Ä¿Ç°ÊÀ½çÉÏ60%µÄþÊÇ´Óº£Ë®ÖÐÌáÈ¡µÄ¡£Ñ§Éú¾ÍÕâ¸ö¿ÎÌâÕ¹¿ªÁËÌÖÂ۵ġ£ÒÑÖªº£Ë®ÌáþµÄÖ÷Òª²½ÖèÈçÏ£º


ѧÉú¾ÍÕâ¸ö¿ÎÌâÌá³öÁËÒÔÏÂÎÊÌ⣺

(Ò»)ÔÚº£Ë®ÌáþµÄ¹ý³ÌÖÐÈçºÎʵÏÖ¶ÔþÀë×ӵĸ»¼¯£¿

ÓÐÈý¸öѧÉúÌá³ö×Ô¼ºµÄ¹Ûµã¡£

ѧÉú¼×µÄ¹Ûµã£ºÖ±½ÓÍùº£Ë®ÖмÓÈë³Áµí¼Á¡£

ѧÉúÒҵĹ۵㣺¸ßμÓÈÈÕô·¢º£Ë®ºó£¬ÔÙ¼ÓÈë³Áµí¼Á¡£

ѧÉú±ûµÄ¹Ûµã£ºÀûÓÃɹÑκóµÄ¿à±ˮ£¬ÔÙ¼ÓÈë³Áµí¼Á¡£¡¡

ÇëÄãÆÀ¼ÛÈý¸öѧÉúÌá³öµÄ¹ÛµãÊÇ·ñÕýÈ·£¨ÌîÊÇ»ò·ñ£©£¬²¢¼òÊöÀíÓÉ¡£

ÊÇ·ñÕýÈ·

¼òÊöÀíÓÉ

ѧÉú¼×µÄ¹Ûµã

ѧÉúÒҵĹ۵ã

ѧÉú±ûµÄ¹Ûµã

(¶þ)ÔÚº£Ë®ÌáþµÄ¹ý³ÌÖÐÈçºÎʵÏÖ¶ÔþÀë×ӵķÖÀ룿

£¨1£©ÎªÁËʹþÀë×Ó³ÁµíÏÂÀ´£¬¼ÓÈëµÄ×ãÁ¿ÊÔ¼Á¢ÙÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¨Ìѧʽ£©¡£

£¨2£©¼ÓÈëµÄ×ãÁ¿ÊÔ¼Á¢ÚÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¨Ìѧʽ£©¡£

£¨3£©ÊÔ´Ó½ÚÔ¼ÄÜÔ´£¬Ìá¸ß½ðÊôþµÄ´¿¶È·ÖÎö£¬ÒÔÏÂÊÊÒ˵Äұþ·½·¨ÊÇ¡¡¡¡¡¡¡¡ ¡£

¡¡¡¡¡¡ A.¡¡ ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ B.

¡¡

¡¡¡¡ C.¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡D.

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

þ¼°ÆäºÏ½ðÊÇÒ»ÖÖÓÃ;ºÜ¹ãµÄ½ðÊô²ÄÁÏ£¬Ä¿Ç°ÊÀ½çÉÏ60%µÄþÊÇ´Óº£Ë®ÖÐÌáÈ¡µÄ£®Ö÷Òª²½ÖèÈçͼ£º

£¨1£©ÎªÁËʹMgSO4ת»¯ÎªMg£¨OH£©2£¬ÊÔ¼Á¢ÙµÄÎïÖÊÀà±ðΪ
¼î
¼î
£¬ÒªÊ¹MgSO4Íêȫת»¯Îª³Áµí£¬¼ÓÈëÊÔ¼Á¢ÙµÄÁ¿Ó¦
¹ýÁ¿
¹ýÁ¿
£®
£¨2£©¼ÓÈëÊÔ¼Á¢Ùºó£¬Äܹ»·ÖÀëµÃµ½Mg£¨OH£©2³ÁµíµÄ·½·¨ÊÇ
¹ýÂË
¹ýÂË
£¬ÔÚʵÑéÊÒÒªÍê³ÉÕâÖÖ²Ù×÷ÐèÒªµÄ²£Á§ÒÇÆ÷³ýÁËÉÕ±­¡¢Â©¶·Í⣬»¹±ØÐëʹÓõÄÒ»ÖÖÒÇÆ÷Ãû³ÆÊÇ
²£Á§°ô
²£Á§°ô
£¬¸ÃÒÇÆ÷µÄÖ÷Òª×÷ÓÃÊÇ
ÒýÁ÷
ÒýÁ÷
£®
£¨3£©ÊÔ¼Á¢ÚµÄÃû³ÆÊÇ
ÑÎËá
ÑÎËá
£®
£¨4£©ÎÞË®MgCl2ÔÚÈÛÈÚ״̬Ï£¬Í¨µçºó»á²úÉúMgºÍCl2£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
MgCl2£¨ÈÛÈÚ£©
 Í¨µç 
.
 
Mg+Cl2¡ü
MgCl2£¨ÈÛÈÚ£©
 Í¨µç 
.
 
Mg+Cl2¡ü
£®
£¨5£©Éú²ú¹ý³ÌÖÐÐèҪѭ»·ÀûÓõÄÎïÖÊÊÇ
Cl2
Cl2
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨¸½¼ÓÌ⣩º£Ë®Öл¯Ñ§×ÊÔ´µÄ×ۺϿª·¢ÀûÓã¬ÒÑÊܵ½¸÷¹úµÄ¸ß¶ÈÖØÊÓ£®Br2ºÍMgµÈÁ½ÖÖµ¥Öʶ¼¿ÉÒÔ´Óº£Ë®ÖÐÌáÈ¡£¬ÏÂͼΪÌáÈ¡ËüÃǵÄÖ÷Òª²½Ö裺

Çë»Ø´ð£º
¢ñ£®´Óº£Ë®ÖÐÌáÈ¡µÄäåÕ¼ÊÀ½çäåÄê²úÁ¿µÄ1/3£¬Ö÷Òª·½·¨¾ÍÊÇÉÏÊöÁ÷³ÌÖеĿÕÆø´µ³ö·¨£®
£¨1£©ÖÆÈ¡Br2ʱµÚÒ»´ÎͨÈëCl2ʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
Cl2+2Br-¨T2Cl-+Br2
Cl2+2Br-¨T2Cl-+Br2
£®
£¨2£©ÎüÊÕËþÖз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
Br2+SO2+2H2O¨T4H++SO42-+2Br-
Br2+SO2+2H2O¨T4H++SO42-+2Br-
£®
ÓÉ£¨1£©¡¢£¨2£©¿ÉÖª£¬SO2¡¢Cl2¡¢Br2 ÈýÖÖÎïÖÊÑõ»¯ÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ
Cl2£¾Br2£¾SO2
Cl2£¾Br2£¾SO2
£®£¨Óá°£¾¡±±íʾ£©
¢ò£®Ã¾¼°ÆäºÏ½ðÊÇÓÃ;ºÜ¹ãµÄ½ðÊô²ÄÁÏ£¬¶øÄ¿Ç°ÊÀ½çÉÏ60%µÄþ¾ÍÊÇ´Óº£Ë®Öа´ÉÏÊöÁ÷³ÌÌáÈ¡µÄ£®
£¨1£©ÉÏÊöÁ÷³ÌÖÐΪÁËʹMgSO4Íêȫת»¯ÎªMg£¨OH£©2£¬ÊÔ¼Á¢Ù¿ÉÒÔÑ¡ÓÃ
NaOH
NaOH
£¨Ð´»¯Ñ§Ê½£©£®
£¨2£©¼ÓÈëÊÔ¼Á¢Úºó·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
Mg£¨OH£©2+2H+¨TMg2++2H2O
Mg£¨OH£©2+2H+¨TMg2++2H2O
£®
£¨3£©²½Öè¢Ù°üÀ¨¼ÓÈÈ¡¢Õô·¢¡¢ÀäÈ´¡¢½á¾§¡¢
¹ýÂË
¹ýÂË
£®
£¨4£©Í¨µçʱÎÞË®MgCl2ÔÚÈÛÈÚ״̬Ï·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
MgCl2£¨ÈÛÈÚ£©
 µç½â 
.
 
Mg+Cl2¡ü
MgCl2£¨ÈÛÈÚ£©
 µç½â 
.
 
Mg+Cl2¡ü
£®
¢ó£®ÉÏÊöÁ÷³ÌÔÚÉú²úäåºÍþµÄͬʱ»¹¿ÉÒÔÖƵÃÆäËû»¯¹¤ÎïÖÊ£¬±ÈÈçÖƱ¸ÄÍ»ð²ÄÁÏÑõ»¯Ã¾ºÍÑÎËᣮÉú²ú·½·¨ÊÇ£º
¢Ù½«ÂÈ»¯Ã¾¾§Ì壨MgCl2?6H2O£©¼ÓÈȵ½523¡æÒÔÉÏ£¬¸Ã¾§Ìå¿ÉÒÔ·Ö½âµÃµ½ÄÍ»ð²ÄÁÏÑõ»¯Ã¾ºÍÁ½ÖÖÆø̬»¯ºÏÎÆäÖÐÒ»ÖÖÆøÌå³£ÎÂÏÂΪÎÞÉ«ÒºÌ壮
¢Ú½«Á½ÖÖÆøÌåÀäÈ´ÖÁÊÒΣ¬ÔÙ¸ù¾ÝÐèÒª£¬¼ÓÈ벻ͬÁ¿µÄË®£¬¾Í¿ÉµÃµ½²»Í¬Å¨¶ÈµÄÑÎËᣮ
£¨1£©MgCl2?6H2OÔÚ523¡æÒÔÉÏ·Ö½âµÄ»¯Ñ§·½³ÌʽÊÇ
MgCl2?6H2O MgO+2 HCl¡ü+5H2O¡ü
MgCl2?6H2O MgO+2 HCl¡ü+5H2O¡ü
£®
£¨2£©ÏÖÓÃ1mol MgCl2?6H2O·Ö½âËùµÃµÄ·Ç¹ÌÌå²úÎïÀ´ÖÆÈ¡ÃܶÈΪ1.19g/cm3µÄÑÎËáÈÜÒº168mL£¬Ðè¼ÓË®
36.9 g
36.9 g
g£¨¾«È·µ½0.1£©£¬¸ÃÑÎËáÖÐÈÜÖʵÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ
11.9
11.9
mol/L£¨¾«È·µ½0.1£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ä¿Ç°ÊÀ½çÉÏ60%µÄþÊÇ´Óº£Ë®ÖÐÌáÈ¡µÄ£®Ñ§Éú¾ÍÕâ¸ö¿ÎÌâÕ¹¿ªÁËÌÖÂÛ£®ÒÑÖªº£Ë®ÌáþµÄÖ÷Òª²½ÖèÈçÏ£º

£¨1£©¹ØÓÚ¼ÓÈëÊÔ¼Á¢Ù×÷³Áµí¼Á£¬Í¬Ñ§ÃÇÌá³öÁ˲»Í¬·½·¨£®ÇëÄã²ÎÓëËûÃǵÄÌÖÂÛ£º
·½·¨ ÊÇ·ñÕýÈ· ¼òÊöÀíÓÉ
·½·¨1£ºÖ±½ÓÍùº£Ë®ÖмÓÈë³Áµí¼Á ²»ÕýÈ· £¨Ò»£©
·½·¨2£º¸ßμÓÈÈÕô·¢º£Ë®ºó£¬ÔÙ¼ÓÈë³Áµí¼Á £¨¶þ£© £¨Èý£©
ÄãÈÏΪ×îºÏÀíµÄÆäËû·½·¨ÊÇ£º£¨ËÄ£©
£¨Ò»£©
º£Ë®ÖÐþÀë×ÓŨ¶ÈС£¬³Áµí¼ÁµÄÓÃÁ¿´ó£¬²»¾­¼Ã
º£Ë®ÖÐþÀë×ÓŨ¶ÈС£¬³Áµí¼ÁµÄÓÃÁ¿´ó£¬²»¾­¼Ã
£»
£¨¶þ£©
²»ÕýÈ·
²»ÕýÈ·
£»
£¨Èý£©
ÄÜÔ´ÏûºÄ´ó£¬²»¾­¼Ã
ÄÜÔ´ÏûºÄ´ó£¬²»¾­¼Ã
£»
£¨ËÄ£©
ÏòÌ«Ñô¹âÕô·¢Å¨ËõºóµÄº£Ë®ÖУ¬¼ÓÈë³Áµí¼Á
ÏòÌ«Ñô¹âÕô·¢Å¨ËõºóµÄº£Ë®ÖУ¬¼ÓÈë³Áµí¼Á
£®
£¨2£©¿òÖмÓÈëµÄÊÔ¼Á¢ÙÓ¦¸ÃÊÇ
Ca£¨OH£©2
Ca£¨OH£©2
£¨Ìî¡°»¯Ñ§Ê½¡±£©£»¼ÓÈëµÄÊÔ¼Á¢ÚÊÇ
HCl
HCl
£¨Ìî¡°»¯Ñ§Ê½¡±£©£»¹¤ÒµÉÏÓÉÎÞË®MgCl2ÖÆȡþµÄ»¯Ñ§·½³ÌʽΪ£º
MgCl2£¨ÈÚÈÜ£©
 Í¨µç 
.
 
Mg+Cl2¡ü
MgCl2£¨ÈÚÈÜ£©
 Í¨µç 
.
 
Mg+Cl2¡ü
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ä¿Ç°ÊÀ½çÉÏ60%µÄþÊÇ´Óº£Ë®ÖÐÌáÈ¡µÄ£®Ñ§Éú¾ÍÕâ¸ö¿ÎÌâÕ¹¿ªÁËÌÖÂ۵ģ®ÒÑÖªº£Ë®ÌáþµÄÖ÷Òª²½ÖèÈçͼËùʾ£º

ѧÉú¾ÍÕâ¸ö¿ÎÌâÌá³öÁËÒÔÏÂÎÊÌ⣺
£¨Ò»£©ÔÚº£Ë®ÌáþµÄ¹ý³ÌÖÐÈçºÎʵÏÖ¶ÔþÀë×ӵĸ»¼¯£¿ÓÐÈý¸öѧÉúÌá³ö×Ô¼ºµÄ¹Ûµã£®
ѧÉú1µÄ¹Ûµã£ºÖ±½ÓÍùº£Ë®ÖмÓÈë³Áµí¼Á£®
ѧÉú2µÄ¹Ûµã£º¸ßμÓÈÈÕô·¢º£Ë®ºó£¬ÔÙ¼ÓÈë³Áµí¼Á£®
ѧÉú3µÄ¹Ûµã£ºÀûÓÃɹÑκóµÄ¿à±ˮ£¬ÔÙ¼ÓÈë³Áµí¼Á£®
ÇëÄãÆÀ¼ÛÈý¸öѧÉúÌá³öµÄ¹ÛµãÊÇ·ñÕýÈ·£¨ÌîÊÇ»ò·ñ£©£¬²¢¼òÊöÀíÓÉ£®
ÊÇ·ñÕýÈ· ¼òÊöÀíÓÉ
ѧÉú1µÄ¹Ûµã
·ñ
·ñ
º£Ë®ÖÐþÀë×ÓŨ¶ÈС£¬³Áµí¼ÁµÄÓÃÁ¿´ó£¬²»ÀûÓÚþÀë×ӵijÁµí
º£Ë®ÖÐþÀë×ÓŨ¶ÈС£¬³Áµí¼ÁµÄÓÃÁ¿´ó£¬²»ÀûÓÚþÀë×ӵijÁµí
ѧÉú2µÄ¹Ûµã
·ñ
·ñ
ÄÜÔ´ÏûºÄ´ó£¬º£Ë®µÄ×ÛºÏÀûÓõͣ¬³É±¾¸ß
ÄÜÔ´ÏûºÄ´ó£¬º£Ë®µÄ×ÛºÏÀûÓõͣ¬³É±¾¸ß
ѧÉú3µÄ¹Ûµã
ÊÇ
ÊÇ
þÀë×Ó¸»¼¯Å¨¶È¸ß£¬³É±¾µÍ
þÀë×Ó¸»¼¯Å¨¶È¸ß£¬³É±¾µÍ
£¨¶þ£©ÔÚº£Ë®ÌáþµÄ¹ý³ÌÖÐÈçºÎʵÏÖ¶ÔþÀë×ӵķÖÀ룿
£¨1£©ÎªÁËʹþÀë×Ó³ÁµíÏÂÀ´£¬¼ÓÈëµÄ×ãÁ¿ÊÔ¼Á¢ÙÊÇ
NaOH
NaOH
£¨Ìѧʽ£©£®
£¨2£©ÊÔ´Ó½ÚÔ¼ÄÜÔ´£¬Ìá¸ß½ðÊôþµÄ´¿¶È·ÖÎö£¬ÊÊÒËÒ±Á¶Ã¾µÄ·½·¨ÊÇ£¨Óû¯Ñ§·½³Ìʽ±íʾ£©
Mg£¨OH£©2+2HCl=MgCl2+2H2O¡¢MgCl2£¨ÈÛÈÚ£©
 µç½â 
.
 
Mg+Cl2¡ü
Mg£¨OH£©2+2HCl=MgCl2+2H2O¡¢MgCl2£¨ÈÛÈÚ£©
 µç½â 
.
 
Mg+Cl2¡ü
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

þ¼°ÆäºÏ½ðÊÇÓÃ;¹ã·ºµÄ½ðÊô²ÄÁÏ£¬Ä¿Ç°ÊÀ½çÉÏ60%µÄþÊÇ´Óº£Ë®ÖÐÌáÈ¡ µÄ£®Ä³Ñ§Ð£¿ÎÍâÐËȤС×é´Óº£Ë®É¹ÑκóµÄÑᣨÖ÷Òªº¬Na+¡¢Mg2+¡¢Cl-¡¢Br-µÈ£©ÖÐÄ£Ä⹤ҵÉú²úÀ´Ìáȡþ£¬Ö÷Òª¹ý³ÌÈçÏ£¬»Ø´ðÏÂÁÐÎÊÌ⣺
¾«Ó¢¼Ò½ÌÍø
£¨1£©´Óת»¯£¨¢ÙµÃµ½µÄMg£¨ OH£©2³ÁµíÖлìÓÐÉÙÁ¿µÄCa£¨ OH£©2£¬³ýÈ¥ÉÙÁ¿Ca£¨ OH£©2µÄ·½·¨ÊÇÏȽ«³Áµí¼ÓÈ뵽ʢÓÐ
 
µÄÉÕ±­ÖУ¬³ä·Ö½Á°èºó¾­
 
£¨Ìî²Ù×÷·½·¨£©¿ÉµÃ´¿¾»µÄMg£¨ OH£©2£¬Ôڴ˲Ù×÷¹ý³ÌÖУ¬²£Á§°ôµÄ×÷ÓÃÊǽÁ°èºÍ
 
£®¾«Ó¢¼Ò½ÌÍø
£¨2£©Ð´³öת»¯¢ÜÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³Ìʽ
 
£®
£¨3£©ÒÑ֪ת»¯¢ÛµÄ·´Ó¦Ô­ÀíÓëÖÆÈ¡ÎÞË®AlCl3ÏàͬÏÂͼÊÇÖÆÈ¡ÎÞË®AlCl3ʵÑé×°ÖÃͼ£®×°ÖÃAÖеÄÁ½ÒºÌå·Ö±ðÊÇŨÁòËáºÍŨÑÎËᣮÇë»Ø´ð£º
¢ÙDΪʲô²»Ö±½Ó¼ÓÈÈÀ´ÖÆÈ¡ÎÞË®AlCl3£¬ÇëÓû¯Ñ§·½³Ìʽ±íʾ£º
 
£»
¢Ú·ÖҺ©¶·ÖÐӦʢװµÄÊÔ¼ÁÊÇ
 
£»
¢ÛÓÉ·ÖҺ©¶·ÏòÉÕÆ¿ÖмÓÊÔ¼ÁʱӦעÒâµÄÊÂÏîÊÇ
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸