ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ÑÇÁòËáÑÎÔÚ¹¤ÒµÉú²úÖЃӹ㷺µÄÓ¦Óã¬Ä³Í¬Ñ§ÔÚʵÑéÖжÔÑÇÁòËáÑεÄÖƱ¸ºÍÐÔÖʽøÐÐ̽¾¿¡£
(1)Cu2SO3¡¤CuSO32H2OÊÇÒ»ÖÖÉîºìÉ«¹ÌÌ壬²»ÈÜÓÚË®ºÍÒÒ´¼£¬100¡æʱ·¢Éú·Ö½â£¬ÆäÖƱ¸ÊµÑé×°ÖÃÈçͼËùʾ¡£
¢ÙÒÇÆ÷XµÄÃû³ÆÊÇ________¡£³£ÎÂÏÂÓÃ×°ÖÃAÖÆÈ¡SO2ʱ£¬ÓýÏŨµÄÁòËá¶ø²»ÓÃÏ¡ÁòËᣬÆäÔÒòÊÇ____________________¡£
¢Ú×°ÖÃCµÄ×÷ÓÃÊÇ________________________¡£
¢Û×°ÖÃBÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________¡£
¢Ü´Ó×°ÖÃBÖлñµÃµÄ¹ÌÌåÐèÏÈÓÃÕôÁóË®³ä·ÖÏ´µÓ£¬ÔÙÕæ¿Õ¸ÉÔ¶ø²»Ö±½ÓÓúæ¸ÉµÄ·½Ê½µÃµ½²úÆ·£¬ÆäÔÒòÊÇ_________________________¡£
(2)ÏòNaHSO3ÈÜÒºÖмÓÈëNaClOÈÜҺʱ£¬·´Ó¦ÓÐÈýÖÖ¿ÉÄܵÄÇé¿ö£º
I.NaHSO3ºÍNaClOÇ¡ºÃ·´Ó¦£»II.NaHSO3¹ýÁ¿£»III.NaClO¹ýÁ¿¡£¼×ͬѧÓûͨ¹ýÏÂÁÐʵÑéÈ·¶¨¸Ã·´Ó¦ÊôÓÚÄÄÒ»ÖÖÇé¿ö£¬ÇëÍê³ÉÏÂ±í£º
ʵÑé²Ù×÷ | Ô¤ÆÚÏÖÏó¼°½áÂÛ |
È¡ÉÏÊö·´Ó¦ºóµÄ»ìºÏÈÜÒºÓÚÊÔ¹Ü AÖУ¬µÎ¼ÓÏ¡ÁòËá | ÈôÓÐÆøÅݲúÉú£¬Ôò_¢Ù__(Ìî¡°I¡±¡°II¡±»ò¡°III¡±£¬ÏÂͬ)³ÉÁ¢£¬ÈôûÓÐÆøÅݲúÉú£¬Ôò_¢Ú___³ÉÁ¢ |
ÁíÈ¡ÉÏÊö·´Ó¦ºóµÄ»ìºÏÈÜÒºÓÚÊÔ¹ÜBÖУ¬µÎ¼Ó¼¸µÎµí·ÛKIÈÜÒº£¬ ³ä·ÖÕñµ´ | ¢Û___£¬ÔòIII³ÉÁ¢ |
(3)ÇëÉè¼Æ¼òµ¥ÊµÑé·½°¸±È½ÏÊÒÎÂÏÂNaHSO3ŨҺÖÐHSO3-µÄµçÀëƽºâ³£ÊýKaÓëË®½âƽºâ³£ÊýKbµÄÏà¶Ô´óС£º________________________¡£
¡¾´ð°¸¡¿ ·ÖҺ©¶· SO2Ò×ÈÜÓÚË®£¬ÓýÏŨµÄÁòËáÓÐÀûÓÚSO2µÄÒݳö ·ÀÖ¹µ¹Îü(»ò×÷°²È«Æ¿) 3Cu2++3SO2+6H2O==Cu2SO3CuSO32H2O¡ý+8H++SO42- ·ÀÖ¹Cu2SO3CuSO32H2O ·¢Éú·Ö½âºÍ±»Ñõ»¯ II I»òIII ÈÜÒº±äΪÀ¶É« ³£ÎÂÏ£¬ÓÃpHÊÔÖ½(»òpH¼Æ)²â¶¨NaHSO3ÈÜÒºµÄpH£¬ÈôPH<7£¬ÔòKa>Kb£»ÈôpH>7£¬ÔòKa<Kb
¡¾½âÎö¡¿(1)¢Ù¸ù¾ÝÒÇÆ÷µÄ½á¹¹¿ÉÖªÒÇÆ÷XµÄÃû³ÆÊÇ·ÖҺ©¶·£»¶þÑõ»¯ÁòÒ×ÈÜÓÚË®£¬ÓÃŨÁòËẬˮÁ¿ÉÙ£¬ÇÒÄÜÎüË®£¬ÓÐÀûÓÚ¶þÑõ»¯ÁòµÄÒç³ö£»
¢Ú×°ÖÃCΪ°²È«Æ¿£¬·Àµ¹Îü£»
¢Û×°ÖÃBÖÐÁòËáÍÓë¶þÑõ»¯ÁòÉú³É²úÎïCu2SO3CuSO32H2OµÄ·´Ó¦£¬ÍÔªËØ»¯ºÏ¼ÛÓнµµÍµ½+1¼Û£¬Ôò¶þÑõ»¯ÁòÖÐÁòÔªËØÉý¼Ûµ½+6¼Û£¬Ôò¸ÃÀë×Ó·½³ÌʽΪ£º3Cu2++3SO2+6H2O=Cu2SO3CuSO32H2O¡ý+8H++SO42-£»
¢ÜÓÉÌâ¿ÉÖª£¬Ð»¸¥ÀÕ¶ûÑÎÊÜÈÈÒ׷ֽ⣬²»Äܺæ¸ÉÊÇΪÁË·ÀÖ¹Cu2SO3CuSO32H2O ·¢Éú·Ö½âºÍ±»Ñõ»¯£»
(2)ÏòNaHSO3ÈÜÒºÖмÓÈëNaClOÈÜҺʱ£¬·´Ó¦ÓÐI£®NaHSO3ºÍNaClOÇ¡ºÃ·´Ó¦£ºNaHSO3+NaClO=NaHSO4+NaCl£»¢ò£®NaClO²»×㣺2NaHSO3+NaClO=Na2SO4+SO2¡ü+H2O+NaCl£»¢ó£®NaClO¹ýÁ¿£ºNaHSO3+NaClO=Na2SO4+NaCl+HClO£»
ÐòºÅ | ʵÑé²Ù×÷ | Ô¤ÆÚÏÖÏó¼°½áÂÛ |
¢Ù | ¼ÌÐøÏòÊÔ¹ÜAÖеμÓäåË®£¬³ä·ÖÕñµ´£® | ÈôÈÜÒºÍÊÉ«£¬½áºÏ²½Öè¢Ù£¬Ôò¢ò³ÉÁ¢£»ÈôÈÜÒº²»ÍÊÉ«£¬½áºÏ²½Öè¢Ù£¬Ôò¢ñ³ÉÁ¢ |
¢Ú | ÁíÈ¡ÉÏÊö»ìºÏÈÜÒºÓÚÊÔ¹ÜBÖУ¬µÎ¼Ó¼¸µÎµí·ÛKIÈÜÒº£¬³ä·ÖÕñµ´£® | ÈÜÒº±äΪÀ¶É«£¬Ôò¢ó³ÉÁ¢£® |
(3)µçÀëÏÔËáÐÔ£¬Ë®½âÏÔ¼îÐÔ£¬Ôò²â¶¨pH¼´¿É£¬ÔòÉè¼ÆʵÑéΪ³£ÎÂÏ£¬ÓÃpHÊÔÖ½£¨»òpH¼Æ£©²â¶¨NaHS03ÈÜÒºµÄpH£¬ÈôpH£¼7£¬ÔòKa£¾Kb£¬ÈôpH£¾7£¬ÔòKa£¼Kb¡£
¡¾ÌâÄ¿¡¿Ï±íÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö,Õë¶Ô±íÖеĢ٢âÖÖÔªËØ£¬ÌîдÏÂÁпհףº
×å ÖÜÆÚ | IA | IIA | ¢óA | IVA | VA | VIA | ¢÷A |
1 | ¢Ù | ||||||
2 | ¢Ú | ¢Û | ¢Ü | ||||
3 | ¢Ý | ¢Þ | ¢ß | ¢à | ¢á | ¢â |
(1)ÉÏÊöÔªËØÖУ¬Ðγɻ¯ºÏÎïÖÖÀà×î¶àµÄÊÇ_____________(ÌîÔªËØ·ûºÅ)¡£
(2)ÔªËآ١¢¢ÜºÍ¢ÝÐγɵĻ¯ºÏÎïµÄµç×ÓʽÊÇ_______£¬¸Ã»¯ºÏÎïÖдæÔڵĻ¯Ñ§¼üÀàÐÍÊÇ_______¡£
(3)¢Ú¡¢¢Û¡¢¢àÈýÖÖÔªËØÔ×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ___________(ÓÃÔªËØ·ûºÅ±íʾ)¡£
(4)¢Ý¡¢¢Þ¡¢¢ßÈýÖÖÔªËØ×îó{¼ÛÑõ»¯Îï¶ÔӦˮ»¯ÎïµÄ¼îÐÔÓÉÇ¿µ½ÈõµÄ˳ÐòÊÇ______________¡£(ÓöÔÓ¦ÎïÖʵĻ¯Ñ§Ê½±íʾ£©
(5)ÔªËآߺ͢á×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÏ໥·´Ó¦µÄÀë×Ó·½³ÌʽΪ___________________¡£
(6)ÄܱȽÏÔªËØ¢áºÍ¢â·Ç½ðÊôÐÔÇ¿ÈõµÄʵÑéÊÂʵÊÇ_________(Ìî×ÖĸÐòºÅ)¡£
a.¢áµÄÇ⻯ÎïµÄËáÐԱȢâµÄÇ⻯ÎïµÄËáÐÔÈõ
b.¢âµÄµ¥ÖÊR2ÓëH2»¯ºÏ±È¢áµÄµ¥ÖÊQÓëH2»¯ºÏÈÝÒ×£¬ÇÒHRµÄÎȶ¨ÐÔ±ÈH2QÇ¿
c.ÔÚ¢áµÄÇ⻯ÎïH2QµÄË®ÈÜÒºÖÐͨÉÙÁ¿¢âµÄµ¥ÖÊR2ÆøÌå¿ÉÖû»³öµ¥ÖÊQ
¡¾ÌâÄ¿¡¿Åð¼°Æ仯ºÏÎïÔÚ²ÄÁÏÖÆÔì¡¢ÓлúºÏ³ÉµÈ·½ÃæÓÃ;·Ç³£¹ã·º¡£»Ø´ðÏÂÁÐÎÊÌ⣺
(1)VB2ÊÇÒ»ÖÖµ¼µçÌմɲÄÁÏ£¬»ù̬·°Ô×ӵļ۵ç×ÓÅŲ¼Í¼Îª_______¡£
(2)B¡¢C¡¢NÈýÖÖÔªËصÚÒ»µçÀëÄÜÓÉСµ½´óµÄ˳ÐòΪ________¡£
(3)ÅðµÄ±»¯ÎïÔÚ¹¤ÒµÖÐÓÐÖØÒª×÷Óã¬ÅðµÄËÄÖÖ±»¯ÎïµÄ·ÐµãÈçϱíËùʾ¡£
BF3 | BCl3 | BBr3 | BI3 | |
·Ðµã/K | 172 | 285 | 364 | 483 |
¢ÙËÄÖÖ±»¯Îï·ÐµãÒÀ´ÎÉý¸ßµÄÔÒòÊÇ__________________¡£
¢ÚÓÃBF3·Ö×ӽṹ½âÊÍ·´Ó¦BF3(g)+NH4F(s)==NH4[BF4] (s)Äܹ»·¢ÉúµÄÔÒò£º____________¡£
ÖƱ¸»·Åð°±ÍéµÄ·½·¨ÈçÏ£º
BCl3¡¢LiBH4ÖÐÅðÔ×ÓµÄÔÓ»¯¹ìµÀÀàÐÍÒÀ´ÎΪ_________£»ÓëB3N3H6»¥ÎªµÈµç×ÓÌåµÄ·Ö×ӵĽṹ¼òʽΪ________________¡£
(4)Á¢·½µª»¯ÅðµÄÈÛµãΪ3000¡æ£¬Æ侧°û½á¹¹ÈçͼËùʾ£¬¾§°û²ÎÊýa=361.5pm¡£
¢ÙÁ¢·½µª»¯ÅðµÄ¾§ÌåÀàÐÍΪ_______________¡£
¢Ú½ôÁÚµÄÁ½¸öÅðÔ×Ó¼äµÄ¾àÀëΪ_______(Áгö¼ÆËãʽ¼´¿É) pm¡£
¢ÛÁ¢·½µª»¯ÅðµÄÃܶÈΪ_____(Áгö¼ÆËãʽ¼´¿É)g©M-3¡£