(1)[2013¡¤ÖØÇìÀí×Û£¬11(2)]µç»¯Ñ§½µ½âNOµÄÔ­ÀíÈçͼËùʾ¡£

¢ÙµçÔ´Õý¼«Îª__________(Ìî¡°A¡±»ò¡°B¡±)£¬Òõ¼«·´Ó¦Ê½Îª

________________________________________________________________________¡£

¢ÚÈôµç½â¹ý³ÌÖÐתÒÆÁË2 molµç×Ó£¬ÔòĤÁ½²àµç½âÒºµÄÖÊÁ¿±ä»¯²î(¦¤m×ó£­¦¤mÓÒ)Ϊ__________ g¡£

(2)[2013¡¤¸£½¨Àí×Û£¬23(2)¢Ú]ÀûÓÃH2S·ÏÆøÖÆÈ¡ÇâÆøµÄ·½·¨ÓжàÖÖ¡£ÆäÖе绯ѧ·¨ÈçͼËùʾ¡£·´Ó¦³ØÖз´Ó¦ÎïµÄÁ÷Ïò²ÉÓÃÆø¡¢ÒºÄæÁ÷·½Ê½£¬ÆäÄ¿µÄÊÇ________£»·´Ó¦³ØÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________¡£·´Ó¦ºóµÄÈÜÒº½øÈëµç½â³Ø£¬µç½â×Ü·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________¡£

(3)[2013¡¤É½¶«Àí×Û£¬28(3)(4)]¢ÙÏÂͼΪµç½â¾«Á¶ÒøµÄʾÒâͼ£¬________(Ìî¡°a¡±»ò¡°b¡±)¼«Îªº¬ÓÐÔÓÖʵĴÖÒø£¬Èôb¼«ÓÐÉÙÁ¿ºì×ØÉ«ÆøÌå²úÉú£¬ÔòÉú³É¸ÃÆøÌåµÄµç¼«·´Ó¦Ê½Îª________¡£

¢ÚΪ´¦ÀíÒøÆ÷±íÃæµÄºÚ°ß(Ag2S)£¬½«ÒøÆ÷½þÓÚÂÁÖÊÈÝÆ÷ÀïµÄʳÑÎË®Öв¢ÓëÂÁ½Ó´¥£¬Ê¹Ag2Sת»¯ÎªAg£¬Ê³ÑÎË®µÄ×÷ÓÃÊÇ________________¡£


´ð°¸¡¡(1)¢ÙA¡¡2NO£«6H2O£«10e£­===N2¡ü£«12OH£­

¢Ú14.4

(2)Ôö´ó·´Ó¦Îï½Ó´¥Ãæ»ý£¬Ê¹·´Ó¦¸ü³ä·Ö

H2S£«2FeCl3===2FeCl2£«S¡ý£«2HCl

2Fe2£«£«2H£«2Fe3£«£«H2¡ü

(3)¢Ùa¡¡2H£«£«NO£«e£­===NO2¡ü£«H2O¡¡¢Ú×öµç½âÖÊÈÜÒº(»òµ¼µç)

½âÎö¡¡(1)¢ÙNOÉú³ÉN2·¢ÉúÁË»¹Ô­·´Ó¦£¬Ó¦ÔÚµç½â³ØµÄÒõ¼«·¢Éú£¬¹ÊAΪԭµç³ØµÄÕý¼«£»Òõ¼«·¢Éú»¹Ô­·´Ó¦£¬¹Êµç¼«·´Ó¦Ê½Îª2NO£«6H2O£«10e£­===N2¡ü£«12OH£­¡£¢ÚÓÉÒõ¼«·´Ó¦Ê½¿ÉÖª£¬Í¨¹ý2 molµç×Ó£¬ÈÜÒº¼õÉÙµÄÖÊÁ¿Îª5.6 g(N2)£¬Í¬Ê±ÓÐ2 mol H£«Í¨¹ýÖÊ×Ó½»»»Ä¤½øÈëÓҲ࣬¹ÊÓÒ²àÈÜÒº¼õÉÙ 3.6 g¡£

Ñô¼«·¢ÉúµÄ·´Ó¦Îª4OH£­£­4e£­===O2¡ü£«2H2O£¬Ã¿Í¨¹ý2 molµç×Ó£¬Éú³É16 g O2£¬Í¬Ê±ÓÐ2 mol H£«Í¨¹ýÖÊ×Ó½»»»Ä¤½øÈëÓҲ࣬ʹ×ó²àÈÜÒºÖÊÁ¿¼õÉÙ18 g£¬¹ÊÁ½²àÈÜÒº¼õÉÙµÄÖÊÁ¿²îΪ14.4 g¡£

(2)²ÉÓÃÆø¡¢ÒºÄæÁ÷·½Ê½µÄÄ¿µÄÊÇÔö´ó·´Ó¦Îï½Ó´¥Ãæ»ý£¬Ê¹·´Ó¦¸ü³ä·Ö¡£·´Ó¦³ØÄÚ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪH2S£«2FeCl3===2FeCl2£«S¡ý£«2HCl¡£½øÈëµç½â³ØµÄÎïÖÊӦΪFeCl2ºÍHCl£¬¸ù¾Ýµç½â³Øͼʾ£¬×ó²àÉú³ÉµÄΪFe3£«£¬ÓÒ²àÉú³ÉµÄΪH2£¬Ôò¿ÉµÃ×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ£º2Fe2£«£«2H£«Í¨µç,2Fe3£«£«H2¡ü¡£

(3)¢Ùµç½â¾«Á¶Òø£¬´ÖÒø×÷Ñô¼«£¬´¿Òø×÷Òõ¼«£¬Òõ¼«ÉϳýÁ˸½×ÅÒøÖ®Í⣬»¹»áÓÐÉÙÁ¿µÄNO2ÆøÌå²úÉú¡£Ô­ÒòÊÇNO£«2H£«£«e£­===NO2¡ü£«H2O¡£

¢ÚÓÃÂÁÖÊÈÝÆ÷ÓëʳÑÎË®´¦ÀíAg2SµÄÔ­ÀíÊÇ£ºÀûÓÃÔ­µç³ØÔ­Àí£¬¸º¼«£º2Al£­6e£­===

2Al3£«£¬Õý¼«£º3Ag2S£«6e£­===6Ag£«3S2£­£¬2Al3£«£«3S2£­£«6H2O===2Al(OH)3¡ý£«3H2S¡ü£¬×Ü·´Ó¦£º2Al£«3Ag2S£«6H2O===2Al(OH)3¡ý£«3H2S¡ü£«6Ag¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


³ýÈ¥ÒÒÍéÖлìÓеÄÉÙÁ¿ÒÒÏ©£¬ÕýÈ·µÄ´¦Àí·½·¨ÊÇ
A.½«ÆøÌåͨÈëË®                        B.½«ÆøÌåͨÈë³ÎÇåʯ»ÒË®

C.½«ÆøÌåͨÈëäåË®                      D.½«ÆøÌåͨÈëÇâÆø½øÐйâÕÕ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


       ÓöèÐԵ缫µç½âÒ»¶¨Å¨¶ÈµÄCuSO4ÈÜÒº£¬Í¨µçÒ»¶Îʱ¼äºó£¬ÏòËùµÃµÄÈÜÒºÖмÓÈë0.1 mol Cu(OH)2ºóÇ¡ºÃ»Ö¸´µ½µç½âÇ°µÄŨ¶ÈºÍpH¡£ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ                (¡¡¡¡)

A£®µç½â¹ý³ÌÖÐÒõ¼«Ã»ÓÐÆøÌåÉú³É

B£®µç½â¹ý³ÌÖÐתÒƵĵç×ÓµÄÎïÖʵÄÁ¿Îª0.4 mol

C£®Ô­CuSO4ÈÜÒºµÄŨ¶ÈΪ0.1 mol¡¤L£­1

D£®µç½â¹ý³ÌÖÐÑô¼«ÊÕ¼¯µ½µÄÆøÌåÌå»ýΪ1.12 L(±ê¿öÏÂ)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÈçͼËùʾ£¬¸÷ÉÕ±­ÖÐÊ¢Óк£Ë®£¬ÌúÔÚÆäÖб»¸¯Ê´µÄËÙ¶ÈÓÉ¿ìµ½ÂýµÄ˳ÐòΪ       (¡¡¡¡)

A£®¢Ú¢Ù¢Û¢Ü¢Ý¢Þ                                   B£®¢Ý¢Ü¢Û¢Ù¢Ú¢Þ

C£®¢Ý¢Ü¢Ú¢Ù¢Û¢Þ                                   D£®¢Ý¢Û¢Ú¢Ü¢Ù¢Þ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


µç½â×°ÖÃÈçͼËùʾ£¬µç½â²ÛÄÚ×°ÓÐKI¼°µí·ÛÈÜÒº£¬ÖмäÓÃÒõÀë×Ó½»»»Ä¤¸ô¿ª¡£ÔÚÒ»¶¨µÄµçѹÏÂͨµç£¬·¢ÏÖ×ó²àÈÜÒº±äÀ¶É«£¬Ò»¶Îʱ¼äºó£¬À¶É«Öð½¥±ädz¡£

ÒÑÖª£º3I2£«6OH£­===IO£«5I£­£«3H2O

ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ                                                                                        (¡¡¡¡)

A£®ÓҲ෢ÉúµÄµç¼«·´Ó¦Ê½£º2H2O£«2e£­===H2¡ü£«2OH£­

B£®µç½â½áÊøʱ£¬ÓÒ²àÈÜÒºÖк¬IO

C£®µç½â²ÛÄÚ·¢Éú·´Ó¦µÄ×Ü»¯Ñ§·½³Ìʽ£ºKI£«3H2OKIO3£«3H2¡ü

D£®Èç¹ûÓÃÑôÀë×Ó½»»»Ä¤´úÌæÒõÀë×Ó½»»»Ä¤£¬µç½â²ÛÄÚ·¢ÉúµÄ×Ü»¯Ñ§·´Ó¦²»±ä

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


       ÏÖ½«ÂÈ»¯ÂÁÈÜÒºÕô¸É×ÆÉÕ²¢ÈÛÈÚºóÓò¬µç¼«½øÐеç½â£¬ÏÂÁÐÓйص缫²úÎïµÄÅжÏÕýÈ·µÄÊÇ                                                                                                                    (¡¡¡¡)

A£®Òõ¼«²úÎïÊÇÇâÆø

B£®Ñô¼«²úÎïÊÇÑõÆø

C£®Òõ¼«²úÎïÊÇÂÁºÍÑõÆø

D£®Ñô¼«²úÎïÖ»ÓÐÂÈÆø

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÈçͼËùʾ£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ                                                                          (¡¡¡¡)

A£®XΪÒõ¼«£¬·¢Éú»¹Ô­·´Ó¦

B£®YΪÕý¼«£¬·¢ÉúÑõ»¯·´Ó¦

C£®YÓëÂËÖ½½Ó´¥´¦ÓÐÇâÆøÉú³É

D£®XÓëÂËÖ½½Ó´¥´¦±äºì

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ijÓлúÎïµÄ½á¹¹¼òʽÈçÓÒͼËùʾ£¬Ôò´ËÓлúÎï¿É·¢Éú·´Ó¦µÄÀàÐÍ¿ÉÄÜÓТÙÈ¡´ú ¢Ú¼Ó³É ¢ÛÏûÈ¥ ¢Üõ¥»¯ ¢ÝË®½â ¢ÞÖÐºÍ ¢ßÑõ»¯ ¢à¼Ó¾Û£¬ÆäÖÐ×éºÏÕýÈ·µÄÊÇ(   )

A£®¢Ù¢Ú¢Û¢Ý¢Þ

B£®¢Ú¢Û¢Ü¢Ý¢Þ

C£®¢Ù¢Ú¢Û¢Ü¢Ý¢Þ

D£®È«²¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÔË®ÂÈþʯ(Ö÷Òª³É·ÖΪMgCl2¡¤6H2O)ΪԭÁÏÉú²ú¼îʽ̼ËáþµÄÖ÷Òª

Á÷³ÌÈçÏ£º

(1)Ô¤°±»¯¹ý³ÌÖÐÓÐMg(OH)2³ÁµíÉú³É£¬ÒÑÖª³£ÎÂÏÂMg(OH)2µÄKsp£½1.8¡Á10£­11£¬ÈôÈÜÒºÖÐc(OH£­)£½3.0¡Á10£­6 mol¡¤L£­1£¬ÔòÈÜÒºÖÐc(Mg2£«)£½________¡£ (2)ÉÏÊöÁ÷³ÌÖеÄÂËҺŨËõ½á¾§£¬ËùµÃÖ÷Òª¹ÌÌåÎïÖʵĻ¯Ñ§Ê½______________________

(3)¸ßÎÂìÑÉÕ¼îʽ̼ËáþµÃµ½MgO¡£È¡¼îʽ̼Ëáþ4.66 g£¬¸ßÎÂìÑÉÕÖÁºãÖØ£¬µÃµ½¹ÌÌå2.00 gºÍ±ê×¼×´¿öÏÂCO2 0.896 L£¬Í¨¹ý¼ÆËãÈ·¶¨¼îʽ̼ËáþµÄ»¯Ñ§Ê½¡£

(4)ÈôÈÈË®½â²»ÍêÈ«£¬ËùµÃ¼îʽ̼ËáþÖн«»ìÓÐMgCO3£¬Ôò²úÆ·ÖÐþµÄÖÊÁ¿·ÖÊý________(Ìî¡°Éý¸ß¡±¡¢¡°½µµÍ¡±»ò¡°²»±ä¡±)¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸