¡¾ÌâÄ¿¡¿Åäλ¼üÊÇÒ»ÖÖÌØÊâµÄ¹²¼Û¼ü£¬¼´¹²Óõç×Ó¶ÔÓÉijÔ×Óµ¥·½ÃæÌṩºÍÁíÒ»Ìṩ¿Õ¹ìµÀµÄÁ£×Ó½áºÏ¡£ÈçNH¾ÍÊÇÓÉNH3(µªÔ×ÓÌṩµç×Ó¶Ô)ºÍH£«(Ìṩ¿Õ¹ìµÀ)ͨ¹ýÅäλ¼üÐγɵġ£¾Ý´Ë£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÏÂÁÐÁ£×ÓÖпÉÄÜ´æÔÚÅäλ¼üµÄÊÇ£¨£¨____£©£©
A£®CO2 B£®H3O£« C£®CH4 D£®[Ag(NH3)2] +
£¨2£©ÏòÁòËáÍÈÜÒºÖеμӰ±Ë®£¬»áÓÐÀ¶É«³Áµí²úÉú£¬¼ÌÐøµÎ¼Ó£¬³ÁµíÈܽ⣬ÈÜÒº±ä³ÉÉîÀ¶É«¡£Çëд³öÆäÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ_______________¡¢________________¡£
£¨3£©Çëд³öÍÔ×ӵĺËÍâµç×ÓÅŲ¼Ê½_________________¡£
£¨4£©Åäλ»¯Ñ§´´Ê¼ÈËά¶ûÄÉ·¢ÏÖ£¬½«¸÷Ϊ1molµÄCoCl3¡¤6NH3(»ÆÉ«)¡¢CoCl3¡¤5NH3(×ϺìÉ«)¡¢CoCl3¡¤4NH3(ÂÌÉ«)¡¢CoCl3¡¤4NH3(×ÏÉ«)ËÄÖÖÅäºÏÎïÈÜÓÚË®£¬¼ÓÈë×ãÁ¿ÏõËáÒøÈÜÒº£¬Éú³ÉÂÈ»¯Òø³Áµí·Ö±ðΪ3mol¡¢2mol¡¢1mol¡¢ºÍ1mol¡£ÒÑÖªÉÏÊöÅäºÏÎïÖÐÅäÀë×ÓµÄÅäλÊý¾ùΪ6¡£Çë¸ù¾ÝʵÑéÊÂʵÓÃÅäºÏÎïµÄÐÎʽд³öËüÃǵĻ¯Ñ§Ê½¡£
¢ÙCoCl3¡¤5NH3________________¡¢¢ÚCoCl3¡¤4NH3£¨×ÏÉ«£©__________________¡£
¡¾´ð°¸¡¿ B D Cu2++2NH3¡¤H2O = Cu(OH)2¡ý+2NH4+ Cu(OH)2+4NH3=[Cu(NH3)4]2++2OH- »ò Cu(OH)2+4NH3¡¤H2O=[Cu(NH3)4]2++2OH-+4H2O 1s22s22p63s23p63d104s1»ò[Ar]3d104s1 [Co(NH3)5Cl]Cl2 [Co(NH3)4Cl2]Cl
¡¾½âÎö¡¿±¾Ìâ·ÖÎö£º±¾ÌâÖ÷Òª¿¼²éÅäºÏÎïµÄ½á¹¹¡£
£¨1£©H2O(ÑõÔ×ÓÌṩµç×Ó¶Ô)ºÍH£«(Ìṩ¿Õ¹ìµÀ)ÐγÉÅäλ¼ü£¬NH3(µªÔ×ÓÌṩµç×Ó¶Ô)ºÍAg(Ìṩ¿Õ¹ìµÀ)ÐγÉÅäλ¼ü£¬¹ÊÑ¡BD¡£
£¨2£©ÆäÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪCu2++2NH3¡¤H2O = Cu(OH)2¡ý+2NH4+¡¢Cu(OH)2+4NH3=[Cu(NH3)4]2++2OH- »ò Cu(OH)2+4NH3¡¤H2O=[Cu(NH3)4]2++2OH-+4H2O¡£
£¨3£©ÍÔ×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p63d104s1»ò[Ar]3d104s1¡£
£¨4£©¢Ù1molCoCl3¡¤5NH3Éú³ÉÂÈ»¯Òø³Áµí2mol£¬ËùÒÔÍâ½çÓÐ2¸öÂÈÀë×Ó£¬ÄÚ½çÓÐ1¸öÂÈÀë×Ó¡¢5¸öNH3£¬ÅäºÏÎïµÄ»¯Ñ§Ê½Îª[Co(NH3)5Cl]Cl2£»¢Ú1molCoCl3¡¤4NH3Éú³ÉÂÈ»¯Òø³Áµí1mol£¬ËùÒÔÍâ½çÓÐ1¸öÂÈÀë×Ó£¬ÄÚ½çÓÐ2¸öÂÈÀë×Ó¡¢4¸öNH3£¬ÅäºÏÎïµÄ»¯Ñ§Ê½Îª[Co(NH3)4Cl2]Cl¡£
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¿¡¿ÈçͼÊÇÔªËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬¢Ù¡«¢áÊÇÔªËØÖÜÆÚ±íÖеIJ¿·ÖÔªËØ¡£
¢ñA | 0 | |||||||
¢òA | ¢óA | ¢ôA | ¢õA | ¢öA | ¢÷A | |||
¡¡ | ¢Ú | ¢Û | ¢Ü | |||||
¢Ý | ¢Þ | ¡¡ | ¢ß | ¢à | ¢á |
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¢ÞµÄÔªËØ·ûºÅÊÇ_______ ,¢ßµÄÔ×ӽṹʾÒâͼ _______________________£¬Ô×Ӱ뾶¢Ý______¢Þ£¨Ìî¡°£¾¡±»ò¡°£¼¡±£©¡£
(2)ÔªËآڢۢܵÄÆø̬Ç⻯ÎïÖÐ×îÎȶ¨µÄÊÇ______________£¨Ìѧʽ£©¡£
(3)ÔªËآݵÄ×î¸ß¼ÛÑõ»¯Îï¶ÔÓ¦µÄË®»¯ÎïÓëÔªËØ¢áµÄÇ⻯ÎïµÄË®ÈÜÒº·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_________________________________¡£
(4)ÔªËآٺ͢áÐγɵĻ¯ºÏÎïµÄµç×ÓʽÊÇ_____________________¡£
(5)ÔªËطǽðÊôÐÔ¢à_______¢á£¨Ìî¡°£¾¡±»ò¡°£¼¡±£©¡£
(6)ÒÑÖªÔªËØ¢àµÄÔ×ÓÐòÊýΪa£¬ÔòÔªËØ¢ÛµÄÔ×ÓÐòÊý¿ÉÄÜΪ___________¡£
A¡¢a + 8 B¡¢a-8 C¡¢a + 2 D¡¢a-2
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¶ÔSO2ÓëCO2˵·¨ÕýÈ·µÄÊÇ
A£®¶¼ÊÇÖ±ÏßÐνṹ
B£®ÖÐÐÄÔ×Ó¶¼²ÉÈ¡spÔÓ»¯¹ìµÀ
C£®SÔ×ÓºÍCÔ×ÓÉ϶¼Ã»Óй¶Եç×Ó
D£®SO2ΪVÐνṹ£¬ CO2ΪֱÏßÐνṹ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿£¨10·Ö£©¸ù¾ÝÏÂÃæµÄ·´Ó¦Â·Ïß¼°Ëù¸øÐÅÏ¢Ìî¿Õ£º
£¨1£©AµÄ½á¹¹¼òʽÊÇ £¬Ãû³ÆÊÇ ¡£
£¨2£©¢ÙµÄ·´Ó¦ÀàÐÍ ¢ÚµÄ·´Ó¦ÀàÐÍ ¡£
£¨3£©·´Ó¦¢ÜµÄ»¯Ñ§·½³Ìʽ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®ÔÚ·´Ó¦ÖУ¬½ðÊôµ¥ÖÊÖ»±íÏÖ»¹ÔÐÔ£¬·Ç½ðÊôµ¥ÖÊÖ»±íÏÖÑõ»¯ÐÔ
B£®Ê§È¥µç×ÓµÄÎïÖÊ×öÑõ»¯¼Á
C£®ÔªËصĵ¥ÖÊ¿ÉÓÉÑõ»¯»ò»¹Ôº¬¸ÃÔªËصĻ¯ºÏÎïÀ´ÖƵÃ
D£®½ðÊôÑôÀë×Ó±»»¹ÔÒ»¶¨µÃµ½½ðÊôµ¥ÖÊ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿NaHSO4ÔÚË®ÈÜÒºÖÐÄܹ»µçÀëH+¡¢Na+ºÍSO42-£¬ÏÂÁжÔÓÚNaHSO4µÄ·ÖÀàÖв»ÕýÈ·µÄÊÇ( )
A.NaHSO4ÊÇÑÎ B.NaHSO4ÊÇËáʽÑÎ
C.NaHSO4ÊÇÄÆÑÎ D.NaHSO4ÊÇËá
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿£¨1£©»ý¼«±£»¤Éú̬»·¾³¿ÉʵÏÖÈËÓë×ÔÈ»µÄºÍг¹²´¦¡£
¢ÙÏÂÁÐ×ö·¨»á¼Ó¾çÎÂÊÒЧӦµÄÊÇ__________£¨Ìî×Öĸ£©¡£
a.Ö²Ê÷ÔìÁÖ b.ȼú¹©Å¯ c.·çÁ¦·¢µç
¢ÚÏÂÁзÀÖΡ°°×É«ÎÛȾ¡±µÄÕýÈ··½·¨ÊÇ_____________£¨Ìî×Öĸ£©¡£
a.ʹÓÿɽµ½âËÜÁÏ b.¶Ìì·ÙÉշϾÉËÜÁÏ c.Ö±½ÓÌîÂñ·Ï¾ÉËÜÁÏ
¢ÛΪ¼õÇá´óÆøÎÛȾ£¬¶à¸ö³ÇÊÐÒѽûֹȼ·ÅÑÌ»¨±¬Öñ¡£¡°½ûֹȼ·ÅÑÌ»¨±¬Öñ¡±µÄ±êʶÊÇ_____£¨Ìî×Öĸ£©¡£
£¨2£©ºÏÀíʹÓû¯Ñ§ÖªÊ¶¿ÉÌá¸ßÈËÃǵÄÉú»îÖÊÁ¿¡£
ijƷÅÆÑÀ¸àµÄ³É·ÖÓиÊÓÍ¡¢É½ÀæËá¼Ø¡¢·ú»¯ÄƵȡ£
¢ÙÔÚÉÏÊöÑÀ¸à³É·ÖÖУ¬ÊôÓÚ·À¸¯¼ÁµÄÊÇ_______________¡£
¢Ú¸ÊÓ͵Ľṹ¼òʽΪ____________£»ÓÍ֬ˮ½â¿ÉÉú³É¸ÊÓͺÍ_____________¡£
¢Û·ú»¯ÄÆ(NaF)¿ÉÓëÑÀ³ÝÖеÄôÇ»ùÁ×Ëá¸Æ[Ca5(PO4)3OH]·´Ó¦£¬Éú³É¸üÄÑÈܵķúÁ×Ëá¸Æ[Ca5(PO4)3F]£¬´Ó¶ø´ïµ½·ÀÖÎÈ£³ÝµÄÄ¿µÄ¡£Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____________________¡£
£¨3£©´´Ð·¢Õ¹²ÄÁϼ¼Êõ¿ÆÍƶ¯ÈËÀàÉç»áµÄ½ø²½¡£
¢Ùʯīϩ£¨¼ûÏÂͼ£©¿ÉÓÃ×÷Ì«ÑôÄܵç³ØµÄµç¼«£¬ÕâÀïÖ÷ÒªÀûÓÃÁËʯīϩµÄ______________ÐÔ¡£
¢Ú»ù´¡¹¤³Ì½¨ÉèÖг£Óõ½Ë®Äà¡¢²£Á§¡¢¸Ö²ÄµÈ¡£Éú³ÉË®ÄàºÍ²£Á§¶¼Óõ½µÄÔÁÏÊÇ__________£»ÔÚ¸Ö²ÄÖÐÌí¼Ó¸õ¡¢ÄøµÈÔªËصÄÄ¿µÄÊÇ___________¡£
¢ÛÐÂÐÍÕ½¶·»ú³£ÓÃÄÉÃ×SiC·ÛÌå×÷ΪÎü²¨²ÄÁÏ¡£¸ßÎÂϽ¹Ì¿ºÍʯӢ·´Ó¦¿ÉÖƵÃSiC£¬Ê¯Ó¢µÄ»¯Ñ§Ê½Îª________________£»¸ßηֽâSi(CH3)2Cl2Ò²¿ÉÖƵÃSiC£¬Í¬Ê±»¹Éú³ÉCH4ºÍÒ»ÖÖ³£¼ûËáÐÔÆøÌ壬д³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º____________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Èç±íΪԪËØÖÜÆÚ±íµÄÒ»²¿·Ö£¬±ê³öÁËA¡ªK¹²Ê®ÖÖÔªËØËùÔÚλÖá£ÇëÓÃÔªËØ·ûºÅ»ò»¯Ñ§Ê½»Ø´ðÏÂÁÐÎÊÌ⣮
Ö÷×å ÖÜÆÚ | ¢ñA | ¢òA | ¢óA | ¢ôA | ¢õA | ¢öA | ¢÷A | 0 |
2 | A | B | ||||||
3 | C | D | E | F | G | |||
4 | H | I | K |
£¨1£©10ÖÖÔªËØÖУ¬»¯Ñ§ÐÔÖÊ×î²»»îÆõÄÊÇ______£¬×î»îÆõĽðÊôÊÇ_____¡£
£¨2£©C¡¢E¡¢FÈýÖÖÔªËØÐγɵļòµ¥Àë×Ӱ뾶×îСµÄÊÇ_______¡£
£¨3£©C¡¢KÐγɵĻ¯ºÏÎïÖл¯Ñ§¼üµÄÀàÐÍÊôÓÚ______________¡£
£¨4£©»¯ºÏÎïC2B2µÄµç×ÓʽΪ__________________£»¸Ã»¯ºÏÎïºÍAB2·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________________________________¡£
£¨5£©EµÄ×î¸ß¼ÛÑõ»¯ÎïÊôÓÚ________________ÐÔÑõ»¯ÎËüÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________________________________________¡£
£¨6£©DµÄµ¥ÖÊÔÚA¡¢BÐγɻ¯ºÏÎïÖÐȼÉյĻ¯Ñ§·½³ÌʽΪ________________________¡£
£¨7£©Óõç×Óʽ±íʾHÓëB×é³ÉµÄH2BÐÍ»¯ºÏÎïµÄÐγɹý³Ì£º_____________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÏÂÁÐÆøÌåÖУ¬²»ÄÜÓÃÏòÉÏÅÅ¿ÕÆø·¨ÊÕ¼¯µÄÊÇ
A. H2 B. CO2 C. SO2 D. O2
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