆÈ¡1gµ¨·¯·ÅÈëAÈÝÆ÷ÖУ¬ÔÙ¼Ó90mLË®½Á°è£¬ÍêÈ«ÈܽâÖƳÉÁòËáÍ­ÈÜÒº£»³ÆÈ¡1gÉúʯ»Ò·ÅÈëBÈÝÆ÷ÖУ¬ÏȼÓÉÙÁ¿Ë®½Á°è£¬Ê¹Éúʯ»Ò±ä³ÉÊìʯ»Ò£¬ÔÙ¼Ó10mLË®½Á°è£¬ÖƳÉʯ»ÒÈ飮½«ÁòËáÍ­ÈÜÒºÂýÂýµØµ¹Èëʯ»ÒÈéÖУ¬Í¬Ê±²»¶Ï½Á°è£¬¼´³É²¨¶û¶àÒº£®
Çë»Ø´ð£º
£¨1£©Ð¡ºìÓõÄAÈÝÆ÷ÄÜ·ñÑ¡ÓÃÌúÖÆÈÝÆ÷£¿
²»ÄÜ
£¬Ô­ÒòÊÇ
ÌúÓëÁòËáÍ­ÔÚÈÜÒºÖз¢ÉúÖû»·´Ó¦
£®
£¨2£©ÅäÖƹý³ÌÖУ¬Ð¡ºìÐèÒªÓõ½ÄÄЩ»¯Ñ§ÒÇÆ÷£¿
Çëд³öÒÇÆ÷µÄÃû³Æ£º
ÍÐÅÌÌìƽ¡¢Á¿Í²¡¢ÉÕ±­¡¢²£Á§°ô
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

11¡¢£¨Ì½¾¿Ì⣩Сºì²Î¼Ó»¯Ñ§¿ÎÍâС×éµÄ»î¶¯£¬»î¶¯µÄÄÚÈÝÊÇÔÚʵÑéÊÒÁ·Ï°ÅäÖÆÅ©Ò©²¨¶û¶àÒº£®ÆäÅäÖƹý³ÌÊÇ£º³ÆÈ¡1 gµ¨·¯·ÅÈëAÈÝÆ÷ÖУ¬ÔÙ¼Ó90 mLË®½Á°è£¬ÍêÈ«ÈܽâÖƳÉÁòËáÍ­ÈÜÒº£»³ÆÈ¡1 gÉúʯ»Ò·ÅÈëBÈÝÆ÷ÖУ¬ÏȼÓÉÙÁ¿Ë®½Á°è£¬Ê¹Éúʯ»Ò±ä³ÉÊìʯ»Ò£¬ÔÙ¼ÓÈë10 mLË®½Á°è£¬ÖƳÉʯ»ÒÈ飮½«ÁòËáÍ­ÈÜÒºÂýÂýµØµ¹Èëʯ»ÒÈéÖУ¬Í¬Ê±²»¶Ï½Á°è£¬¼´³É²¨¶û¶àÒº£®ÇëÍê³ÉÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öÅäÖƹý³ÌÖÐÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ù
CaO+H2O¨TCa£¨OH£©2
£¬¢Ú
CuSO4+Ca£¨OH£©2¨TCu£¨OH£©2¡ý+CaSO4
£®
£¨2£©Ð¡ºìÓõÄAÈÝÆ÷ÄÜ·ñÑ¡ÓÃÌúÖÆÈÝÆ÷£¿
²»ÄÜ
£¬ÀíÓÉÊÇ
ÒòΪÌú»áÓëÁòËáÍ­ÈÜÒº·¢Éú·´Ó¦£¬µ¼ÖÂÌúÖÆÈÝÆ÷¸¯Ê´ºÍũҩʧЧ
£®
£¨3£©ÅäÖƹý³ÌÖУ¬Ð¡ºìÐèÒªÓõ½ÄÄЩÖ÷Òª»¯Ñ§ÒÇÆ÷£¿Çëд³öÒÇÆ÷µÄÃû³Æ£º
ÍÐÅÌÌìƽ
¡¢
Á¿Í²
¡¢
ÉÕ±­
¡¢
²£Á§°ô
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º³õÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

18¡¢ÔĶÁÏÂÁжÌÎÄ£º
Сºì²Î¼Ó»¯Ñ§¿ÎÍâС×é»î¶¯£¬Á·Ï°ÅäÖÆÅ©Ò©²¨¶û¶àÒº£®ÆäÅäÖƹý³ÌÊÇ£º³ÆÈ¡2gµ¨·¯·ÅÈëAÈÝÆ÷ÖУ¬ÔÙ¼ÓË®180mL£¬½Á°èºóÍêÈ«ÈܽâÖƳÉÁòËáÍ­ÈÜÒº£»ÔÚBÈÝÆ÷ÖмÓ2gÉúʯ»Ò£¬ÏȼÓÉÙÁ¿Ë®½Á°è£¬Ê¹Éúʯ»Ò±ä³ÉÊìʯ»Ò£¬ÔÙ¼Ó20mLË®½Á°è£¬Åä³Éʯ»ÒÈ飮½«ÁòËáÍ­ÈÜÒºÂýÂýµØµ¹Èëʯ»ÒÈéÖУ¬Í¬Ê±²»¶Ï½Á°è£¬¼´³ÉÅ©Ò©²¨¶û¶àÒº£®
¸ù¾ÝÒÔÉÏÅäÖƹý³Ì»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð¡ºìÑ¡ÓõÄAÈÝÆ÷ÊÇÇáÇÉ¡¢ÄÍÓõÄÂÁÖÆÈÝÆ÷£¬ÄãÈÏΪ¶ÔÂð£¿ÎªÊ²Ã´£¿
£¨2£©Ð´³öBÈÝÆ÷ÄÚ·¢Éú»¯Ñ§·´Ó¦·½³Ìʽ£®
£¨3£©ÔÚÅäÖƹý³ÌÖУ¬Ð¡ºìÐèÒªÓõ½ÄÄЩÒÇÆ÷£¿Çëд³öÒÇÆ÷Ãû³Æ£¿

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸