ÏÂͼËùʾµÄÎïÖÊת»¯¹ØϵÖУ¬¸÷ÎïÖʾùΪ¶ÌÖÜÆÚÖ÷×åÔªËØ×é³ÉµÄµ¥ÖÊ»ò»¯ºÏÎï¡£ÒÑÖª£ºA¡¢C¡¢D¡¢F¡¢K¾ùΪµ¥ÖÊ£»C¡¢E¡¢F¡¢G¡¢K³£ÎÂÏÂÊÇÆøÌ壬ÇÒKΪ¹¤ÒµÉÏÖÆÔìƯ°×·ÛµÄÔ­ÁÏÖ®Ò»£»JΪ°×É«³ÁµíÇÒ¼ÈÄÜÈÜÓÚBµÄË®ÈÜÒº£¬ÓÖÄÜÈÜÓÚEµÄË®ÈÜÒº£»B¡¢G¿ÉÒÔʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£»·´Ó¦¢ÜÊǹ¤ÒµÖÆ»¯·ÊµÄÖØÒª·´Ó¦Ö®Ò»¡££¨Í¼Öв¿·Ö·´Ó¦Ìõ¼þ¼°ÎïÖÊδÁгö£©

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©BµÄµç×ÓʽΪ              ¡£
£¨2£©HµÄ»¯Ñ§Ê½Îª              ¡£
£¨3£©Ð´³ö·´Ó¦¢ÜµÄ»¯Ñ§·½³Ìʽ                          ¡£
£¨4£©Ð´³ö·´Ó¦¢ÝµÄÀë×Ó·½³Ìʽ                           ¡£

£¨1£©
£¨2£© NaAlO2
£¨3£©N2+3H22NH3
£¨4£©Al3++3NH3+3H2O=Al(OH)3¡ý+3NH4+ »ò Al3++3NH3¡¤H2O=Al(OH)3¡ý+3NH4+

½âÎöÊÔÌâ·ÖÎö£ºKΪ¹¤ÒµÉÏÖÆÔìƯ°×·ÛµÄÔ­ÁÏÖ®Ò»£¬¿ÉÖªkΪÂÈÆø£¬B¡¢G¿ÉÒÔʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬ËµÃ÷Ë®ÈÜÒº³Ê¼îÐÔ£¬GΪ°±Æø£¬CΪÇâÆø£¬FΪµªÆø£»JΪ°×É«³ÁµíÇÒ¼ÈÄÜÈÜÓÚBµÄË®ÈÜÒº£¬ÓÖÄÜÈÜÓÚEµÄË®ÈÜÒº£¬ËµÃ÷jΪÇâÑõ»¯ÂÁ£»AΪÄÆ£¬BΪÇâÑõ»¯ÄÆ¡£DΪÂÁ£¬HΪƫÂÁËáÄÆ£¬IΪÂÈ»¯ÂÁ¡£¢Å BµÄµç×ÓʽΪ¡£¢Æ HµÄ»¯Ñ§Ê½Îª    NaAlO2¡£
¢Ç д³ö·´Ó¦¢ÜµÄ»¯Ñ§·½³ÌʽN2+3H22NH3 ¡£¢È д³ö·´Ó¦¢ÝµÄÀë×Ó·½³ÌʽAl3++3NH3+3H2O=Al(OH)3¡ý+3NH4+ »ò Al3++3NH3¡¤H2O=Al(OH)3¡ý+3NH4+¡£
¿¼µã£º±¾Ì⿼²éÎÞ»úÁ÷³ÌͼÖÐÎïÖʵÄÍƶϡ£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

A¡«G¸÷ÎïÖʼäµÄ¹ØϵÈçÏÂͼËùʾ£¬ÆäÖÐB¡¢DΪÆøÌåµ¥ÖÊ¡£ÔòÏÂÁÐ˵·¨´íÎóµÄÊÇ

A£®Èô·´Ó¦¢ÙÔÚ³£ÎÂϽøÐУ¬Ôò1 mol AÔÚ·´Ó¦ÖÐÄÜתÒÆ1 molµç×Ó
B£®·´Ó¦¢ÚµÄÀë×Ó·½³ÌʽΪMnO2£«4H£«£«2Cl£­Mn2£«£«2H2O£«Cl2¡ü
C£®ÐÂÅäÖƵÄFÈÜÒºÒ»°ãÐèÒª¼ÓÈëÌúмºÍÏ¡ÑÎËᣬǰÕßÓÃÓÚ·ÀÖ¹Fe2+±»¿ÕÆøÑõ»¯³ÉFe3+£¬ºóÕß¿ÉÒÖÖÆFe2+µÄË®½â
D£®ÒÑÖªCµÄŨÈÜÒºÔÚ´ß»¯¼Á´æÔÚµÄÌõ¼þϼÓÈÈ£¬ÄÜÓëB·´Ó¦Éú³ÉD£¬ÓÉ´Ë¿ÉÒÔÍƶÏBµÄÑõ»¯ÐÔ±ÈMnO2Ç¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÔÚÏÂͼת»¯¹ØϵÖУ¬¹ÌÌå¼×µÄÑæÉ«·´Ó¦³Ê»ÆÉ«£¬MΪ³£¼ûµÄÒºÌåÎïÖÊ£¬IΪһ³£¼û½ðÊô£¬ËáGÊÇÖØÒªµÄ»¯¹¤²úÆ·ºÍ»¯¹¤Ô­ÁÏ£»¹ÌÌåHÄܹ»ÈܽâÔÚAÈÜÒººÍËáGÖУ¬ÇÒHΪÁ¼ºÃµÄÄÍ»ð²ÄÁÏ£¨Í¼Öв¿·Ö²úÎïûÓÐÁгö£©¡£

£¨1£©¹ÌÌå¼×ÓëÒºÌåM·´Ó¦µÄ·½³ÌʽΪ______¡£AÈÜÒºÓë¹ÌÌåH·´Ó¦µÄÀë×Ó·½³ÌʽΪ______¡£
£¨2£©¹ÌÌåÒҵĻ¯Ñ§Ê½Îª_______¡£ÒºÌåMµÄµç×ÓʽΪ_______¡£
£¨3£©·´Ó¦¢Ù¡«¢ßÖÐÊôÓÚÑõ»¯»¹Ô­·´Ó¦µÄΪ_______£¨Ìîд·´Ó¦ÐòºÅ£©¡£
£¨4£©ÈôIÓëCµÄÏ¡ÈÜÒº²»·´Ó¦£¬Ö»ÄÜÓëGµÄŨÈÜÒºÔÚ¼ÓÈÈÌõ¼þÏ·´Ó¦£¬Ôò·´Ó¦¢ßµÄ»¯Ñ§·½³ÌʽΪ_______¡£
£¨5£©ÈôÓÉ»ÆÌú¿ó(FeS2)ÓëÆøÌåB·´Ó¦À´Éú²úÆøÌåE£¬ÇÒÿÉú³É1 mol E·Å³ö426.5 kJµÄÈÈÁ¿£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ_______ ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂͼÊÇÎÞ»úÎïA¡«MÔÚÒ»¶¨Ìõ¼þϵÄת»¯¹Øϵ£¨²¿·Ö²úÎï¼°·´Ó¦Ìõ¼þδÁгö£©¡£ÆäÖУ¬IÊÇÓɵÚÈýÖÜÆÚÔªËØ×é³ÉµÄµ¥ÖÊÖÐÈÛµã×î¸ßµÄ½ðÊô£¬KÊÇÒ»ÖÖºì×ØÉ«ÆøÌå¡£

Ìáʾ£º4FeS2£«11O2¸ßÎÂ,2Fe2O3£«8SO2
ÇëÌîдÏÂÁпհףº
£¨1£©ÔÚÖÜÆÚ±íÖУ¬×é³Éµ¥ÖÊGµÄÔªËØλÓÚµÚ________ÖÜÆÚ________×å¡£
£¨2£©ÔÚ·´Ó¦¢ßÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ________¡£
£¨3£©ÔÚ¢Ú¡¢¢Û¡¢¢Þ¡¢¢áÖмÈÊôÓÚ»¯ºÏ·´Ó¦ÓÖÊôÓÚ·ÇÑõ»¯»¹Ô­·´Ó¦µÄÊÇ________£¨ÌîÐòºÅ£©¡£
£¨4£©·´Ó¦¢ÜµÄÀë×Ó·½³ÌʽÊÇ_____________________________________¡£
£¨5£©½«»¯ºÏÎïDÓëKNO3¡¢KOH¹²ÈÛ£¬¿ÉÖƵÃÒ»ÖÖ¡°ÂÌÉ«¡±»·±£¸ßЧ¾»Ë®¼ÁK2FeO4£¨¸ßÌúËá¼Ø£©£¬Í¬Ê±»¹Éú³ÉKNO2ºÍH2O¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÈçͼÖеÄËùÓÐÎïÖʾùÓɶÌÖÜÆÚÔªËØ×é³É¡£ÆäÖмס¢ÒÒ¡¢±û¡¢¶¡Îªµ¥ÖÊ£¬³£ÎÂϼס¢ÒÒΪÎÞÉ«ÆøÌ壬¶¡Îª»ÆÂÌÉ«ÆøÌå¡£±ûÊdz£¼û½ðÊô£¬¹ã·ºÓÃÓÚº½Ìì¡¢º½¿Õ¹¤Òµ¡£¹¤ÒµÉÏ´Óº£Ë®ÖÐÌáÈ¡GºóÔÙͨ¹ýµç½âGÖÆÈ¡±û£¬Í¬Ê±µÃµ½¸±²úÎﶡ¡£A¡¢EµÄ·Ö×ÓÖоùº¬10¸öµç×Ó£¬AÊÇÒ»ÖÖÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌ壬EÔÚ³£ÎÂÏÂÊÇÎÞÉ«ÎÞζµÄÈÜÒº¡£¿òͼÖи÷ÎïÖÊת»¯ËùÉæ¼°µÄÌõ¼þ¾ùÒÑÊ¡ÂÔ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)·´Ó¦¢Ù¡«¢ÝÖÐÊôÓÚÑõ»¯»¹Ô­·´Ó¦µÄÊÇ_________(ÌîÐòºÅ)¡£
(2)»­³ö±ûµÄÔ­×ӽṹʾÒâͼ_________£¬Ð´³öAµÄµç×Óʽ_________£¬Ð´³öA¡¢BµÄµÈµç×ÓÌå(Ô­×ÓÊýºÍµç×ÓÊý¾ùÏàµÈµÄÁ£×Ó) _________¡¢_________ (Óû¯Ñ§Ê½±íʾ)¡£
(3)CµÄË®ÈÜÒºÏÔËáÐÔ£¬ÓÃÀë×Ó·½³Ìʽ½âÊÍ_______________________________
_______________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÔªËؼ°Æ仯ºÏÎïÔÚÉú²ú¡¢Éú»îÖоßÓй㷺µÄÓÃ;¡£
¢ñ.¸õ»¯Ñ§·á¸»¶à²Ê.
£¨1£©ÔÚ³£ÎÂÏ£¬¸õÄÜ»ºÂýÓëÏ¡ÁòËá·´Ó¦£¬Éú³ÉÀ¶É«ÈÜÒº¡£ ÓëÍ­Ïà±È£¬Æä½ðÊô»îÆÃÐÔ      £¨Ìî¡°Ç¿¡±»ò¡°Èõ¡±£©£»
¢ÆCr( OH)3ºÍAl( OH)3ÀàËÆ£¬Ò²ÊÇÁ½ÐÔÇâÑõ»¯ÎÔÚË®ÖдæÔÚËáʽºÍ¼îʽµçÀëƽºâ£¬ÆäËáʽµçÀë·½³ÌʽÊÇ                          £»
¢Ç¹¤ÒµÉϾ»»¯´¦Àí¸õÎÛȾ·½·¨Ö®Ò»ÊÇ£º½«º¬K2Cr2O7ËáÐÔ·ÏË®·ÅÈ˵ç½â²ÛÄÚ£¬¼ÓÈëÊÊÁ¿µÄNaCl£¬ÒÔFeºÍʯīΪµç¼«½øÐеç½â¡£¾­¹ýÒ»¶Îʱ¼äºó£¬Éú³ÉCr(OH)3ºÍFe(OH)3³Áµí³ýÈ¥(ÒÑÖªKsP[ Fe(OH)3]=4.0¡Á10-38£¬KsP[Cr(OH)3]=6.0¡Ál0-31)¡£ÒÑÖªµç½âºóµÄÈÜÒºÖÐc( Fe3+)Ϊ2.0¡Á10£­13mol/L£¬ÔòÈÜÒºÖÐc(Cr3+)Ϊ            mol/L¡£
¢ò.ÎïÖÊA¡«HÓÐÈçͼËùʾת»¯¹Øϵ(²¿·ÖÉú³ÉÎïδÁгö)¡£A¡¢E¡¢F¡¢G¾ùΪÆøÌ壬DΪ½ðÊôµ¥ÖÊ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©AµÄµç×ÓʽΪ        DµÄ»¯Ñ§Ê½          £¬CÈÜÒºµÄÃû³ÆÊÇ                   ¡£
£¨2£©·´Ó¦¢ÙµÄ»¯Ñ§·½³ÌʽΪ                              £»
·´Ó¦¢ÛµÄÀë×Ó·½³ÌʽΪ                                  ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÔÚÏÂͼת»¯¹ØϵÖУ¬ÒÑÖªB¡¢D¶¼Êǵ­»ÆÉ«¹ÌÌ壬Çë»Ø´ðÏÂÁÐÎÊÌâ¡£

¢Åд³öÏÂÁÐÎïÖʵĻ¯Ñ§Ê½£ºB              £¬G              ¡£
¢Æд³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
A¡úB£º               £»E¡úF£º                   £»B¡úC£º                  ¡£
¢Ç½«¹ýÁ¿ÆøÌåEͨÈëµ½ÏÂÁи÷×éÈÜÒººó£¬Àë×ÓÈÔÄÜ´óÁ¿¹²´æµÄÊÇ         ¡£ 
A£®Ba2+¡¢Ca2+¡¢Cl£­             B£®OH£­¡¢CO32£­¡¢Na+
C£®Ca2+¡¢ClO£­¡¢Cl£­             D£®H+¡¢Fe3+¡¢NO3£­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÔÚÏÂͼËùʾµÄÎïÖÊת»¯¹ØϵÖУ¬AÊǺ£Ë®Öк¬Á¿×î·á¸»µÄÑΣ¬BÊdz£¼ûµÄÎÞÉ«ÒºÌ壬GµÄË®ÈÜÒºÊÇÒ»ÖÖ³£ÓõÄƯ°×¼Á£¬FÊǵؿÇÖк¬Á¿×î¶àµÄ½ðÊôÔªËØ¡££¨·´Ó¦ÖÐÉú³ÉµÄË®ºÍ²¿·Ö·´Ó¦Ìõ¼þδÁгö£©
      
£¨1£©»­³öAÖÐÒõÀë×ӵĽṹʾÒâͼ         ¡£
£¨2£©·´Ó¦¢ÚÔÚµãȼÌõ¼þϵÄÏÖÏóÊÇ         ¡£
£¨3£©Hת»¯ÎªFµÄÇâÑõ»¯Îï×îºÃÑ¡ÔñµÄÊÔ¼ÁÊÇ         ¡£
£¨4£©·´Ó¦¢ÛµÄÀë×Ó·½³ÌʽΪ         ¡£
£¨5£©·´Ó¦¢ÜµÄÀë×Ó·½³ÌʽΪ         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÖÐѧ³£¼û»¯Ñ§·´Ó¦·½³ÌʽΪ£ºA+B¡úX+Y+H2O£¨Î´Åäƽ£¬·´Ó¦Ìõ¼þÂÔÈ¥£©£¬ÆäÖУ¬A¡¢BµÄÎïÖʵÄÁ¿Ö®±ÈΪ1:4¡£Çë»Ø´ð£º
£¨1£©ÈôYΪ»ÆÂÌÉ«ÆøÌ壬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ                               £¬BÌåÏÖ³öµÄ»¯Ñ§ÐÔÖÊÓР                                         
£¨2£©ÈôAΪ³£¼ûµÄ·Ç½ðÊôµ¥ÖÊ£¬BµÄÈÜҺΪijŨËᣬ·´Ó¦Ìõ¼þΪ¼ÓÈÈ£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                        
£¨3£©ÈôAΪij²»»îÆõĽðÊôµ¥ÖÊ£¬ÊµÑéÊÒ³£Óø÷´Ó¦À´ÖƱ¸Ä³ÖÖÄÜÐγÉËáÓêµÄÆøÌ壬¸Ã·´Ó¦ÖÐÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ           
£¨4£©ÈôAΪ³£¼ûµÄ½ðÊôµ¥ÖÊ£¬³£ÎÂÏÂAÔÚBµÄŨÈÜÒºÖС°¶Û»¯¡±£¬ÇÒA¿ÉÈÜÓÚXÈÜÒºÖС£
¢ÙAÔªËØÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ                  
¢Úº¬amolXµÄÈÜÒºÈܽâÁËÒ»¶¨Á¿Aºó£¬ÈôÈÜÒºÖÐÁ½ÖÖ½ðÊôÑôÀë×ÓµÄÎïÖʵÄÁ¿Ç¡ºÃÏàµÈ£¬Ôò±»»¹Ô­µÄXµÄÎïÖʵÄÁ¿ÊÇ           

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸