¡¾ÌâÄ¿¡¿NiÔªËØÔÚÉú²ú¡¢Éú»îÖÐÓÐ׏㷺µÄÓ¦Ó᣻شðÏÂÁÐÎÊÌâ:
(1)»ù̬NiÔ×Ó¼Û²ãµç×ÓµÄÅŲ¼Ê½Îª_______¡£
(2)¿Æѧ¼ÒÔÚÑо¿½ðÊô¿óÎïÖÊ×é·ÖµÄ¹ý³ÌÖУ¬·¢ÏÖÁËCu-Ni-FeµÈ¶àÖÖ½ðÊô»¥»¯Îȷ¶¨Ä³ÖÖ½ðÊô»¥»¯ÎïÊǾ§Ì廹ÊǷǾ§Ìå×î¿É¿¿µÄ¿Æѧ·½·¨ÊǶԹÌÌå½øÐÐ______¡£
(3)NiÄÜÓëÀà±ËØ(SCN)2·´Ó¦Éú³ÉNi(SCN)2¡£Ni(SCN)2ÖУ¬µÚÒ»µçÀëÄÜ×î´óµÄÔªËØÊÇ____£»(SCN)2·Ö×ÓÖУ¬ÁòÔ×ÓµÄÔÓ»¯·½Ê½ÊÇ___£¬¦Ò¼üºÍ¦Ð¼üÊýÄ¿Ö®±ÈΪ_____¡£
(4)[Ni(NH3)6](NO3)2ÖУ¬²»´æÔڵĻ¯Ñ§¼üΪ_____(Ìî±êºÅ)¡£
a£®Àë×Ó¼ü b£®½ðÊô¼ü c£®Åäλ¼ü d£®Çâ¼ü
(5)½âÊÍCO±ÈN2µÄÈ۷еã¸ßµÄÔÒò___¡£
(6)ÄøºÏ½ð´¢ÇâµÄÑо¿ÒÑÈ¡µÃºÜ´ó½øÕ¹¡£
¢Ùͼ¼×ÊÇÒ»ÖÖÄø»ùºÏ½ð´¢ÇâºóµÄ¾§°û½á¹¹Ê¾Òâͼ¡£¸ÃºÏ½ð´¢Çâºó£¬º¬1 mol LaµÄºÏ½ð¿ÉÎü¸½H2µÄÊýĿΪ___¡£
¢ÚMg2NiH4ÊÇÒ»ÖÖÖüÇâµÄ½ðÊôÇ⻯Îï¡£ÔÚMg2NiH4¾§°ûÖУ¬NiÔ×ÓÕ¼¾ÝÈçͼÒҵĶ¥µãºÍÃæÐÄ£¬Mg2+´¦ÓÚÒÒͼ°Ë¸öСÁ¢·½ÌåµÄÌåÐÄ¡£Mg2+λÓÚNiÔ×ÓÐγɵÄ___£¨Ìî¡°°ËÃæÌå¿Õ϶¡±»ò¡°ËÄÃæÌå¿Õ϶¡±£©¡£Èô¾§ÌåµÄÃܶÈΪd g/cm3£¬Mg2NiH4µÄĦ¶ûÖÊÁ¿ÎªM g/mol£¬ÔòMg2+ºÍNiÔ×ÓµÄ×î¶Ì¾àÀëΪ___nm(Óú¬d¡¢M¡¢NAµÄ´úÊýʽ±íʾ)¡£
¡¾´ð°¸¡¿3d84s2 X-ÉäÏßÑÜÉäʵÑé N sp3ÔÓ»¯ 5£º4 b ¶þÕßͬÊôÓÚ·Ö×Ó¾§Ì壬ÇÒÏà¶Ô·Ö×ÓÖÊÁ¿Ïàͬ£¬COµÄ¼«ÐÔ´óÓÚN2£¬ËùÒÔ·¶µÂ»ªÁ¦¸ü´ó£¬È۷еã¸ü¸ß 3NA ËÄÃæÌå¿Õ϶ ¡Á107
¡¾½âÎö¡¿
(1)NiÊÇ28ºÅÔªËØ£¬¸ù¾Ý¹¹ÔìÔÀí¿ÉµÃÆäºËÍâµç×ÓÅŲ¼Ê½£¬½ø¶ø¿ÉµÃÆä¼Û²ãµç×ÓµÄÅŲ¼Ê½£»
(2)¸ù¾ÝX-ÉäÏßÑÜÉäʵÑéÈ·¶¨¾§ÌåµÄ´æÔÚ£»
(3)ÔªËصķǽðÊôÐÔԽǿ£¬Æäµç¸ºÐÔ¾ÍÔ½´ó£»(SCN)2·Ö×ӽṹ¼òʽΪN¡ÔC-S-S-C¡ÔN£¬ÆäÖÐSÔ×Ó¼Û²ãµç×Ó¶Ô¸öÊýÊÇ4ÇÒº¬ÓÐ2¸ö¹Âµç×Ó¶Ô£¬¸ù¾Ý¼Û²ãµç×Ó¶Ô»¥³âÀíÂÛÅжÏÁòÔ×ÓµÄÔÓ»¯·½Ê½£»¹²¼Ûµ¥¼üΪ¦Ò¼ü£¬¹²¼ÛË«¼üÖк¬ÓÐÒ»¸ö¦Ò¼ü¡¢Ò»¸ö¦Ð¼ü£¬¹²¼ÛÈý¼üÖк¬ÓÐ1¸ö¦Ò¼ü¡¢2¸ö¦Ð¼ü£»
(4)¸ÃÅäλ»¯ºÏÎïÊôÓÚÀë×Ó»¯ºÏÎº¬ÓÐÀë×Ó¼ü£¬ÔÚÂçÀë×ÓÖк¬Åäλ¼ü¡¢¹²¼Û¼ü£¬Çâ¼ü²»ÊÇ»¯Ñ§¼ü£»
(5)¸ù¾Ý¼«ÐÔ·Ö×ӵķÖ×Ó¼ä×÷ÓÃÁ¦±È·Ç¼«ÐÔ·Ö×ӵķÖ×ÓÖ®¼ä×÷ÓÃÁ¦´ó·ÖÎö£»
(6)¢Ù¸ù¾Ý¾§°ûÖÐÔ×Ó·Ö̯·¨¼ÆËãLa¡¢NiºÍH2ÊýÄ¿£¬Ð´³ö»¯Ñ§Ê½¼´¿É½â´ð£»
¢ÚNiÔ×ÓÕ¼¾Ý¾§°ûµÄ¶¥µãºÍÃæÐÄ£¬½«¾§°û·Ö³É8¸öСÕý·½Ì壬Mg2+´¦ÓÚÒÒͼ°Ë¸öСÁ¢·½ÌåµÄÌåÐÄ£¬¼´Î»ÓÚNiÔ×ÓÐγɵÄËÄÃæÌå¿Õ϶ÄÚ£¬ËùÒÔMg2+ºÍNiÔ×ÓµÄ×î¶Ì¾àÀëΪ¾§°ûÌå¶Ô½ÇÏßµÄËÄ·ÖÖ®Ò»£¬¸ù¾ÝÃܶȹ«Ê½¼ÆË㾧°ûµÄ±ß³¤¼´¿É½â´ð¡£
(1)28ºÅÔªËØNiµÄºËÍâµç×ÓÅŲ¼Ê½ÊÇ1s22s22p63s23p63d84s2£¬Ôڲμӷ´Ó¦Ê±£¬NiÔ×ÓµÄ×îÍâ²ãµÄ4sµç×ӺʹÎÍâ²ãµÄ3dµç×Ó¶¼¿ÉÄÜ·¢Éú±ä»¯£¬Òò´ËÆä¼Û²ãµç×ÓµÄÅŲ¼Ê½ÊÇ3d84s2£»
(2)È·¶¨Ä³ÖÖ½ðÊô»¥»¯ÎïÊǾ§Ì廹ÊǷǾ§Ìå×î¿É¿¿µÄ¿Æѧ·½·¨ÊǶԹÌÌå½øÐÐX-ÉäÏßÑÜÉäʵÑ飻
(3)ÔÚNi(SCN)2ÖÐÉæ¼°µ½µÄÔªËØÓÐNi¡¢S¡¢C¡¢NËÄÖÖÔªËØ£¬ÆäÖÐNiÊǽðÊôÔªËØ£¬µç¸ºÐÔ×îС£¬ÔÚÈýÖַǽðÊôÔªËØS¡¢C¡¢NÖУ¬ÔªËطǽðÊôÐÔ×îÇ¿µÄÔªËØÊÇN£¬ËùÒÔNÔªËصĵ縺ÐÔ×î´ó£»(SCN)2·Ö×ӽṹ¼òʽΪN¡ÔC-S-S-C¡ÔN£¬SÔ×Ó¼Û²ãµç×Ó¶Ô¸öÊýÊÇ4ÇÒº¬ÓÐ2¸ö¹Âµç×Ó¶Ô£¬¸ù¾Ý¼Û²ãµç×Ó¶Ô»¥³âÀíÂÛ¿ÉÖªÁòÔ×ÓµÄÔÓ»¯·½Ê½ÊÇsp3ÔÓ»¯£»ÓÉÓÚ¹²¼Ûµ¥¼üΪ¦Ò¼ü£¬¹²¼ÛË«¼üÖк¬ÓÐÒ»¸ö¦Ò¼ü¡¢Ò»¸ö¦Ð¼ü£¬¹²¼ÛÈý¼üÖк¬ÓÐ1¸ö¦Ò¼ü¡¢2¸ö¦Ð¼ü£¬Ôò(SCN)2·Ö×ÓÖк¬ÓеĦҼüÊýÄ¿ÊÇ5¸ö£¬¦Ð¼üÊýÄ¿ÊÇ4¸ö£¬ËùÒÔ¦Ò¼üºÍ¦Ð¼üÊýÄ¿Ö®±ÈΪ5£º4£»
(4)[Ni(NH3)6](NO3)2ÊÇÀë×Ó»¯ºÏÎÅäÀë×Ó[Ni(NH3)6]2+ÓëNO3-ͨ¹ýÀë×Ó¼ü½áºÏ£»ÔÚÅäÀë×Ó[Ni(NH3)6]2+ÖУ¬ÖÐÐÄNi2+Àë×ÓÓëÅäλ¼üNH3Ö®¼äͨ¹ýÅäλ¼ü½áºÏ£¬ÔÚÅäλÌåNH3ÖУ¬N¡¢HÁ½ÖÖÔªËصÄÔ×ÓÖ®¼äͨ¹ý¹²¼Û¼üN-H½áºÏ£¬ÎïÖÊÖв»º¬½ðÊô¼ü£¬Çâ¼ü²»ÊÇ»¯Ñ§¼ü£¬ËùÒÔ²»´æÔڵĻ¯Ñ§¼üÑ¡ÏîΪb£»
(5) COºÍN2¶¼ÊôÓÚ·Ö×Ó¾§Ì壬ÇÒÏà¶Ô·Ö×ÓÖÊÁ¿Ïàͬ£¬ÓÉÓÚCOÊǼ«ÐÔ·Ö×Ó£¬N2ÊǷǼ«ÐÔ·Ö×Ó£¬COµÄ¼«ÐÔ´óÓÚN2£¬ËùÒÔ·¶µÂ»ªÁ¦¸ü´ó£¬Òò´ËCOÈ۷еã±ÈN2¸ü¸ß£»
(6)¢Ù¾§°ûÖУ¬La¸öÊýΪ8¡Á=1£»Ni¸öÊýΪ8¡Á+1=5£»H2¸öÊýΪ2¡Á+8¡Á=3£¬ÔòÎïÖÊ»¯Ñ§Ê½ÎªLaNi5(H2)3£¬ËùÒÔ1 mol LaµÄºÏ½ð¿ÉÎü¸½3 mol H2£¬º¬ÓеÄH2µÄÊýĿΪ3NA£»
¢ÚNiÔ×ÓÕ¼¾Ý¾§°ûµÄ¶¥µãºÍÃæÐÄ£¬½«¾§°û·Ö³É8¸öСÕý·½Ì壬Mg2+´¦ÓÚÒÒͼ°Ë¸öСÁ¢·½ÌåµÄÌåÐÄ£¬¼´Î»ÓÚNiÔ×ÓÐγɵÄËÄÃæÌå¿Õ϶ÄÚ£¬ËùÒÔMg2+ºÍNiÔ×ÓµÄ×î¶Ì¾àÀëxΪ¾§°ûÌå¶Ô½ÇÏßµÄËÄ·ÖÖ®Ò»£¬É辧°û±ß³¤Îªa cm£¬Ôòx=a cm£¬¾§°ûÖÐÓÐ8¸öMg2+£¬¼´¾§°ûÖÐÓÐ4¸ö¡°Mg2NiH4¡±£¬¾§°ûÖÊÁ¿m==da3g£¬a= cm£¬¹Êx=a cm=¡Á cm=¡Á¡Á107nm¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿±½¶¡Ëᵪ½æÊǵª½æÀ࿹°©Ò©µÄ´ú±íÎÆäºÏ³É·ÏßÈçͼËùʾ£º
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)·´Ó¦¢ÙËùÐèµÄÊÔ¼ÁºÍÌõ¼þÊÇ______________£¬BÖеĹÙÄÜÍÅÃû³ÆÊÇ______________
(2)CµÄ½á¹¹¼òʽΪ______________¡£
(3)д³ö¾ßÓб½»·½á¹¹£¬¼ÈÄÜ·¢ÉúÒø¾µ·´Ó¦ÓÖÄÜ·¢ÉúË®½â·´Ó¦µÄDµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ______________¡£(²»¿¼ÂÇÁ¢ÌåÒì¹¹£¬Ö»Ðèд³ö3¸ö)
(4)¢ÚµÄ·´Ó¦ÀàÐÍÊÇ______________¡£
(5)д³öFµ½GµÄ·´Ó¦·½³Ìʽ______________¡£
(6)Éè¼ÆÓɱ½ºÍÖƱ¸µÄºÏ³É·Ïß(ÎÞ»úÊÔ¼ÁÈÎÑ¡)¡£______________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ËûÃ×°ÍÂÞÍ¡(I)¿ÉÓÃÓÚÖÎÁƼ±ÐÔ°×Ѫ²¡£¬ÆäºÏ³ÉÑо¿¾ßÓÐÖØÒªÒâÒ壬ºÏ³É·ÏßÈçͼËùʾ¡£
ÒÑÖª£ºi.HCCH+
ii.R¡ªNO2R¡ªNH2
iii.R1¡ªNH2++HCl
(1)AÖйÙÄÜÍÅÃû³ÆÊÇ_________¡£
(2)BµÄ½á¹¹¼òʽÊÇ________¡£
(3)D¡úEµÄ»¯Ñ§·½³ÌʽÊÇ__________¡£
(4)ÊÔ¼ÁaÊÇ_________¡£
(5)ÒÑÖªHÔںϳÉIµÄͬʱ£¬»¹Éú³É¼×´¼£¬G¡úHËù¼ÓÎïÖÊLµÄ½á¹¹¼òʽÊÇ______¡£
(6)BµÄÒ»ÖÖͬ·ÖÒì¹¹Ìå·ûºÏÏÂÁÐÌõ¼þ£¬Æä½á¹¹¼òʽÊÇ________¡£
¢ÙÄÜ·¢ÉúÒø¾µ·´Ó¦
¢ÚºË´Å¹²ÕñÇâÆ×Ö»ÓÐÁ½×éÎüÊÕ·å
(7)D¡úEµÄ¹ý³ÌÖÐÓжàÖÖ¸±²úÎÆäÖÐÊôÓڸ߷Ö×Ó»¯ºÏÎïµÄ½á¹¹¼òʽÊÇ_______¡£
(8)Ò²ÊǺϳÉËûÃ×°ÍÂÞÍ¡(I)µÄÒ»ÖÖÔÁÏ£¬ºÏ³É·ÏßÈçͼËùʾ¡£ÀûÓÃÌâÖÐËù¸øÐÅÏ¢£¬Öмä²úÎïµÄ½á¹¹¼òʽÊÇ_______________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÉèNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ
A. 6.02¡Á1023¾ÍÊÇ°¢·ü¼ÓµÂÂÞ³£Êý
B. 1 molË®ÖеÄÇâÔ×ÓÊýĿΪ2NA
C. °¢·ü¼ÓµÂÂÞ³£Êý¸öÁ£×ÓµÄÎïÖʵÄÁ¿ÊÇ1 mol
D. 1 mol °±ÆøËùº¬Ô×ÓÊýԼΪ2.408¡Á1024¸ö
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿±íʾÈÜÒºÖÐŨ¶ÈµÄ·½·¨Í¨³£ÓÐÁ½ÖÖ£ºÈÜÒºÖÐÈÜÖʵÄÖÊÁ¿·ÖÊý(w)ºÍÎïÖʵÄÁ¿Å¨¶È(c)£¬Òò´ËÔÚÅäÖÆÈÜҺʱ£¬¸ù¾Ý²»Í¬µÄÐèÒª£¬Óв»Í¬µÄÅäÖÆ·½·¨£¬ÇëÍê³ÉÌî¿Õ¡£
(1)ÓÃ10%(ÃܶÈΪ1.01 g¡¤cm£3)µÄÇâÑõ»¯ÄÆÈÜÒºÅäÖƳÉ27.5 g 2% µÄÇâÑõ»¯ÄÆÈÜÒº¡£
¢Ù¼ÆË㣺Ðè________g 10%(ÃܶÈΪ1.01 g¡¤cm3)µÄÇâÑõ»¯ÄÆÈÜÒº£¬ÆäÌå»ýΪ________mL£¬Ðè¼Ó________mLË®(¦ÑË®£½1 g¡¤cm£3)½øÐÐÏ¡ÊÍ¡£
¢ÚÁ¿È¡£ºÓÃ________mLÁ¿Í²È¡10% ÇâÑõ»¯ÄÆ£¬Á¿È¡Ê±ÊÓÏßÒª¸úÁ¿Í²________±£³Öˮƽ£¬È»ºóµ¹ÈëÉÕ±ÀÓÃ________mLÁ¿Í²Á¿È¡ÕôÁóˮҲעÈëÉÕ±Àï¡£
¢ÛÈܽ⣺ÓÃ________½«ÉÏÊöÈÜÒº½Á°è¾ùÔÈ£¬¼´µÃ27.5 g 2% µÄÇâÑõ»¯ÄÆÈÜÒº¡£
(2)ÓÃ98%(ÃܶÈΪ1.84 g¡¤cm£3)µÄŨÁòËáÏ¡ÊͳÉ3 mol¡¤L£1µÄÏ¡ÁòËá100 mL£¬»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÐèҪȡŨÁòËá________mL£»
¢ÚÅäÖƲÙ×÷¿É·Ö½â³ÉÈçϼ¸²½£¬ÒÔÏÂÕýÈ·µÄ²Ù×÷˳ÐòÊÇ__________________________(Ìî×Öĸ£¬ÏÂͬ)¡£
A£®ÏòÈÝÁ¿Æ¿ÖÐ×¢ÈëÉÙÁ¿ÕôÁóË®£¬¼ì²éÊÇ·ñ©ˮ
B£®ÓÃÉÙÁ¿ÕôÁóˮϴµÓÉÕ±¼°²£Á§°ô£¬½«ÈÜҺעÈëÈÝÁ¿Æ¿£¬²¢Öظ´²Ù×÷Á½´Î
C£®ÓÃÒÑÀäÈ´µÄÏ¡ÁòËá×¢ÈëÒѼì²é²»Â©Ë®µÄÈÝÁ¿Æ¿ÖÐ
D£®¸ù¾Ý¼ÆË㣬ÓÃÁ¿Í²Á¿È¡Ò»¶¨Ìå»ýµÄŨÁòËá
E£®½«Å¨ÁòËáÑØÉÕ±±ÚÂýÂý×¢ÈëÊ¢ÓÐÕôÁóË®µÄСÉÕ±ÖУ¬²¢²»¶ÏÓò£Á§°ô½Á°è
F£®¸ÇÉÏÈÝÁ¿Æ¿Èû×Ó£¬Õñµ´£¬Ò¡ÔÈ
G£®ÓýºÍ·µÎ¹ÜµÎ¼ÓÕôÁóË®£¬Ê¹ÈÜÒº°¼ÃæÇ¡ºÃÓë¿Ì¶ÈÏßÏàÇÐ
H£®¼ÌÐøÍùÈÝÁ¿Æ¿ÖÐСÐĵؼÓÕôÁóË®£¬Ê¹ÒºÃæ½Ó½ü¿Ì¶ÈÏß
(3)ʵÑéÊÒÐèÅäÖÆ1 mol¡¤L£1µÄÇâÑõ»¯ÄÆÈÜÒººÍ1 mol¡¤L£1µÄÁòËáÈÜÒº¸÷100mL¡£
¢ÙÒªÅäÖÆÇâÑõ»¯ÄÆÈÜÒº£¬ÔÚÓÃÍÐÅÌÌìƽ³ÆÈ¡ÇâÑõ»¯ÄƹÌÌåʱ£¬Ììƽ¶ÁÊýΪ________¡£
A£®4.0 g¡¡¡¡¡¡¡¡¡¡B£®4.00 g¡¡¡¡¡¡¡¡ C£®£¾4.0 g
¢ÚÔÚÅäÖÆÇâÑõ»¯ÄÆÈÜÒººÍÁòËáÈÜÒºµÄ¸÷²½²Ù×÷ÖУ¬ÓÐÃ÷ÏÔ²»Í¬µÄÊÇ__________¡£
A£®³ÆÁ¿»òÁ¿È¡¡¡¡¡¡¡B£®Èܽâ»òÏ¡ÊÍ ¡¡C£®ÒÆÒº¡¢Ï´µÓ¡¡¡¡D£®¶¨ÈÝ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¹è¡¢Õà(32Ge£¬ÈÛµã 937¡æ)ºÍïØ(31Ga)¶¼ÊÇÖØÒªµÄ°ëµ¼Ìå²ÄÁÏ£¬ÔÚº½¿Õº½Ìì²â¿Ø¡¢ºËÎïÀí̽²â¡¢¹âÏËͨѶ¡¢ºìÍâ¹âѧ¡¢Ì«ÑôÄܵç³Ø¡¢»¯Ñ§´ß»¯¼Á¡¢ÉúÎïҽѧµÈÁìÓò¶¼Óй㷺¶øÖØÒªµÄÓ¦Óá£ÕàÓë¹èÊÇͬÖ÷×åÔªËØ¡£
(1)¹èÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ________¡£
(2)¹èºÍÕàÓëÂÈÔªËض¼ÄÜÐγÉÂÈ»¯Îï RCl4(R´ú±íSiºÍGe)£¬´ÓÔ×ӽṹ½Ç¶È½âÊÍÔÒò _______¡£
(3)×ÔÈ»½ç¿óʯÖÐÕàŨ¶È·Ç³£µÍ£¬Òò´Ë´ÓÕà¼Ó¹¤·ÏÁÏ(º¬ÓÎÀë̬Õà)ÖлØÊÕÕàÊÇÒ»Öַdz£ÖØÒªµÄ·½·¨¡£ÈçͼÊÇÒ»ÖÖÌáÈ¡ÕàµÄÁ÷³Ì£º
¢ÙNaClO ÈÜÒº½þÈ¡º¬Õà·ÏÁÏÖеÄÕàʱ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____£»ÎªÁ˼ÓNaClO ÈÜÒº½þÈ¡º¬Õà·ÏÁϵÄËÙÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÓÐ_____¡£
¢Ú²Ù×÷ 1 ºÍ²Ù×÷ 2 ÊÇ____¡£
¢Û GeO2 µÄÈÛµãΪ 1086¡æ£¬ÀûÓÃÇâÆø»¹ÔGeO2£¬Ã¿Éú³É 146kg µÄÕà·Å³ö akJ µÄÈÈÁ¿£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ_______¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁз´Ó¦µÄÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ
A.ÏòCa(ClO)2ÈÜÒºÖÐͨÈë¹ýÁ¿CO2ÖÆ´ÎÂÈË᣺2ClO + H2O + CO2=2HClO +
B.[Ag(NH3)2]OHÓë½ÏŨÑÎËá·´Ó¦Éú³ÉAgCl£º[Ag(NH3)2]+ + OH + 3H+ + Cl=AgCl¡ý+2+ H2O
C.Cl2ÓëÈȵÄNaOHÈÜÒº·´Ó¦ÖÆÈ¡NaClO3£º2Cl2 + 6OH3Cl ++ 3H2O
D.ÏòËáÐÔKMnO4ÈÜÒºÖÐͨÈëSO2£º2+ 5SO2 + 4OH=2Mn2+ + 5+ 2H2O
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿²ÝËáÑÇÌú¾§Ìå(FeC2O4¡¤2H2O£¬M=180g/mol)³Êµ»ÆÉ«,¿ÉÓÃ×÷ɹÖÆÀ¶Í¼¡£Ä³ÊµÑéС×é¶ÔÆä½øÐÐÁËһϵÁÐ̽¾¿¡£
I.´¿¾»²ÝËáÑÇÌú¾§ÌåÈÈ·Ö½â²úÎïµÄ̽¾¿
(1)ÆøÌå²úÎï³É·ÖµÄ̽¾¿¡£Ð¡×é³ÉÔ±²ÉÓÃÈçÏÂ×°ÖÿÉÖظ´Ñ¡ÓÃ)½øÐÐʵÑ飺
¢ÙEÖÐÊ¢×°¼îʯ»ÒµÄÒÇÆ÷Ãû³ÆΪ_________¡£
¢Ú°´ÕÕÆøÁ÷´Ó×óµ½Óҵķ½Ïò£¬ÉÏÊö×°ÖõĽӿÚ˳ÐòΪa¡úg¡úf¡ú_____¡úβÆø´¦Àí×°ÖÃ(ÒÇÆ÷¿ÉÖظ´Ê¹ÓÃ)¡£
¢ÛʵÑéÇ°ÏÈͨÈëÒ»¶Îʱ¼äN2£¬ÆäÄ¿µÄΪ__________________¡£
¢ÜʵÑéÖ¤Ã÷ÁËÆøÌå²úÎïÖк¬ÓÐCO£¬ÒÀ¾ÝµÄʵÑéÏÖÏóΪ_____________¡£
(2)С×é³ÉÔ±Éè¼ÆʵÑéÖ¤Ã÷ÁËAÖзֽâºóµÄ¹ÌÌå³É·ÖΪFeO£¬Ôò²ÝËáÑÇÌú¾§Ìå·Ö½âµÄ»¯Ñ§·½³ÌʽΪ____________________¡£
(3)ɹÖÆÀ¶Í¼Ê±£¬ÒÔK3[Fe(CN)6]ÈÜҺΪÏÔÉ«¼Á£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______________¡£
¢ò.²ÝËáÑÇÌú¾§ÌåÑùÆ·´¿¶ÈµÄ²â¶¨
¹¤ÒµÖƵõIJÝËáÑÇÌú¾§ÌåÖг£º¬ÓÐFeSO4ÔÓÖÊ£¬²â¶¨Æä´¿¶ÈµÄ²½ÖèÈçÏ£º
²½Öè1£º³ÆÈ¡mg²ÝËáÑÇÌú¾§ÌåÑùÆ·²¢ÈÜÓÚÏ¡HSO4ÖУ¬Åä³É250mLÈÜÒº£»
²½Öè2£ºÈ¡ÉÏÊöÈÜÒº25.00mL£¬Óà cmol/L KMnO4±ê×¼ÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ±ê×¼ÒºV1mL£»
²½Öè3£ºÏò·´Ó¦ºóÈÜÒºÖмÓÈëÊÊÁ¿Ð¿·Û£¬³ä·Ö·´Ó¦ºó£¬¼ÓÈëÊÊÁ¿Ï¡H2SO4£¬ÔÙÓà cmol/L KMnO4±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ±ê×¼ÒºV2mL¡£
(4)²½Öè2Öеζ¨ÖÕµãµÄÏÖÏóΪ______________£»²½Öè3ÖмÓÈëп·ÛµÄÄ¿µÄΪ_______¡£
(5)²ÝËáÑÇÌú¾§ÌåÑùÆ·µÄ´¿¶ÈΪ________£»Èô²½Öè1ÅäÖÆÈÜҺʱ²¿·ÖFe2+±»Ñõ»¯±äÖÊ£¬Ôò²â¶¨½á¹û½«____(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°²»±ä¡±)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿íÚÊÇ·¢Õ¹¸ß¿Æ¼¼²úÒµ¡¢¹ú·ÀÓë¼â¶Ë¼¼Êõ²»¿É»òȱµÄÔÁÏ¡£H2TeO3ÊÇÒ»ÖֱȲÝËáËáÐÔÈõµÄ¶þÔªËᣬ¹¤ÒµÉϳ£ÓÃÍÑô¼«Äà[Ö÷Òª³É·ÖÊÇíÚ»¯ÑÇÍ(Cu2Te)£¬º¬ÉÙÁ¿µÄAg¡¢Au]»ØÊÕíÚ£¬Æ乤ÒÕÁ÷³ÌÈçÏ£º
ÒÑÖª£ºCuC2O4µÄKspΪ2.2¡Á10-8£»Àë×ÓŨ¶ÈСÓÚ1¡Á10-5mol/Lʱ£¬¼´Àë×ÓÍêÈ«³Áµí¡£
(1)Cu2TeÖÐTeµÄ»¯ºÏ¼ÛÊÇ___¡£
(2)ÂËÔüµÄ³É·ÖÊÇ___£¬ÂËÒº¢ÙÖк¬ÓеÄÑõ»¯Ëá½þʱÑõ»¯²úÎïΪ____¡£Ñõ»¯Ëá½þʱζȹý¸ß»áʹíڵĽþ³öÂʽµµÍ£¬ÔÒòÊÇ_______¡£
(3)ÈôҪʹCu2+ÍêÈ«³Áµí£¬Ó¦¿ØÖÆC2O42-µÄŨ¶È²»µÍÓÚ_____¡£
(4)»¹Ô·´Ó¦µÄÀë×Ó·½³ÌʽΪ______________¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com