ÌâÄ¿ÄÚÈÝ

ÈçÏÂͼװÖÃÖУ¬bµç¼«ÓýðÊô MÖƳɣ¬a¡¢c¡¢dΪʯīµç¼«£¬½ÓͨµçÔ´£¬½ðÊôM³Á»ýÓÚb¼«£¬Í¬Ê±a¡¢dµç¼«ÉϲúÉúÆøÅÝ¡£ÊԻشð£º

£¨1£©aΪ        ¼«£¬c¼«µÄµç¼«·´Ó¦Ê½Îª                             ¡£
£¨2£©µç½â½øÐÐÒ»¶Îʱ¼äºó£¬ÕÖÔÚc¼«ÉϵÄÊÔ¹ÜÖÐÒ²ÄÜÊÕ¼¯µ½µÄÆøÌ壬´Ëʱc¼«Éϵĵ缫·´Ó¦Ê½Îª          ¡£
£¨3£©µ±d¼«ÉÏÊÕ¼¯µ½44.8mLÆøÌ壨±ê×¼×´¿ö£©Ê±Í£Ö¹µç½â£¬a¼«ÉϷųöÁËÆøÌåµÄÎïÖʵÄÁ¿Îª       £¬Èôbµç¼«ÉϳÁ»ý½ðÊôMµÄÖÊÁ¿Îª0.432g£¬Ôò´Ë½ðÊôµÄĦ¶ûÖÊÁ¿Îª                        ¡£
£¨1£©Ñô£¨1·Ö£©£»   2I¡ªÒ»2e¡ª£½I2£¨2·Ö£©
£¨2£©40H¡ª¡ª4e¡ª£½2H2O+O2¡ü £¨2·Ö£©
£¨3£©0£®001 mol£¨1·Ö£©£»108g£¯mol  £¨2·Ö£©

ÊÔÌâ·ÖÎö£ºÓɵç½âÔ­Àí¿ÉµÃ£º½ðÊôM³Á»ýÓÚb¼«£¬ËµÃ÷bÊÇÒõ¼«£¬ÔòaÊÇÑô¼«£¬cÊÇÑô¼«£¬dÊÇÒõ¼«£¬£¨1£©ÒòaÊÇÑô¼«£¬ÈÜÒºÖеÄÒõÀë×ӷŵ磬¸ù¾ÝÀë×ӵķŵç˳Ðò£¬¿ÉÖªÊÇ2I--2e-=I2£»£¨2£©ÔÚBÉÕ±­ÖУ¬ cÊÇÑô¼«£¬ÈÜÒºÖеÄÒõÀë×ӷŵ磬¼´2I--2e-=I2£¬I2Óöµ½µí·ÛÄÜʹµí·Û±äÀ¶£¬I-·ÅµçÍê±Ïºó£¬½Ó×ÅÊÇOH-·Åµç£º4OH--4e=2H2O+O2¡ü£¬c¼«ÉϵÄÊÔ¹ÜÖÐÊÕ¼¯µ½µÄÆøÌåΪÑõÆø£¬£¨3£©dµç¼«ÉÏÊÕ¼¯µÄ44.8mlÆøÌ壨±ê×¼×´¿ö£©ÊÇÇâÆø£¬a¼«ÉÏÊÕ¼¯µ½µÄÆøÌåÊÇÑõÆø£¬¸ù¾ÝתÒƵç×ÓÊýÏàµÈÖª£¬ÑõÆøºÍÇâÆøµÄÌå»ýÖ®±ÈÊÇ1£º2£¬dµç¼«ÉÏÊÕ¼¯µÄ44.8mlÆøÌåÇâÆø£¬Ôòaµç¼«ÉÏÊÕ¼¯µ½22.4mLÑõÆø£»ÎïÖʵÄÁ¿Îª0.01mol£¬dµç¼«ÉÏÎö³öµÄÇâÆøµÄÎïÖʵÄÁ¿=£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿ÊÇ0.04mol£¬ÏõËáÑÎÖÐMÏÔ+1¼Û£¬ËùÒÔµ±×ªÒÆ0.04molµç×ÓʱÎö³ö0.04mol½ðÊôµ¥ÖÊ£¬M=
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÏÂͼΪϸ¾úұͭºÍ»ð·¨Ò±Í­µÄÖ÷ÒªÁ÷³Ì¡£

(1) ÁòËáÍ­ÈÜÒºÒ»°ã³Ê________(Ìî¡°Ëᡱ¡¢¡°¼î¡±»ò¡°ÖС±)ÐÔ£¬Ô­ÒòÊÇ________    (ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£Ð´³öµç½âÁòËáÍ­ÈÜÒºµÄ»¯Ñ§·½³Ìʽ£º______________(µç½â¹ý³ÌÖУ¬Ê¼ÖÕÎÞÇâÆø²úÉú)¡£
(2)ϸ¾úÒ±½ðÓÖ³Æ΢ÉúÎï½þ¿ó£¬Êǽü´úʪ·¨Ò±½ð¹¤ÒµÉϵÄÒ»ÖÖй¤ÒÕ¡£Ï¸¾úұͭÓë»ð·¨Ò±Í­Ïà±È£¬ÓŵãΪ________________(д³öÒ»µã¼´¿É)¡£

(3) ÓöèÐԵ缫·Ö±ðµç½âŨµÄÂÈ»¯Í­ÈÜÒººÍÁòËáÍ­ÈÜÒº¡£µç½âŨµÄÂÈ»¯Í­ÈÜҺʱ·¢ÏÖÒõ¼«ÓнðÊôÍ­Éú³É£¬Í¬Ê±Òõ¼«¸½½ü»á³öÏÖ×غÖÉ«ÈÜÒº¡£¶øµç½âÁòËáÍ­ÈÜҺʱ£¬Ã»ÓÐ×غÖÉ«ÈÜÒºÉú³É¡£ÏÂÃæÊǹØÓÚ×غÖÉ«ÈÜÒº³É·ÖµÄ̽¾¿£º
¢ÙÓÐͬѧÈÏΪ£¬Òõ¼«¸½½ü³öÏÖµÄ×غÖÉ«ÈÜÒºÊÇÂÈÆø·´Ó¦µÄ½á¹û£¬ÄãÈÏΪËûµÄ²Â²âÊÇ·ñÕýÈ·£¿________(Ìî¡°ÕýÈ·¡±»ò¡°²»ÕýÈ·"£©£¬Ô­ÒòÊÇ__________
×ÊÁÏ1:
Ò»°ã¾ßÓлìºÏ¼Û̬(Ö¸»¯ºÏÎïÖÐͬһԪËØ´æÔÚÁ½ÖÖ²»Í¬µÄ»¯ºÏ¼Û£¬ÈçFe3O4ÖÐµÄ FeÔªË÷)µÄÎïÖʵÄÑÕÉ«±Èµ¥Ò»¼Û̬µÄÎïÖʵÄÑÕÉ«ÒªÉî¡£
×ÊÁÏ2:
CuCl΢ÈÜÓÚË®£¬ÄÜÈÜÓÚŨÑÎËá¡£
¢Ú²ÂÏ룺×غÖÉ«ÈÜÒºÖпÉÄܺ¬ÓеÄÀë×ÓÊÇ________(Ìî3ÖÖÖ÷ÒªÀë×Ó·ûºÅ£©¡£
¢ÛÑéÖ¤²ÂÏ룺Íê³ÉʵÑé·½°¸(ÅäÖÆ×غÖÉ«ÈÜÒº£©¡£
È¡ÉÙÁ¿________¹ÌÌåÓÚÊÔ¹ÜÖУ¬¼ÓÈë________ʹÆäÈܽ⣬ÔÙ¼ÓÈë________ÈÜÒº£¬¹Û²ìÏÖÏó¡£
¢ÜÒÑÖªµç½âÇ°£¬UÐιÜÖмÓÈëÁË________100mL0.5 mol . L£­1 CuCl2ÈÜÒº£¬µç½â½á–cʱµç·ÖÐÒ»¹²×ªÒÆÁË 0.03 molµç×Ó£¬ÇÒÒõ¼«Éú³É0. 64 gÍ­£¬ÔòÐγɵĵͼÛÑôÀë×ÓµÄÎïÖʵÄÁ¿Îª________mol¡£
¹¤ÒµÉÏÒÔ»ÆÍ­¿óΪԭÁÏ£¬²ÉÓûð·¨ÈÛÁ¶¹¤ÒÕÉú²ú´Ö¸Ö¡£
£¨1£©¸Ã¹¤ÒÕµÄÖмä¹ý³Ì»á·¢Éú·´Ó¦£º2Cu2O+Cu2S6Cu+SO2¡ü£¬·´Ó¦µÄÑõ»¯¼ÁÊÇ         ¡£
£¨2£©»ð·¨ÈÛÁ¶µÄ´ÖÍ­º¬ÔÓÖʽ϶ࡣij»¯Ñ§Ñо¿ÐÔѧϰС×éÔÚʵÑéÊÒÌõ¼þÏÂÓÃCuSO4ÈÜÒº×÷µç½âÒºÀ´ÊµÏÖ´ÖÍ­µÄÌá´¿£¬²¢¶Ôµç½âºóÈÜÒº½øÐо»»¯³ýÔӺͺ¬Á¿²â¶¨¡£
ʵÑéÒ» ´ÖÍ­µÄÌá´¿´ÖÍ­Öк¬ÓÐÉÙÁ¿µÄп¡¢Ìú¡¢Òø¡¢½ðµÈ½ðÊôºÍÉÙÁ¿¿óÎïÔÓÖÊ£¨ÓëËá²»·´Ó¦£©£¬µç½âʱ´ÖÍ­Ó¦ÓëµçÔ´µÄ        ¼«ÏàÁ¬£¬Òõ¼«Éϵĵ缫·´Ó¦Ê½Îª              ¡£
ʵÑé¶þ µç½âºóÈÜÒºµÄ¾»»¯³ýÔÓÔÚ¾«Á¶Í­µÄ¹ý³ÌÖУ¬µç½âÒºÖÐc£¨Cu2+£©Öð½¥Ï½µ£¬c£¨Fe2+£©¡¢c£¨Zn2+£©»áÖð½¥Ôö´ó£¬ËùÒÔÐ趨ʱ³ýÈ¥ÆäÖеÄFe2+¡¢Zn2+¡£¼×ͬѧ²Î¿¼Ï±íµÄÊý¾Ý£¬Éè¼ÆÁËÈçÏ·½°¸£º

ÊÔ¼ÁaÊÇ               £¨Ìѧʽ£©£¬ÆäÄ¿µÄÊÇ             £»¸Ã·½°¸Äܹ»³ýÈ¥µÄÔÓÖÊÀë×ÓÊÇ           £¨ÌîÀë×Ó·ûºÅ£©¡£
ʵÑéÈý µç½âºóÈÜÒºÀë×Óº¬Á¿µÄ²â¶¨
ÒÒͬѧÉè¼ÆÁËÈçÏ·½°¸£º

Ôò100mLÈÜÒºÖÐCu2+µÄŨ¶ÈΪ         mol¡¤L-1£¬Fe2+µÄŨ¶ÈΪ         mol¡¤L-1¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø