ÌâÄ¿ÄÚÈÝ

£¨17·Ö£©H2O2ÊÇÒ»ÖÖÂÌÉ«Ñõ»¯»¹Ô­ÊÔ¼Á£¬ÔÚ»¯Ñ§Ñо¿ÖÐÓ¦Óù㷺¡£
£¨1£©Ä³Ð¡×éÄâÔÚͬŨ¶ÈFe3+µÄ´ß»¯Ï£¬Ì½¾¿H2O2Ũ¶È¶ÔH2O2·Ö½â·´Ó¦ËÙÂʵÄÓ°Ïì¡£ÏÞÑ¡ÊÔ¼ÁÓëÒÇÆ÷£º30% H2O2¡¢0.1mol?L-1Fe2(SO4)3¡¢ÕôÁóË®¡¢×¶ÐÎÆ¿¡¢Ë«¿×Èû¡¢Ë®²Û¡¢½º¹Ü¡¢²£Á§µ¼¹Ü¡¢Á¿Í²¡¢Ãë±í¡¢ºãÎÂˮԡ²Û¡¢×¢ÉäÆ÷
¢Ùд³ö±¾ÊµÑéH2O2·Ö½â·´Ó¦·½³Ìʽ²¢±êÃ÷µç×ÓתÒƵķ½ÏòºÍÊýÄ¿£º____________
¢ÚÉè¼ÆʵÑé·½°¸£ºÔÚ²»Í¬H2O2Ũ¶ÈÏ£¬²â¶¨  ____________________________________£¨ÒªÇóËù²âµÃµÄÊý¾ÝÄÜÖ±½ÓÌåÏÖ·´Ó¦ËÙÂÊ´óС£©¡£
¢ÛÉè¼ÆʵÑé×°Öã¬Íê³ÉÏÂͼµÄ×°ÖÃʾÒâͼ¡£

¢Ü²ÎÕÕϱí¸ñʽ£¬ÄⶨʵÑé±í¸ñ£¬ÍêÕûÌåÏÖʵÑé·½°¸£¨ÁгöËùÑ¡ÊÔ¼ÁÌå»ý¡¢Ðè¼Ç¼µÄ´ý²âÎïÀíÁ¿ºÍËùÄⶨµÄÊý¾Ý£»Êý¾ÝÓÃ×Öĸ±íʾ£©¡£

£¨2£©ÀûÓÃͼ21£¨a£©ºÍ21£¨b£©ÖеÄÐÅÏ¢£¬°´Í¼21£¨c£©×°Öã¨Á¬ÄܵÄA¡¢BÆ¿ÖÐÒѳäÓÐNO2ÆøÌ壩½øÐÐʵÑé¡£¿É¹Û²ìµ½BÆ¿ÖÐÆøÌåÑÕÉ«±ÈAÆ¿ÖеÄ_________£¨Ìî¡°É»ò¡°Ç³¡±£©£¬ÆäÔ­ÒòÊÇ______________________________________________________¡£
£¨1£©¢Ù
¢ÚÉú³ÉÏàͬÌå»ýµÄÑõÆøËùÐèµÄʱ¼ä
¢Û
¢Ü²â¶¨·´Ó¦Ê±¼ä
 
H2O2µÄÌå»ý(mL)
0.1mol?L-1Fe2(SO4)3µÄÌå»ý£¨mL£©
¼ÓÈëÕôÁóË®µÄÌå»ý£¨mL£©
Éú³ÉO2µÄÌå»ý(mL)
·´Ó¦Ê±¼ä£¨min£©
ʵÑé1
b
a  
c
d
 
ʵÑé2
c
a
b
d
 
£¨2£©ÉÒòΪ¹ýÑõ»¯Çâ·Ö½âÊÇ·ÅÈÈ·´Ó¦£¬2NO2(g)N2O4(g)Ò²ÊÇ·ÅÈÈ·´Ó¦£¬ËùÒÔBƿζȸßÓÚAÆ¿£¬Î¶ÈÉý¸ß£¬Æ½ºâÄæÏòÒƶ¯£¬¶þÑõ»¯µªÅ¨¶ÈÔö´ó£¬ÑÕÉ«¼ÓÉî¡£

ÊÔÌâ·ÖÎö£º£¨1£©¢Ù¹ýÑõ»¯ÇâÔÚÁòËáÌú×÷´ß»¯¼ÁÌõ¼þÏ·ֽâÉú³ÉË®ºÍÑõÆø£¬¹ýÑõ»¯Çâ¼È×÷Ñõ»¯¼ÁÓÖ×÷»¹Ô­¼Á£¬»¯Ñ§·½³Ìʽ¼°µç×ÓתÒÆ·½ÏòºÍÊýĿΪ
¢Ú·´Ó¦ËÙÂÊÊǵ¥Î»Ê±¼äÄÚÎïÖÊŨ¶ÈµÄ¸Ä±äÁ¿£¬ËùÒԲⶨ²»Í¬Å¨¶ÈµÄ¹ýÑõ»¯Çâ¶Ô·Ö½âËÙÂʵÄÓ°Ï죬Ðè²â¶¨Ïàͬʱ¼äÄÚ£¬²úÉúÑõÆøµÄÌå»ýµÄ¶àÉÙ£¬»òÉú³ÉÏàͬÌå»ýµÄÑõÆøËùÐèʱ¼äµÄ¶àÉÙ£»
¢ÛÀûÓÃÅÅË®Á¿Æø·¨£¬ÊÕ¼¯Ò»¶¨Ìå»ýµÄO2£¬ÐèҪˮ²Û¡¢Á¿Í²¡¢µ¼Æø¹Ü£¬Á¿Í²ÄÚÊ¢ÂúË®µ¹¿ÛÔÚË®²ÛÖУ¬×°ÖÃÈçͼ
¢Ü
 
H2O2µÄÌå»ý(mL)
0.1mol?L-1Fe2(SO4)3µÄÌå»ý£¨mL£©
¼ÓÈëÕôÁóË®µÄÌå»ý£¨mL£©
Éú³ÉO2µÄÌå»ý(mL)
·´Ó¦Ê±¼ä£¨min£©
ʵÑé1
b
a
c
d
 
ʵÑé2
c
a
b
d
 
£¨2£©ÓÉͼa¿ÉÖª¹ýÑõ»¯ÇâµÄ·Ö½â·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬ÓÉͼb¿ÉÖª¶þÑõ»¯µª×ª»¯ÎªËÄÑõ»¯¶þµªµÄ·´Ó¦Ò²ÊÇ·ÅÈÈ·´Ó¦£¬ËùÒÔ×°ÖÃcÖУ¬ÓÒ²àÉÕ±­BƿζȸßÓÚ×ó²àÉÕ±­AÆ¿£¬¶øζÈÉý¸ß£¬Ê¹Æ½ºâ2NO2(g)N2O4(g)ÄæÏò½øÐУ¬¶þÑõ»¯µªÅ¨¶ÈÔö´ó£¬BÆ¿µÄÑÕÉ«±ÈAÆ¿µÄÑÕÉ«Éî¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨16·Ö£©MnO2Óë¹ýÁ¿¹ÌÌåKOHºÍKClO3ÔÚ¸ßÎÂÏ·´Ó¦£¬Éú³ÉÃÌËá¼Ø£¨K2MnO4£©ºÍKCl£¬K2MnO4ÈÜÒº¾­Ëữºóת»¯ÎªMnO2ºÍKMnO4¡£Ò»ÖÖMnO2ÖƱ¸KMnO4µÄ·½·¨ÈçÏ£º

£¨1£©Ð´³ö¡°ÈÜ»¯Ñõ»¯¡±µÄ»¯Ñ§·´Ó¦·½³Ìʽ£º                       £»
£¨2£©´ÓÈÜÒº¢ÚÖеõ½KMnO4´Ö¾§ÌåµÄ·½·¨ÊÇ                       £»
£¨3£©ÉÏÊöÁ÷³ÌÖпÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊÊÇ                       £»
£¨4£©ÀíÂÛÉÏ£¨Èô²»¿¼ÂÇÖƱ¸¹ý³ÌÖеÄËðʧÓëÎïÖÊÑ­»·£©1mol MnO2¿ÉÖƵĠ  mol KMnO4£»
£¨5£©ÒÑÖª293Kʱ´×Ëá¼ØµÄÈܽâ¶ÈΪ217g£¬ÁòËá¼ØµÄÈܽâ¶ÈΪ11.1g£¬Çë½âÊÍËữ¹ý³ÌÖвÉÓô×Ëá¶ø²»²ÉÓÃÑÎËá»òÁòËáµÄÔ­Òò£º
¢Ù²»²ÉÓÃÑÎËáµÄÔ­Òò£º                                  £»
¢Ú²»²ÉÓÃÁòËáµÄÔ­Òò£º                                  ¡£
£¨6£©²ÉÓÃÉÏÊöÁ÷³ÌÖƱ¸KMnO4²úÂʲ¢²»¸ß£¬ÎªÌá¸ß²úÂÊ£¬»¹¿ÉÒÔ²ÉÓÃÒÔÏÂÁ½ÖÖ·½·¨£º
¢ÙÔÚK2MnO4ÈÜÒºÖÐͨÈëÂÈÆø£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ                      £»
¢Úµç½âK2MnO4ÈÜÒº£¬µç½âµÄ×Ü·´Ó¦Àë×Ó·½³ÌʽΪ                ¡£
£¨14·Ö£©Ä³¾­¼Ã¿ª·¢Çø½«îÑÒ±Á¶³§ÓëÂȼ¡¢¼×´¼³§×é³ÉÁËÒ»¸ö²úÒµÁ´£¨ÈçͼËùʾ£©£¬´ó´óµØÌá¸ßÁË×ÊÔ´ÀûÓÃÂÊ£¬¼õÉÙÁË»·¾³ÎÛȾ¡£

ÇëÌîдÏÂÁпհףº
£¨1£©Ð´³öîÑÌú¿ó¾­ÂÈ»¯µÃµ½ËÄÂÈ»¯îѵĻ¯Ñ§·½³Ìʽ£º                     ¡£
£¨2£©ÓÉCOºÍH2ºÏ³É¼×´¼ÊÇ·ÅÈȵģ¬·½³ÌʽÊÇ£ºCO(g)£«2H2(g)CH3OH(g)¡£
¢ÙÒÑÖª¸Ã·´Ó¦ÔÚ300¡æʱµÄ»¯Ñ§Æ½ºâ³£ÊýΪ0.27£¬¸ÃζÈϽ«2 mol CO¡¢3 mol H2ºÍ2 mol CH3OH³äÈëÈÝ»ýΪ2 LµÄÃܱÕÈÝÆ÷ÖУ¬´Ëʱ·´Ó¦½«     £¨Ìî¡°ÏòÕý·´Ó¦·½Ïò½øÐС±¡¢¡°ÏòÄæ·´Ó¦·½Ïò½øÐС±»ò¡°´¦ÓÚƽºâ״̬¡±£©¡£
¢ÚÏÂͼ±íʾºÏ³É¼×´¼·´Ó¦´ïµ½Æ½ºâºó£¬Ã¿´ÎÖ»¸Ä±äζȡ¢Ñ¹Ç¿¡¢´ß»¯¼ÁÖеÄijһÌõ¼þ£¬·´Ó¦ËÙÂʦÔÓëʱ¼ätµÄ¹Øϵ¡£ÆäÖбíʾƽºâ»ìºÏÎïÖеļ״¼µÄº¬Á¿×î¸ßµÄÒ»¶Îʱ¼äÊÇ       ¡£Í¼ÖÐt3ʱ¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇ        ¡£

£¨3£©Ä³Í¬Ñ§Éè¼ÆÁËÒ»¸ö¼×´¼È¼Áϵç³Ø£¬²¢Óøõç³Øµç½â200mL¡ª¶¨Å¨¶ÈNaClÓëCuSO4»ìºÏÈÜÒº£¬Æä×°ÖÃÈçͼ£º

¢Ùд³ö¼×ÖÐͨÈë¼×´¼ÕâÒ»¼«µÄµç¼«·´Ó¦Ê½                                 ¡£
¢ÚÀíÂÛÉÏÒÒÖÐÁ½¼«ËùµÃÆøÌåµÄÌå»ýËæʱ¼ä±ä»¯µÄ¹ØϵÈç±ûͼËùʾ£¨ÆøÌåÌå»ýÒÑ»»Ëã³É±ê×¼×´¿öϵÄÌå»ý£©£¬Ð´³öÔÚt1ºó,ʯīµç¼«Éϵĵ缫·´Ó¦Ê½                £¬Ô­»ìºÏÈÜÒºÖÐNaClµÄÎïÖʵÄÁ¿Å¨¶ÈΪ            mol/L¡££¨¼ÙÉèÈÜÒºÌå»ý²»±ä£©
þÊǺ£Ë®Öк¬Á¿½Ï¶àµÄ½ðÊô£¬Ã¾¡¢Ã¾ºÏ½ð¼°ÆäþµÄ»¯ºÏÎïÔÚ¿ÆѧÑо¿ºÍ¹¤ÒµÉú²úÖÐÓÃ;·Ç³£¹ã·º¡£
£¨1£©Mg2NiÊÇÒ»ÖÖ´¢ÇâºÏ½ð£¬ÒÑÖª£º
Mg(s) + H2(g)=MgH2(s) ¡÷H1=£­74.5kJ¡¤mol£­1
Mg2Ni(s) + 2H2(g)=Mg2NiH4(s) ¡÷H2=£­64.4kJ¡¤mol£­1
Mg2Ni(s)+2MgH2(s) = 2Mg(s)+Mg2NiH4(s)  ¡÷H3
Ôò¡÷H3 =               kJ¡¤mol£­1¡£
£¨2£©¹¤ÒµÉÏ¿ÉÓõç½âÈÛÈÚµÄÎÞË®ÂÈ»¯Ã¾»ñµÃþ¡£ÆäÖÐÂÈ»¯Ã¾ÍÑË®Êǹؼü¹¤ÒÕÖ®Ò»£¬Ò»ÖÖÕýÔÚÊÔÑéµÄÂÈ»¯Ã¾¾§ÌåÍÑË®µÄ·½·¨ÊÇ£ºÏȽ«MgCl2¡¤6H2Oת»¯ÎªMgCl2¡¤NH4Cl¡¤nNH3£¨ï§Ã¾¸´ÑΣ©,È»ºóÔÚ700¡æÍÑ°±µÃµ½ÎÞË®ÂÈ»¯Ã¾£¬ÍÑ°±·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ               £»µç½âÈÛÈÚÂÈ»¯Ã¾£¬Òõ¼«µÄµç¼«·´Ó¦Ê½Îª               ¡£
£¨3£©´¢Çâ²ÄÁÏMg(AlH4)2ÔÚ110-200¡ãCµÄ·´Ó¦Îª£ºMg(AlH4)2=MgH2 +2A1+3H2¡üÿÉú³É27gAlתÒƵç×ÓµÄÎïÖʵÄÁ¿Îª               ¡£

£¨4£©¹¤ÒµÉÏÓÃMgC2O4¡¤2H2OÈÈ·Ö½âÖƳ¬Ï¸MgO£¬ÆäÈÈ·Ö½âÇúÏßÈçͼ¡£
ͼÖиô¾ø¿ÕÆøÌõ¼þÏÂB¡úC·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ               ¡£

£¨5£©Ò»ÖÖÓлúþ»¯ºÏÎï¿ÉÓÃÓÚÖÆÔì¹âѧԪ¼þµÄÍ¿²¼Òº£¬»¯Ñ§Ê½¿É±íʾΪ£º£¬Ëü¿É·¢ÉúÈçÏ·´Ó¦£º
ROHÓëBµÄºË´Å¹²ÕñÇâÆ×ÈçÏÂͼ:

ROHÓÉC¡¢H¡¢O¡¢FËÄÖÖÔªËØ×é³ÉµÄº¬·úÓлúÎ·Ö×ÓÖÐÖ»ÓÐ1¸öÑõÔ­×Ó£¬ËùÓзúÔ­×Ó»¯Ñ§»·¾³Ïàͬ£¬Ïà¶Ô·Ö×ÓÖÊÁ¿Îª168£¬ÔòROHµÄ½á¹¹¼òʽΪ               £» BµÄ½á¹¹¼òʽΪ               ¡£

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

¾«Ó¢¼Ò½ÌÍø