ÌâÄ¿ÄÚÈÝ

Ò»ÖÖÒÔ»ÆÍ­¿óºÍÁò»ÇΪԭÁÏÖÆÈ¡Í­ºÍÆäËû²úÎïµÄй¤ÒÕ£¬Ô­ÁϵÄ×ÛºÏÀûÓÃÂʽϸߣ®ÆäÖ÷ÒªÁ÷³ÌÈçÏ£º

×¢£º·´Ó¦¢òµÄÀë×Ó·½³ÌʽΪCu2++CuS+4Cl-=2[CuCl2]-+S

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©·´Ó¦¢ñµÄ²úÎïΪ____________£¨Ìѧʽ£©£®

£¨2£©·´Ó¦¢óµÄÀë×Ó·½³ÌʽΪ____________£»

£¨3£©Ò»¶¨Î¶ÈÏ£¬ÔÚ·´Ó¦¢óËùµÃµÄÈÜÒºÖмÓÈëÏ¡ÁòËᣬ¿ÉÒÔÎö³öÁòËáÍ­¾§Ì壬Æä¿ÉÄܵÄÔ­ÒòÊÇ____________£»

£¨4£©·´Ó¦¢ôÔÚ¸ßÎÂÌõ¼þϽøÐУ¬»¯Ñ§·½³ÌʽÊÇ____________£»

£¨5£©Ä³ÁòË᳧Ϊ²â¶¨·´Ó¦¢ôËùµÃÆøÌåÖÐSO2µÄÌå»ý·ÖÊý£¬È¡280mL£¨ÒÑÕÛËã³É±ê×¼×´¿ö£©ÆøÌåÑùÆ·Óë×ãÁ¿Fe2£¨SO4£©3ÈÜÒºÍêÈ«·´Ó¦ºó£¬ÓÃŨ¶ÈΪ0.02000mol•L-1µÄK2Cr2O7±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄK2Cr2O7ÈÜÒº25.00mL£®ÒÑÖª£ºCr2O72-+Fe2++H+¡úCr3++Fe3++H2O£¨Î´Å䯽£©

¢ÙSO2ͨÈëFe£¨SO4£©3ÈÜÒºÖУ¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ____________£»

¢Ú·´Ó¦¢ôËùµÃÆøÌåÖÐSO2µÄÌå»ý·ÖÊýΪ____________¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

îÑ¡¢¸õ¡¢Ìú¡¢Äø¡¢Í­µÈ½ðÊô¼°Æä»¯ºÏÎïÔÚ¹¤ÒµÉÏÓÐÖØÒªÓÃ;¡£

£¨1£©îÑÌúºÏ½ðÊÇîÑϵ´¢ÇâºÏ½ðµÄ´ú±í£¬¸ÃºÏ½ð¾ßÓзÅÇâζȵ͡¢¼Û¸ñÊÊÖеÈÓŵ㡣

¢ÙTiµÄ»ù̬ԭ×Ó¼Ûµç×ÓÅŲ¼Ê½Îª________________¡£

¢ÚFeµÄ»ù̬ԭ×Ó¹²ÓÐ________ÖÖ²»Í¬Äܼ¶µÄµç×Ó¡£

£¨2£©ÖƱ¸CrO2Cl2µÄ·´Ó¦ÎªK2Cr2O7£«3CCl4===2KCl£«2CrO2Cl2£«3COCl2¡ü¡£

¢ÙÉÏÊö»¯Ñ§·½³ÌʽÖзǽðÊôÔªËØµç¸ºÐÔÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ__________(ÓÃÔªËØ·ûºÅ±íʾ)¡£

¢ÚCOCl2·Ö×ÓÖÐËùÓÐÔ­×Ó¾ùÂú×ã8µç×Ó¹¹ÐÍ£¬COCl2·Ö×ÓÖЦҼüºÍ¦Ð¼üµÄ¸öÊý±ÈΪ_____£¬ÖÐÐÄÔ­×ÓµÄÔÓ»¯·½Ê½Îª________¡£

£¨3£©NiO¡¢FeOµÄ¾§Ìå½á¹¹¾ùÓëÂÈ»¯ÄƵľ§Ìå½á¹¹Ïàͬ£¬ÆäÖÐNi2£«ºÍFe2£«µÄÀë×Ó°ë¾¶·Ö±ðΪ690pmºÍ780pm¡£ÔòÈ۵㣺NiO________(Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±)FeO¡£

£¨4£©NiºÍLaµÄºÏ½ðÊÇĿǰʹÓù㷺µÄ´¢Çâ²ÄÁÏ£¬¾ßÓдóÈÝÁ¿¡¢¸ßÊÙÃü¡¢Ä͵ÍεÈÌØµã£¬ÔÚÈÕ±¾ºÍÖйúÒÑʵÏÖÁ˲úÒµ»¯¡£¸ÃºÏ½ðµÄ¾§°û½á¹¹ÈçͼËùʾ¡£

¢Ù¸Ã¾§ÌåµÄ»¯Ñ§Ê½Îª________________¡£

¢ÚÒÑÖª¸Ã¾§°ûµÄĦ¶ûÖÊÁ¿ÎªM g¡¤mol£­1£¬ÃܶÈΪd g¡¤cm£­3¡£ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬Ôò¸Ã¾§°ûµÄÌå»ýÊÇ________ cm3(Óú¬M¡¢d¡¢NAµÄ´úÊýʽ±íʾ)¡£

¢Û¸Ã¾§ÌåµÄÄÚ²¿¾ßÓпÕ϶£¬ÇÒÿ¸ö¾§°ûµÄ¿Õ϶Öд¢´æ6¸öÇâÔ­×ӱȽÏÎȶ¨¡£ÒÑÖª£ºa£½511 pm£¬c£½397 pm£»±ê×¼×´¿öÏÂÇâÆøµÄÃܶÈΪ8.98¡Á10£­5 g¡¤cm£­3£»´¢ÇâÄÜÁ¦£½¡£ÈôºöÂÔÎüÇâǰºó¾§°ûµÄÌå»ý±ä»¯£¬Ôò¸Ã´¢Çâ²ÄÁϵĴ¢ÇâÄÜÁ¦Îª_______¡£

¼×´¼ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬ÓֿɳÆÎªÈ¼ÁÏ¡£¹¤ÒµÉÏÀûÓÃºÏ³ÉÆø(Ö÷Òª³É·ÖΪCO¡¢CO2ºÍH2)ÔÚ´ß»¯¼ÁµÄ×÷ÓÃϺϳɼ״¼£¬·¢ÉúµÄÖ÷Òª·´Ó¦ÈçÏ£º

¢ÙCO(g)+2H2(g)CH3OH(g) ¡÷H

¢ÚCO2(g)+3H2(g)CH3OH(g)+H2O(g) ¡÷H2=-58kJ•mol-1

¢ÛCO2(g)+H2(g)CO(g)+H2O(g) ¡÷H3=+41kJ•mol-1

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÑÖª·´Ó¦¢ÙÖеÄÏà¹ØµÄ»¯Ñ§¼ü¼üÄÜÊý¾ÝÈçÏ£º

Ôòx=____________£»

£¨2£©ÈôT¡æÊ±½«6molCO2ºÍ8molH2³äÈë2LÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦¢Ú£¬²âµÃH2µÄÎïÖʵÄÁ¿ËæÊ±¼ä±ä»¯¹ØÏµÈçͼÖÐ״̬I(ͼÖÐʵÏß)Ëùʾ¡£Í¼ÖÐÊý¾ÝA(1£¬6)´ú±íÔÚ1minʱH2µÄÎïÖʵÄÁ¿ÊÇ6mol¡£

¢ÙT¡æÊ±×´Ì¬IÌõ¼þÏ£¬0~3minÄÚCH3OHµÄƽ¾ù·´Ó¦ËÙÂÊv=_______mol/(L¡¤min)£¬Æ½ºâ³£ÊýK=_____£»

¢ÚÆäËûÌõ¼þ²»±äʱ£¬½ö¸Ä±äijһÌõ¼þºó²âµÃH2µÄÎïÖʵÄÁ¿ËæÊ±¼ä±ä»¯ÈçͼÖÐ״̬¢òËùʾ£¬Ôò¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇ_____£»

¢ÛÆäËûÌõ¼þ²»±ä£¬½ö¸Ä±äζÈʱ£¬²âµÃH2µÄÎïÖʵÄÁ¿ËæÊ±¼ä±ä»¯ÈçͼÖÐ״̬¢óËùʾ£¬Ôò״̬¢ó¶ÔÓ¦µÄζÈ_______(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±)T¡æ£»

¢ÜÈô״̬¢òµÄƽºâ³£ÊýΪK2£¬×´Ì¬¢óµÄƽºâ³£ÊýΪK3£¬ÔòK2_______(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±)K3£»

¢ÝÒ»¶¨Î¶ÈÏÂͬ£¬´Ë·´Ó¦ÔÚºãÈÝÈÝÆ÷ÖнøÐУ¬ÄÜÅжϸ÷´Ó¦´ïµ½»¯Ñ§Æ½ºâÒÀ¾ÝµÄÊÇ_______¡£

a£®ÈÝÆ÷ÖÐѹǿ²»±ä

b£®¼×´¼ºÍË®ÕôÆûµÄÌå»ý±È±£³Ö²»±ä

c£®vÕý(H2)=3vÄæCH3OH)

d£®2¸öC=O¶ÏÁѵÄͬʱÓÐ6¸öH¡ªH¶ÏÁÑ

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

¾«Ó¢¼Ò½ÌÍø