ÌâÄ¿ÄÚÈÝ

Ìú¼°Æä»¯ºÏÎïÓ¦Óù㷺¡£
£¨1£©ÈýÂÈ»¯ÌúÊÇÒ»ÖÖË®´¦Àí¼Á£¬¹¤ÒµÖƱ¸ÎÞË®ÈýÂÈ»¯Ìú¹ÌÌåµÄ²¿·ÖÁ÷³ÌÈçÏÂͼ£º

¢Ù¼ìÑ鸱²úÆ·Öк¬ÓÐXʱ£¬Ñ¡ÓõÄÊÔ¼ÁÊÇ   (ÌîÏÂÁи÷ÏîÖÐÐòºÅ)¡£
a£®NaOHÈÜÒº    b£®KSCNÈÜÒº  c£®ËáÐÔKMnO4ÈÜÒº  d¡¢Ìú·Û
¢ÚÔÚÎüÊÕËþÖУ¬Éú³É¸±²úÆ·FeCl¡£µÄÀë×Ó·½³ÌʽΪ
£¨2£©¸ßÌúËá¼Ø(K2FeO4)Ò²ÊÇÒ»ÖÖÓÅÁ¼µÄË®´¦Àí¼Á£¬¹¤ÒµÉÏ£¬¿ÉÓÃÌú×÷Ñô¼«£¬µç½âKOHÈÜÒºÖÆ±¸¸ßÌúËá¼Ø¡£µç½â¹ý³ÌÖУ¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª       £»µç½âÒ»¶Îʱ¼äºó£¬ÈôÑô¼«ÖÊÁ¿¼õÉÙ28 g£¬ÔòÔڴ˹ý³ÌÖУ¬Òõ¼«Îö³öµÄÆøÌåÔÚ±ê×¼×´¿öϵÄÌå»ýΪ   L¡£
£¨3£©Áò»¯ÑÇÌú³£ÓÃÓÚ¹¤Òµ·ÏË®µÄ´¦Àí¡£
¢ÙÄãÈÏΪ£¬ÄÜ·ñÓÃÁò»¯ÑÇÌú´¦Àíº¬Cd2+µÄ¹¤Òµ·ÏË®?   (Ìî¡°ÄÜ¡±»ò¡°·ñ¡±)¡£Çë¸ù¾Ý³ÁµíÈÜ½âÆ½ºâµÄÔ­Àí½âÊÍÄãµÄ¹Ûµã(ÓñØÒªµÄÎÄ×ÖºÍÀë×Ó·½³Ìʽ˵Ã÷)£º                                           (ÒÑÖª£º25¡æÊ±£¬ÈܶȻý³£ÊýKsp(FeS)=6£®310-18¡¢Ksp(CdS)=3£®610-29)
¢Ú¹¤ÒµÉÏ´¦Àíº¬Cd2+·ÏË®»¹¿ÉÒÔ²ÉÓüÓ̼ËáÄÆµÄ·½·¨£¬·´Ó¦ÈçÏ£º2Cd2++2CO32-+H2O=Cd2(OH)2CO3+A¡£ÔòAµÄ»¯Ñ§Ê½Îª        ¡£

£¨17·Ö£©
£¨1£©£¨5·Ö£©¢Ùc£¨2·Ö£©¢Ú2Fe2++Cl2=2Fe3++2Cl?£¨3·Ö£©
£¨2£©£¨5·Ö£©Fe¡ª6e?+8OH?=FeO42?+4H2O£¨3·Ö£©33.6£¨2·Ö£©
£¨3£©£¨7·Ö£©¢ÙÄÜ£¨1·Ö£©CdS±ÈFeS¸üÄÑÈÜ£¬¿É·¢Éú³Áµíת»¯Cd2+(aq)+FeS(s)=CdS(s)+Fe2+(aq)£¨3·Ö£©
¢ÚCO2£¨3·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¢ÙÎüÊÕ¼ÁXµÄ×÷ÓÃÊÇÎüÊÕδ·´Ó¦µÄCl2£¬Éú³ÉFeCl3£¬ËùÒÔXΪFeCl2£¬¼ìÑéFeCl3ÈÜÒºÖк¬ÓÐFeCl2µÄÊÔ¼ÁÊÇËáÐÔKMnO4ÈÜÒº£¬¹ÊcÏîÕýÈ·¡£
¢ÚFe×÷Ñô¼«Ê§È¥µç×ÓÉú³ÉFeO42?£¬ÓÃKOHÈÜÒº×öµç½âÖÊÈÜÒº£¬OH?²Î¼Ó·´Ó¦£¬ËùÒÔÑô¼«µç¼«·½³ÌʽΪ£ºFe¡ª6e?+8OH?=FeO42?+4H2O£»Òõ¼«ºÍÑô¼«µÃʧµç×Ó×ÜÊýÏàµÈ£¬ËùÒÔ¶ÔÓ¦¹ØÏµÎª£ºFe ~ 6e? ~ 3H2£¬Ôòn(H2)=3n(Fe)=3¡Á28g¡Â56g/mol=1.5mol£¬±ê×¼×´¿öϵÄÌå»ýΪ33.6L¡£
£¨3£©¢ÙÒòΪKsp(CdS)СÓÚKsp(FeS)£¬ËùÒÔCdS±ÈFeS¸üÄÑÈÜ£¬¿É·¢Éú³Áµíת»¯Cd2+(aq)+FeS(s)=CdS(s)+Fe2+(aq)£¬Òò´ËÄÜÓÃÁò»¯ÑÇÌú´¦Àíº¬Cd2+µÄ¹¤Òµ·ÏË®¡£
¢Ú¸ù¾ÝÔªËØÊØºã£¬¿ÉÖªAΪCO2¡£
¿¼µã£º±¾Ì⿼²é»¯Ñ§¹¤ÒÕÁ÷³ÌµÄ·ÖÎö¡¢µç½âÔ­Àí¼°¼ÆËã¡¢Àë×Ó·½³ÌʽµÄÊéд¡¢³ÁµíÈÜ½âÆ½ºâ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ij¿óÔüµÄ³É·ÖΪCu2O¡¢Al2O3¡¢Fe2O3¡¢SiO2£¬¹¤ÒµÉÏÓøÿóÔü»ñȡͭºÍµ¨·¯µÄ²Ù×÷Á÷³ÌÈçÏ£º

ÒÑÖª£º ¢ÙCu2O +2 H+="Cu" + Cu2++H2O
¢Ú²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÎªÁ˼ӿ췴ӦIµÄËÙÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ                            (д2µã)¡£
£¨2£©¹ÌÌå»ìºÏÎïAÖеijɷÖÊÇ                             ¡£
£¨3£©·´Ó¦IÍê³Éºó£¬ÌúÔªËØµÄ´æÔÚÐÎʽΪ   (ÌîÀë×Ó·ûºÅ)£»Ð´³öÉú³É¸ÃÀë×ÓµÄÀë×Ó·½³Ìʽ           ¡£
£¨4£©²Ù×÷1Ö÷Òª°üÀ¨£º          ¡¢            ¡¢       ¡£Ï´µÓCuSO4?5H2O´Ö²úÆ·²»ÄÜÓôóÁ¿Ë®Ï´£¬¶øÓñùˮϴµÓ¡£Ô­ÒòÊÇ                                          ¡£
£¨5£©ÓöèÐԵ缫µç½âµ¨·¯ÈÜÒºÒ»¶Îʱ¼ä£¬¼ÓÈË0£®1 molµÄCu(OH)2¿É»Ö¸´ÈÜÒºÔ­¿ö(Ũ¶È¡¢³É·Ö)£¬Ôòµç½âÊ±×ªÒÆµç×ÓµÄÎïÖʵÄÁ¿Îª                            £®¡£
£¨6£©ÓÃNaClOµ÷pH£¬Éú³É³ÁµíBµÄͬʱÉú³ÉÒ»ÖÖ¾ßÓÐÆ¯°××÷ÓõÄÎïÖÊ£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ           ¡£

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

¾«Ó¢¼Ò½ÌÍø