ÌâÄ¿ÄÚÈÝ

ÈçͼÊÇ´«Í³µÄ¹¤ÒµÉú²ú½ðÊôÂÁµÄ»ù±¾Á÷³Ìͼ¡£

Çë»Ø´ð£º
£¨1£©½ðÊôÂÁ³£ÓÃÀ´ÖÆÔìÈÝÆ÷£¬µ«ÂÁÖÆ²Í¾ß²»ÒËÓÃÀ´ÕôÖó»ò³¤Ê±¼ä´æ·ÅËáÐÔ¡¢¼îÐÔµÈʳÎд³öÂÁÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º_____________________________________¡£
ʯÓÍÁ¶ÖƺÍúµÄ¸ÉÁó²úÆ·________(ÌîÎïÖÊÃû³Æ)×÷µç½âÂÁµÄÒõ¼«ºÍÑô¼«²ÄÁÏ¡£
£¨2£©ÔÚÒ±Á¶¹ý³ÌÖУ¬Ñô¼«²ÄÁÏÐèÒª¶¨ÆÚ½øÐиü»»£¬Ô­ÒòÊǸü«²ÄÁϲ»¶Ï±»ÏûºÄ£¬²úÉúÕâÖÖÏÖÏóµÄÔ­ÒòÊÇ____________________________________(Óû¯Ñ§·½³Ìʽ±íʾ)¡£
£¨3£©¹¤ÒµÉÏͨ¹ýµç½âÈÛÈÚµÄMgCl2ÖÆÈ¡½ðÊôþ£¬µç½â·´Ó¦·½³ÌʽΪ________________¡£Ã¾ºÍÂÁ¶¼ÊÇ»îÆÃ½ðÊô£¬ÎªÊ²Ã´ÔÚµç½âÒ±Á¶¹ý³ÌÖУ¬Ò»¸öÓÃÂÈ»¯Îһ¸öÓÃÑõ»¯ÎÇë˵Ã÷ÀíÓÉ¡£________________________________________________¡£
£¨1£©2Al£«2OH£­£«2H2O=2AlO2-£«3H2¡ü¡¡Ê¯Ä«(»ò̼)
£¨2£©2C£«O22CO
£¨3£©MgCl2(ÈÛÈÚ)Mg£«Cl2¡ü¡¡MgOµÄÈܵãÌ«¸ß£¬¶øMgCl2µÄÈÛµã½ÏµÍ£¬ÈÛ»¯Ê±MgCl2ÄÜ·¢ÉúµçÀë¶øµ¼µç£»AlCl3Êǹ²¼Û»¯ºÏÎÈÛ»¯Ê±²»ÄÜ·¢ÉúµçÀë
£¨1£©AlÓëÇ¿¼îÈÜÒº×÷Óõõ½Æ«ÂÁËáÑÎÓëÇâÆø£¬¹Ê·´Ó¦µÄÀë×Ó·½³ÌʽΪ2Al£«2OH£­£«2H2O=2AlO2-£«3H2¡ü¡£
£¨2£©µç½â¹ý³ÌÖÐÑô¼«²úÉúÑõÆø£¬ÔÚ¸ßÎÂÏÂÑõÆøÓëµç¼«²ÄÁÏ(ʯī)·´Ó¦Éú³ÉCO¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¶þ¼×ÃÑ(CH3OCH3)ÊÇÎÞÉ«ÆøÌ壬¿É×÷ΪһÖÖÐÂÐÍÄÜÔ´¡£ÓÉºÏ³ÉÆø(×é³ÉΪH2¡¢COºÍÉÙÁ¿µÄCO2)Ö±½ÓÖÆ±¸¶þ¼×ÃÑ£¬ÆäÖеÄÖ÷Òª¹ý³Ì°üÀ¨ÒÔÏÂËĸö·´Ó¦£º
¼×´¼ºÏ³É·´Ó¦£º
(¢¡)CO(g)£«2H2(g)=CH3OH(g)    ¦¤H1£½£­90.1 kJ¡¤mol£­1
(¢¢)CO2(g)£«3H2(g)=CH3OH(g)£«H2O(g)    ¦¤H2£½£­49.0 kJ¡¤mol£­1
Ë®ÃºÆø±ä»»·´Ó¦£º
(¢£)CO(g)£«H2O(g)=CO2(g)£«H2(g)   ¦¤H3£½£­41.1 kJ¡¤mol£­1
¶þ¼×ÃѺϳɷ´Ó¦£º
(¢¤)2CH3OH(g)=CH3OCH3(g)£«H2O(g)    ¦¤H4£½£­24.5 kJ¡¤mol£­1
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)Al2O3ÊÇºÏ³ÉÆøÖ±½ÓÖÆ±¸¶þ¼×ÃÑ·´Ó¦´ß»¯¼ÁµÄÖ÷Òª³É·ÖÖ®Ò»¡£¹¤ÒµÉÏ´ÓÂÁÍÁ¿óÖÆ±¸½Ï¸ß´¿¶ÈAl2O3µÄÖ÷Òª¹¤ÒÕÁ÷³ÌÊÇ____________________________________________(ÒÔ»¯Ñ§·½³Ìʽ±íʾ)¡£
(2)·ÖÎö¶þ¼×ÃѺϳɷ´Ó¦(¢¤)¶ÔÓÚCOת»¯ÂʵÄÓ°Ïì
________________________________________________________________________¡£
(3)ÓÐÑо¿ÕßÔÚ´ß»¯¼Á(º¬Cu­Zn­Al­OºÍAl2O3)¡¢Ñ¹Ç¿Îª5.0 MPaµÄÌõ¼þÏ£¬ÓÉH2ºÍCOÖ±½ÓÖÆ±¸¶þ¼×ÃÑ£¬½á¹ûÈçͼËùʾ¡£ÆäÖÐCOת»¯ÂÊËæÎ¶ÈÉý¸ß¶ø½µµÍµÄÔ­ÒòÊÇ_________________________________________________¡£

(4)¶þ¼×ÃÑÖ±½ÓȼÁÏµç³Ø¾ßÓÐÆô¶¯¿ì¡¢Ð§ÂʸߵÈÓŵ㣬ÆäÄÜÁ¿ÃܶȸßÓÚ¼×´¼Ö±½ÓȼÁÏµç³Ø(5.93 kW¡¤h¡¤kg£­1)¡£Èôµç½âÖÊΪËáÐÔ£¬¶þ¼×ÃÑÖ±½ÓȼÁÏµç³ØµÄ¸º¼«·´Ó¦Îª__________
_____________________£¬Ò»¸ö¶þ¼×ÃÑ·Ö×Ó¾­¹ýµç»¯Ñ§Ñõ»¯£¬¿ÉÒÔ²úÉú________________¸öµç×ӵĵçÁ¿£»¸Ãµç³ØµÄÀíÂÛÊä³öµçѹΪ1.20 V£¬ÄÜÁ¿ÃܶÈE£½_______________(ÁÐʽ¼ÆËã¡£ÄÜÁ¿Ãܶȣ½µç³ØÊä³öµçÄÜ/ȼÁÏÖÊÁ¿£¬1 kW¡¤h£½3.6¡Á106 J)¡£
ÓÉÂÁÍÁ¿ó£¨Ö÷Òª³É·ÖÊÇAl2O3£©Á¶ÖÆÂÁµÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçÏ£º

£¨1£©µç½âÉú³ÉµÄÂÁÔÚÈÛÈÚÒºµÄ       £¨Ìî¡°Éϲ㡱»ò¡°Ï²㡱£©£¬µç½âʱ²»¶ÏÏûºÄµÄµç¼«ÊÇ       £¨Ìî¡°Òõ¼«¡±»ò¡°Ñô¼«¡±£©¡£
£¨2£©Ð´³öͨÈë¹ýÁ¿¶þÑõ»¯Ì¼Ëữʱ·´Ó¦µÄÀë×Ó·½³Ìʽ                                                                       
                                                                       ¡£
£¨3£©µç½âÖÆ±¸ÂÁʱ£¬Ðè¼ÓÈë±ù¾§Ê¯£¨Na3AlF6£©£¬Æä×÷ÓÃÊÇ           £¬¹¤ÒµÉÏ¿ÉÒÔÓ÷ú»¯ÇâÆøÌå¡¢ÇâÑõ»¯ÂÁºÍ´¿¼îÔÚ¸ßÎÂÌõ¼þÏ·¢Éú·´Ó¦À´ÖÆÈ¡±ù¾§Ê¯£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                                                       ¡£
£¨4£©ÉÏÊö¹¤ÒÕËùµÃÂÁÖÐÍùÍùº¬ÓÐÉÙÁ¿FeºÍSiµÈÔÓÖÊ£¬¿ÉÓõç½â·½·¨½øÒ»²½Ìá´¿£¬¸Ãµç½â³ØµÄÒõ¼«²ÄÁÏÊÇ       £¨Ìѧʽ£©£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª                                                                       ¡£
£¨5£©¶Ô½ðÊôÖÆÆ·½øÐп¹¸¯Ê´´¦Àí£¬¿ÉÑÓ³¤ÆäʹÓÃÊÙÃü¡£
¢Ù¿ØÖÆÒ»¶¨Ìõ¼þ½øÐеç½â£¨¼ûͼ£©£¬´ËʱÂÁ±íÃæ¿ÉÐγÉÄÍËáµÄÖÂÃÜÑõ»¯Ä¤£¬Æäµç¼«·´Ó¦Ê½Îª                                 £»

¢Ú¸Ö²Ä¶ÆÂÁºó£¬ÄÜ·ÀÖ¹¸Ö²Ä¸¯Ê´£¬ÆäÔ­ÒòÊÇ                                                                                                                                               ¡£
ij»¯Ñ§ÐËȤС×éÀûÓÃ·ÏÆúÂÁ¿ó£¨º¬CuO¡¢Al2O3¼°SiO2£©£¬Ä£Ä⹤ҵÉÏÌáÈ¡ÂÁµÄ¹¤ÒÕ£¬Éè¼ÆÈçÏÂͼËùʾµÄ¼òµ¥²Ù×÷Á÷³Ì£º

ÒÑÖª£º²¿·ÖÑôÀë×ÓÒÔÇâÑõ»¯ÎïÐÎʽ³ÁµíʱÈÜÒºµÄpH¼ûÏÂ±í£º
³ÁµíÎï



¿ªÊ¼³Áµí
2.3
7.5
3.4
ÍêÈ«³Áµí
3.2
9.7
4.4
 
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÂËÔüÖ÷Òª³É·ÖµÄ»¯Ñ§Ê½Îª______________________¡£
£¨2£©×ÆÉÕʱ£¬Óõ½¶àÖÖ¹èËáÑÎÖʵÄÒÇÆ÷£¬³ý²£Á§°ô¡¢¾Æ¾«µÆ¡¢ÄàÈý½ÇÍ⣬»¹ÓÐ___________£¨ÌîÒÇÆ÷Ãû³Æ£©¡£
£¨3£©ÈÜÒºYÖÐÒª¼ÓÈëÉÔ¹ýÁ¿Ô­ÁÏA£¬Ô­ÁÏAµÄ»¯Ñ§Ê½ÊÇ_________£¬²½Öè¢Ý·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_______________________¡£
£¨4£©²Ù×÷Á÷³ÌÖТ۵ÄÀë×Ó·½³ÌʽΪ_______________________¡£
£¨5£©ÂÁµç³ØÐÔÄÜÓÅÔ½£¬Al¡ªAg2Oµç³Ø¿ÉÓÃ×÷ˮ϶¯Á¦µçÔ´£¬ÆäÔ­ÀíÈçͼËùʾ£º

Çëд³ö¸Ãµç³ØÕý¼«·´Ó¦Ê½_________________£»³£ÎÂÏ£¬Óøû¯Ñ§µçÔ´ºÍ¶èÐԵ缫µç½â300mLÂÈ»¯ÄÆÈÜÒº£¨¹ýÁ¿£©£¬ÏûºÄ27mg Al£¬Ôòµç½âºóÈÜÒºµÄpH£½_________£¨²»¿¼ÂÇÈÜÒºÌå»ýµÄ±ä»¯£©¡£
£¨6£©ÒÔÈÛÈÚÑÎΪµç¶ÆÒº¿ÉÒÔÔڸֲıíÃæ¶ÆÂÁ£¬¶ÆÂÁµç½â³ØÖУ¬¸Ö²ÄΪ________¼«£»¶ÆÂÁºóÄÜ·ÀÖ¹¸Ö²Ä¸¯Ê´£¬ÆäÔ­ÒòÊÇ____________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø