[»¯Ñ§Ñ¡ÐÞ¡ª¡ª2£º»¯Ñ§Óë¼¼Êõ]£¨15·Ö£©
½«º£Ë®µ­»¯ºÍÓëŨº£Ë®×ÊÔ´»¯½áºÏÆðÀ´ÊÇ×ÛºÏÀûÓú£Ë®µÄÖØҪ;¾¶Ö®Ò»¡£Ò»°ãÊÇÏȽ«º£Ë®µ­»¯»ñµÃµ­Ë®£¬ÔÙ´ÓÊ£ÓàµÄŨº£Ë®ÖÐͨ¹ýһϵÁй¤ÒÕÌáÈ¡ÆäËû²úÆ·¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÏÂÁиĽøºÍÓÅ»¯º£Ë®×ÛºÏÀûÓù¤ÒÕµÄÉèÏëºÍ×ö·¨¿ÉÐеÄÊÇ            £¨ÌîÐòºÅ£©¡£
¢ÙÓûìÄý·¨»ñÈ¡µ­Ë®          ¢ÚÌá¸ß²¿·Ö²úÆ·µÄÖÊÁ¿ 
¢ÛÓÅ»¯ÌáÈ¡²úÆ·µÄÆ·ÖÖ        ¢Ü¸Ä½ø¼Ø£®ä壮þµÄÌáÈ¡¹¤ÒÕ
£¨2£©²ÉÓá°¿ÕÆø´µ³ö·¨¡±´ÓŨº£Ë®Öдµ³öBr2£¬²¢Óô¿¼îÎüÊÕ¡£¼îÎüÊÕäåµÄÖ÷Òª·´Ó¦ÊÇ£ºBr2+Na2CO3+H2NaBr + NaBrO3+6NaHCO3£¬ÎüÊÕ1mol Br2ʱתÒƵĵç×ÓΪ        mol¡£
£¨3£©º£Ë®ÌáþµÄÒ»¶Î¹¤ÒÕÁ÷³ÌÈçÏÂͼ£º

Ũº£Ë®µÄÖ÷Òª³É·ÖÈçÏ£º

¸Ã¹¤ÒÕ¹ý³ÌÖУ¬ÍÑÁò½×¶ÎÖ÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪ                             £¬²úÆ·2µÄ»¯Ñ§Ê½Îª       £¬1LŨº£Ë®×î¶à¿ÉµÃµ½²úÆ·2µÄÖÊÁ¿Îª     g¡£
£¨4£©²ÉÓÃʯīÑô¼«£®²»Ðâ¸ÖÒõ¼«µç½âÈÛÈÚµÄÂÈ»¯Ã¾£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                 £»µç½âʱ£¬ÈôÓÐÉÙÁ¿Ë®´æÔÚ»áÔì³É²úƷþµÄÏûºÄ£¬Ð´³öÓйط´Ó¦µÄ»¯Ñ§·½³Ìʽ                        ¡£

£¨1£©¢Ú¢Û¢Ü      
£¨2£©5/3
£¨3£©Ca2£« +SO42¡ª====CaSO4¡ý   Mg(OH)2    69.6
£¨4£©MgCl2Mg + Cl2¡ü
Mg + 2H2OMg(OH)2 + H2¡ü

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÓûìÄý·¨Ö»ÄܳýÈ¥º£Ë®ÖеÄÐü¸¡Î²»ÄÜ»ñÈ¡µ­Ë®£¬¹ÊÉèÏëºÍ×ö·¨¿ÉÐеÄÊǢڢۢܡ££¨2£©ÀûÓû¯ºÏ¼ÛÉý½µ·¨ÅäƽÑõ»¯»¹Ô­·½³Ìʽ£º3Br2+6Na2CO3+3H2O  ==== 5NaBr + NaBrO3+ 6NaHCO3£¬ÔòÎüÊÕ1mol Br2ʱתÒƵĵç×ÓΪ5/3mol¡££¨3£©¸ù¾ÝŨº£Ë®µÄ³É·Ö¼°¹¤ÒÕÁ÷³ÌÖª£¬ÍÑÁò½×¶ÎΪÓøÆÀë×Ó³ýȥŨº£Ë®ÖеÄÁòËá¸ù£¬Ö÷Òª·´Ó¦µÄÀë×Ó·½³ÌʽΪCa2£« +SO42¡ª====CaSO4¡ý£»ÓÉÌâ¸øÁ÷³Ìͼ֪£¬²úÆ·2µÄ»¯Ñ§Ê½ÎªMg(OH)2£¬1LŨº£Ë®º¬Ã¾Àë×Ó28.8g£¬ÎïÖʵÄÁ¿Îª1.2mol£¬¸ù¾ÝþԪËØÊغãÖª£¬×î¶à¿ÉµÃµ½Mg(OH)21.2mol£¬ÖÊÁ¿Îª69.6g¡££¨4£©²ÉÓÃʯīÑô¼«¡¢²»Ðâ¸ÖÒõ¼«µç½âÈÛÈÚµÄÂÈ»¯Ã¾£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪMgCl2Mg + Cl2¡ü£»µç½âʱ£¬ÈôÓÐÉÙÁ¿Ë®´æÔÚ£¬¸ßÎÂÏÂþÓëË®·¢Éú·´Ó¦»áÔì³É²úƷþµÄÏûºÄ£¬Óйط´Ó¦µÄ»¯Ñ§·½³ÌʽΪMg + 2H2OMg(OH)2 + H2¡ü¡£
¿¼µã£º¿¼²é»¯Ñ§Óë¼¼Êõ£¬º£Ë®µÄ×ÛºÏÀûÓá£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓÉÏÂÁÐÎïÖÊÒ±Á¶ÏàÓ¦½ðÊôʱͨ³£ÓÃÈȷֽⷨµÄÊÇ

A£®NaCl  B£®HgO C£®Cu2D£®Al2O3 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÃæÁоٵÄÄÜÔ´ÖÐÊôÓÚ¶þ´ÎÄÜÔ´µÄÊÇ

A£®µçÄÜ¡¢ÕôÆû B£®µçÄÜ¡¢Ë®ÄÜ
C£®ÕôÆû¡¢·çÄÜ D£®Ãº¡¢Ê¯ÓÍ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªMgO¡¢MgCl2µÄÈÛµã·Ö±ðΪ2800¡æ¡¢604¡æ£¬½«MgO¡¢MgCl2¼ÓÈÈÈÛÈÚºóͨµçµç½â£¬¶¼¿ÉµÃµ½½ðÊôþ¡£º£Ë®Öк¬ÓÐMgCl2£¬¹¤ÒµÉÏ´Óº£Ë®ÖÐÌáȡþ£¬ÕýÈ·µÄ·½·¨ÊÇ £¨   £©

A£®º£Ë®Mg(OH)2Mg 
B£®º£Ë®MgCl2ÈÜÒºMgCl2ÈÛÈÚMg 
C£®º£Ë®Mg(OH)2MgOMg 
D£®º£Ë®Mg(OH)2MgCl2ÈÜÒº MgCl2ÈÛÈÚMg 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¡¾Ñ¡ÐÞ2£º»¯Ñ§Óë¼¼Êõ¡¿£¨15·Ö£©
¿ÕÆø´µ³ö·¨¹¤ÒÕ£¬ÊÇÄ¿Ç°¡°º£Ë®Ìáä塱µÄ×îÖ÷Òª·½·¨Ö®Ò»¡£Æ乤ÒÕÁ÷³ÌÈçÏ£º

£¨1£©²½Öè¢ÜµÄÀë×Ó·´Ó¦·½³ÌʽΪ                                           ¡£
£¨2£©äå΢ÈÜÓÚË®£¬²½Öè¢àÖÐäåÕôÆøÀäÄýºóµÃµ½ÒºäåÓëäåË®µÄ»ìºÏÎËüÃǵÄÏà¶ÔÃܶÈÏà²î½Ï´ó¡£·ÖÀë³öÒºäåµÄʵÑéÊÒ·½·¨Îª            ¡£
£¨3£©²½Öè¢ÞÈç¹ûÔÚʵÑéÊÒÖнøÐУ¬ÐèÓõ½µÄ²£Á§ÒÇÆ÷ÓР                       ¡£
£¨4£©¹¤ÒµÉú²ú²»Ö±½ÓÕôÁóº¬äåµÄº£Ë®µÃµ½Òºä壬¶øÒª¾­¹ý¡°¿ÕÆø´µ³ö¡¢SO2ÎüÊÕ¡¢ÂÈ»¯¡±£¬Ô­ÒòÊÇ                                                             ¡£
£¨5£©¿à±ˮ»¹¿ÉÒÔÓÃÓÚÖÆÈ¡½ðÊôþ£¬Óû¯Ñ§·½³Ìʽ±íʾ´Ó¿à±ˮÖÆÈ¡½ðÊôþµÄ·´Ó¦Ô­Àí           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

º£ÑóÊÇÒ»¸ö·á¸»µÄ×ÊÔ´±¦¿â£¬Í¨¹ýº£Ë®µÄ×ÛºÏÀûÓÿɻñµÃÐí¶àÎïÖʹ©ÈËÀàʹÓá£
£¨1£©º£Ë®ÖÐÑεĿª·¢ÀûÓãº
¢Ùº£Ë®ÖÆÑÎÄ¿Ç°ÒÔÑÎÌ﷨ΪÖ÷£¬½¨ÑÎÌï±ØÐëÑ¡ÔÚÔ¶Àë½­ºÓÈ˺£¿Ú£¬¶à·çÉÙÓ꣬³±Ï«Âä²î´óÇÒÓÖƽ̹¿Õ¿õµÄº£Ì²¡£Ëù½¨ÑÎÌï·ÖΪÖüË®³Ø¡¢Õô·¢³ØºÍ      ³Ø¡£
¢ÚÄ¿Ç°¹¤ÒµÉϲÉÓñȽÏÏȽøµÄÀë×Ó½»»»Ä¤µç½â²Û·¨½øÐÐÂȼҵÉú²ú£¬ÔÚµç½â²ÛÖÐÑôÀë×Ó½»»»Ä¤Ö»ÔÊÐíÑôÀë×Óͨ¹ý£¬×èÖ¹ÒõÀë×ÓºÍÆøÌåͨ¹ý£¬Çë˵Ã÷ÂȼîÉú²úÖÐÑôÀë×Ó½»»»Ä¤µÄ×÷Óãº
                                                     £¨Ð´Ò»µã¼´¿É£©¡£
£¨2£©µçÉøÎö·¨ÊǽüÄêÀ´·¢Õ¹ÆðÀ´µÄÒ»ÖֽϺõĺ£Ë®µ­»¯¼¼Êõ£¬ÆäÔ­ÀíÈçͼËùʾ¡£ÆäÖоßÓÐÑ¡ÔñÐÔµÄÒõÀë×Ó½»»»Ä¤ºÍÑôÀë×Ó½»»»Ä¤Ïà¼äÅÅÁС£Çë»Ø´ðÏÂÃæµÄÎÊÌ⣺

¢Ùº£Ë®²»ÄÜÖ±½ÓͨÈ˵½Òõ¼«ÊÒÖУ¬ÀíÓÉÊÇ                   ¡£
¢ÚA¿ÚÅųöµÄÊÇ         £¨Ìî¡°µ­Ë®¡±»ò¡°Å¨Ë®¡±£©¡£
£¨3£©Óÿà±£¨º¬Na+¡¢K+¡¢Mg2+¡¢Cl-¡¢Br-µÈÀë×Ó£©¿ÉÌáÈ¡ä壬ÆäÉú²úÁ÷³ÌÈçÏ£º

¢ÙÈôÎüÊÕËþÖеÄÈÜÒºº¬Br03-£¬ÔòÎüÊÕËþÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ£º                  ¡£
¢Úͨ¹ý¢ÙÂÈ»¯ÒÑ»ñµÃº¬Br2µÄÈÜÒº£¬ÎªºÎ»¹Ðè¾­¹ý´µ³ö¡¢ÎüÊÕ¡¢ËữÀ´ÖØлñµÃº¬Br2µÄÈÜÒº£¿
¢ÛÏòÕôÁóËþÖÐͨÈëË®ÕôÆø¼ÓÈÈ£¬¿ØÖÆζÈÔÚ90¡æ×óÓÒ½øÐÐÕôÁóµÄÔ­ÒòÊÇ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Ò±Á¶½ðÊô³£ÓÃÒÔϼ¸ÖÖ·½·¨£º¢ÙÒÔC¡¢CO»òH2×ö»¹Ô­¼Á»¹Ô­¡¡¢ÚÒԽϻîÆýðÊôNa¡¢MgµÈ»¹Ô­¡¡¢ÛÀûÓÃÂÁÈÈ·´Ó¦Ô­Àí»¹Ô­¡¡¢Üµç½â·¨¡¡¢ÝÈȷֽⷨ
ÏÂÁнðÊô¸÷²ÉÓÃÄÄÖÖ·½·¨»¹Ô­×î¼Ñ¡£
£¨1£©Fe¡¢Zn¡¢CuµÈÖеȻîÆýðÊô________¡£
£¨2£©Na¡¢Mg¡¢AlµÈ»îÆûò½Ï»îÆýðÊô________¡£
£¨3£©Hg¡¢AgµÈ²»»îÆýðÊô________¡£
£¨4£©V¡¢Cr¡¢Mn¡¢WµÈ¸ßÈÛµã½ðÊô________¡£
£¨5£©K¡¢Rb¡¢Cs¡¢TiµÈ½ðÊôͨ³£»¹Ô­·½·¨ÊÇ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚÈÛÈÚ״̬Ï£¬NaÓëKCl´æÔÚ¿ÉÄæ·´Ó¦£ºNa£«KClNaCl£«K£¬Í¨¹ýµ÷Õûζȣ¬¿ÉÀûÓýðÊôNaÀ´ÖÆÈ¡K¡£

ÎïÖÊ
K
Na
KCl
NaCl
È۵㣨¡æ£©
63.6
97.8
770
801
·Ðµã£¨¡æ£©
774
883
1 500
1 413
 
¸ù¾ÝÉϱíµÄÈÛµãºÍ·Ðµã£¬È·¶¨ÓÉNaÓëKCl·´Ó¦ÖÆÈ¡KµÄºÏÊÊζÈΪ£¨ £©
A£®770¡æ        B£®801¡æ
C£®850¡æ        D£®770¡æ¡«801¡æ

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸