ºã¶¨µÄËÙÂÊv0=2 m/sÏòÓÒÔÈËÙÔ˶¯¡£Á½¸öÍêȫһÑùµÄ»¬¿éP¡¢QÓÉÇáÖʵ¯»ÉÏàÁ¬½Ó£¬ÓÃÒ»ÇáÉþ°ÑÁ½»¬¿éÀÖÁ×î½ü£¬Ê¹µ¯»É´¦ÓÚ×î´óѹËõ״̬±Á½ô£¬Çá·ÅÔÚ´«ËÍ´øµÄ×î×ó¶Ë¡£¿ªÊ¼Ê±P¡¢QÒ»Æð´Ó¾²Ö¹¿ªÊ¼Ô˶¯£¬t1=3 sºóͻȻÇáÉþ¶Ï¿ª£¬ºÜ¶Ìʱ¼äÄÚµ¯»ÉÉ쳤ÖÁ±¾ÉíµÄ×ÔÈ»³¤¶È£¨²»¿¼Âǵ¯»ÉµÄ³¤¶ÈµÄÓ°Ï죩£¬´Ëʱ»¬¿éQµÄËÙ¶È´óС¸ÕºÃÊÇPµÄËÙ¶È´óСµÄÁ½±¶£¬·½ÏòÏà·´¡£ÒÑÖª»¬¿éµÄÖÊÁ¿ÊÇm=0.2 kg£¬»¬¿éÓë´«ËÍ´øÖ®¼äµÄ¶¯Ä¦²Á
2
ÒòÊýÊǦÌ=0.1£¬ÖØÁ¦¼ÓËÙ¶Èg=10 m/s¡£Çó£º
£¨1£©µ¯»É´¦ÓÚ×î´óѹËõ״̬ʱ£¬µ¯»ÉµÄµ¯ÐÔÊÆÄÜ£» £¨2£©Á½»¬¿éÂ䵨µÄʱ¼ä²î£» £¨3£©Á½»¬¿éÂ䵨µã¼äµÄ¾àÀë¡£
26£®£¨14·Ö£©Áòõ£ÂÈ(SO2Cl2)ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤ÊÔ¼Á£¬ÊµÑéÊҺϳÉÁòõ£ÂȵÄʵÑé×°ÖÃÈçÏÂͼËùʾ£º
ÒÑÖª£º¢Ù SO2(g)+Cl2(g)
SO2Cl2(l) ¦¤H£½£97kJ/mol£»
¢Ú Áòõ£Âȳ£ÎÂÏÂΪÎÞɫҺÌ壬ÈÛµãΪ-54.1¡æ£¬·ÐµãΪ69.1¡æ£¬ÔÚ³±Êª¿ÕÆøÖС°·¢ÑÌ¡±£» ¢Û 100¡æÒÔÉÏ»ò³¤Ê±¼ä´æ·ÅÁòõ£Âȶ¼Ò׷ֽ⣬Éú³É¶þÑõ»¯ÁòºÍÂÈÆø¡£ »Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Áòõ£ÂÈÔÚ³±Êª¿ÕÆøÖС°·¢ÑÌ¡±µÄÔÒòÊÇ_________£¨Óû¯Ñ§·½³Ìʽ±íʾ£©¡£ £¨2£©¼ìÑé×°ÖÃGÆøÃÜÐԵķ½·¨ÊÇ____________________¡£
£¨3£©×°ÖÃAÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________________¡£
Àí¿Æ×ÛºÏ µÚ9Ò³£¨¹²20Ò³£©
£¨4£©HµÄ×÷ÓÃÊÇ______________£¬ÆäÖÐË®Á÷µÄ·½ÏòÊÇ___¡ú___£¨Ìîa»òb£©¡£ £¨5£©×°ÖÃC¡¢FÖеÄÊÔ¼ÁÏàͬ£¬Ê¢·ÅµÄÊÔ¼ÁΪ_______¡£ £¨6£©¸Ã×°ÖôæÔÚµÄÒ»´¦È±ÏÝÊÇ_____________¡£
£¨7£©Èô½«SO2¡¢Cl2°´ÕÕÒ»¶¨±ÈÀýͨÈëË®ÖУ¬ÇëÉè¼Æ¼òµ¥ÊµÑéÑéÖ¤¶þÕßÊÇ·ñÇ¡ºÃÍêÈ«·´Ó¦(¼òÒªÃèÊöʵ
Ñé²½Öè¡¢ÏÖÏóºÍ½áÂÛ)£º _______ ¡£ ÒÇÆ÷×ÔÑ¡£»¹©Ñ¡ÔñÊÔ¼Á£ºµÎ¼Ó·Ó̪µÄÇâÑõ»¯ÄÆÈÜÒº¡¢ÂÈ»¯ÑÇÌúÈÜÒº¡¢ÁòÇ軯¼ØÈÜÒº¡¢Æ·ºìÈÜÒº¡£
27£®£¨15·Ö£©ÖظõËáÄÆÔÚÖÆ¸ï¹¤Òµ¡¢Ó¡Ë¢¹¤Òµ¡¢µç¶Æ¹¤ÒµµÈÓÐÖØÒªÓÃ;£¬¸õÌú¿óµÄÖ÷Òª³É·Ö¿É±íʾΪFeO?Cr2O3£¬
»¹º¬ÓÐAl2O3¡¢Fe2O3µÈÔÓÖÊ£¬ÒÔÏÂÊÇÒÔ¸õÌú¿óΪÔÁÏÖÆ±¸ÖظõËáÄÆ£¨Na2Cr2O7£©µÄÁ÷³Ìͼ£º
£¨1£©ìÑÉÕǰӦ½«¸õÌú¿ó³ä·Ö·ÛË飬ÆäÄ¿µÄÊÇ_________¡£×ÆÉÕÊÇÖÐѧ»¯Ñ§Öг£ÓõIJÙ×÷·½·¨£¬ÈçÔÚʵ
ÑéÊÒÖн«¸õÌú¿óºÍ̼ËáÄÆ¹ÌÌå»ìºÏÎïׯÉÕ£¬ÒÔϸ÷ʵÑéÒÇÆ÷Öв»ÐèÒªµÄÊÇ_____¡£ a£®ÌÕ´ÉÛáÛö b£®ÌúÛáÛö c£®Èý½Å¼Ü d£®ÄàÈý½Ç £¨2£©×ªÒ¤Öз¢ÉúµÄÖ÷Òª·´Ó¦Îª£º
¢ÙNa2CO3+Al2O3
¸ßΠ2NaAlO2+CO2¡ü£»
¸ßΠ¢Ú__FeO?Cr2O3+__Na2CO3+__O2
__Na2CrO4+__Fe2O3+__ ¡£
ÇëÅ䯽·´Ó¦¢ÚµÄ»¯Ñ§·½³Ìʽ¡£
£¨3£©Éú³É¹ÌÌåYµÄÀë×Ó·½³ÌʽΪ_______________¡£
£¨4£©ÏòĸҺÖмÓÈëŨÁòËᣬ°Ñ¸õËáÄÆ×ª»¯ÎªÖظõËáÄÆ£¬¾Á½´ÎÕô·¢£¬Ö÷ÒªÊdzýÈ¥___(Ìѧʽ)£¬ÀäÈ´
ÖÁ30¡«40¡æµÃµ½²úÆ·¾§Ì塣ϴµÓ¸Ã¾§ÌåµÄÈܼÁ×îºÃÊÇ__________(ÌîÐòºÅ)¡£ a£®ÕôÁóË® b£®ÎÞË®ÒÒ´¼ c£®75%ÒÒ´¼ÈÜÒº
£¨5£©ÎªÁ˲ⶨʵÑéÖÆµÃµÄ²úÆ·ÖÐNa2Cr2O7µÄº¬Á¿£¬³ÆÈ¡ÑùÆ·0.140gÖÃÓÚ×¶ÐÎÆ¿ÖУ¬¼Ó50mLË®£»ÔÙ
¼ÓÈë2gKI£¨¹ýÁ¿£©¼°ÉÔ¹ýÁ¿µÄÏ¡ÁòËáÈÜÒº£¬Ò¡ÔÈ£¬°µ´¦·ÅÖÃ10min£»È»ºó¼Ó150mLÕôÁóË®²¢¼ÓÈë3mL 0.5%µí·ÛÈÜÒº£»ÓÃ0.1000mol/L Na2S2O3±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄNa2S2O3±ê×¼ÈÜÒº30.00mL¡££¨¼Ù¶¨ÔÓÖʲ»²Î¼Ó·´Ó¦£¬ÒÑÖª£ºCr2O72-+6I-+14H+£½2Cr3++3I2+7H2O£¬I2+2S2O32-£½2I-+S4O62-£©
¢ÙÖÕµãʵÑéÏÖÏóÊÇ________¡£
¢Ú¸Ã²úÆ·ÖÐNa2Cr2O7µÄ´¿¶ÈΪ____ ¡££¨ÒÔÖÊÁ¿·ÖÊý±íʾ£©
¢ÛÈôµÎ¶¨¹ÜÔڵζ¨Ç°¸©ÊÓ¶ÁÊý£¬µÎ¶¨½áÊøºóÑöÊÓ¶ÁÊý£¬²âµÃÑùÆ·µÄ´¿¶È½« £¨Ìî¡°Æ«¸ß¡±¡¢
Àí¿Æ×ÛºÏ µÚ10Ò³£¨¹²20Ò³£©
»ò¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©¡£
28£®£¨14·Ö£©¼×´¼ÊÇÖØÒªµÄ»¯¹¤ÔÁÏ£¬ÓÖ¿É×÷ΪȼÁÏ¡£¹¤ÒµÉÏÀûÓÃºÏ³ÉÆø£¨Ö÷Òª³É·ÖΪCO¡¢CO2ºÍH2£©ÔÚ
´ß»¯¼ÁµÄ×÷ÓÃϺϳɼ״¼£¬·¢ÉúµÄÖ÷·´Ó¦ÈçÏ£º ¢ÙCO(g)+2H2(g)¢ÚCO2(g)+3H2(g)¢ÛCO2(g)+H2(g)»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑÖª·´Ó¦¢ÙÖÐÏà¹ØµÄ»¯Ñ§¼ü¼üÄÜÊý¾ÝÈçÏ£º
»¯Ñ§¼ü H£H C£O C 343 1076 O H£O C£H 465 x CH3OH(g) ¦¤H1
CH3OH£¨g£©+H2O(g) ¦¤H£½£58 kJ/mol CO(g)+H2O(g) ¦¤H£½£«41 kJ/mol
E/£¨kJ¡¤mol-1£© 436 Ôòx£½_________¡£ £¨2£©Èô½«lmol CO2ºÍ2mol H2³äÈëÈÝ»ýΪ2LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬ÔÚÁ½ÖÖ²»Í¬Î¶ÈÏ·¢Éú·´Ó¦¢Ú¡£²â
µÃCH3OHµÄÎïÖʵÄÁ¿ËæÊ±¼äµÄ±ä»¯ÈçͼËùʾ¡£
¢ÙÇúÏßI¡¢¢ò¶ÔÓ¦µÄƽºâ³£Êý´óС¹ØÏµÎªKI_____K¢ò(Ìî¡°£¾¡±»ò¡°£½¡±»ò¡°£¼¡±)£» ¢ÚÒ»¶¨Î¶ÈÏ£¬ÄÜÅжϸ÷´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ_____¡£
a£®ÈÝÆ÷ÖÐѹǿ²»±ä b£®¼×´¼ºÍË®ÕôÆøµÄÌå»ý±È±£³Ö²»±ä
c£®vÕý£¨H2£©£½3vÄæ£¨CH3OH£© d£®2¸öC£½O¶ÏÁѵÄͬʱÓÐ6¸öH¡ªH¶ÏÁÑ ¢ÛÈô5minºó·´Ó¦´ïµ½Æ½ºâ״̬£¬H2µÄת»¯ÂÊΪ90%£¬ÔòÓÃCO2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ_________£¬¸ÃζÈÏÂµÄÆ½ºâ³£ÊýΪ_______£»ÈôÈÝÆ÷ÈÝ»ý²»±ä£¬ÏÂÁдëÊ©¿ÉÔö¼Ó¼×´¼²úÂʵÄÊÇ________¡£ a£®ËõС·´Ó¦ÈÝÆ÷µÄÈÝ»ý b£®Ê¹ÓúÏÊʵĴ߻¯¼Á c£®³äÈëHe d£®°´Ô±ÈÀýÔÙ³äÈëCO2ºÍH2
£¨3£©ÒÔ¼×´¼ÎªÈ¼ÁÏ£¬ÑõÆøÎªÑõ»¯¼Á£¬KOHÈÜҺΪµç½âÖÊÈÜÒº£¬¿ÉÖÆ³ÉȼÁÏµç³Ø¡£ÒÔ´Ëµç³Ø×÷µçÔ´£¬ÔÚ
ʵÑéÊÒÖÐÄ£ÄâÂÁÖÆÆ·±íÃæ¡°¶Û»¯¡±´¦Àí¹ý³Ì£¨×°ÖÃÈçͼËùʾ£©¡£ÆäÖÐÎïÖÊbÊÇ_____ £¬Ñô¼«µç¼«·´Ó¦Îª____ ¡£
Àí¿Æ×ÛºÏ µÚ11Ò³£¨¹²20Ò³£©
29£®£¨9·Ö£©ÈçͼΪijÂÌɫֲÎïÒ¶Èâϸ°ûÄÚ·¢ÉúµÄ²¿·Ö´úл£¬ÆäÖТ١«¢à±íʾ²»Í¬ÉúÀí¹ý³Ì£»Ï±íÁгöÁËÔÚ
²»Í¬¹âÕÕÇ¿¶ÈÓëζÈÌõ¼þϸÃÖ²ÎïËùÔÚ»·¾³ÖÐÑõÆøµÄ±ä»¯Á¿¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Í¼ÖеÄ_____________£¨ÌîÊý×ÖÐòºÅ£©¹ý³ÌÖ»ÄÜ·¢ÉúÔÚÉúÎïĤÉÏ¡£¸ÃÖ²ÎïÒ¶Èâϸ°ûÔÚ0 klx¹âÕÕÌõ
¼þÏÂÄÜ·¢Éúͼ¼×ÖеÄ__________£¨ÌîÊý×ÖÐòºÅ£©¹ý³Ì¡£
£¨2£©CO2Ũ¶ÈÊôÓÚ±íÖÐËùʾʵÑéÖеÄ__________±äÁ¿¡£Ö²Îï¹âºÏ×÷Óùý³ÌÖÐCO2Ũ¶ÈͻȻ½µµÍ½«µ¼ÖÂ
¶Ìʱ¼äÄÚÒ¶ÂÌÌåÖÐC3º¬Á¿½µµÍ£¬½áºÏͼ¼×ÖТݢ޹ý³Ì·ÖÎöÔÒò£º____________________¡£ £¨3£©¸ÃÖ²ÎïÔÚ10¡æ¡¢l0 klxµÄÌõ¼þÏÂÅàÑø4Сʱ¹²²úÉúÑõÆø__________mg¡£Èô´ËÖ²ÎïÔÚ20¡æ¡¢5 klx
µÄÌõ¼þÏÂÅàÑø4£®8СʱºóתÈë10¡æ¡¢ºÚ°µ»·¾³ÖмÌÐøÅàÑøl9.2Сʱ£¬ÔÚ´Ë24СʱÖиÃÖ²Îï²»ÄÜÕý³£Éú³¤£¬ÆäÔÒòΪ__________________________________________________¡£
30£®£¨10·Ö£©Çë»Ø´ðÏÂÁÐÓйØÉñ¾µ÷½ÚµÄÎÊÌ⣺
ͼ1 ͼ2
£¨1£©Éñ¾µ÷½ÚµÄ½á¹¹»ù´¡ÊÇ__________¡£ÔÚÉñ¾µ÷½ÚµÄ½á¹¹»ù´¡ÖУ¬Ð§Ó¦Æ÷ÊÇÖ¸________________¡£ £¨2£©Í¼1ÖТ١«¢Ü±íʾijÖÖÒ©Æ·×è¶ÏÉñ¾³å¶¯´«µ¼µÄ¿ÉÄÜλµã¡£Î´Ê¹ÓøÃҩƷʱ£¬ÈôijÈ˵ÄÊÖÊÜÉË£¬
Ôò»áÔÚ__________ÐγÉÍ´¾õ¡£Ê¹ÓøÃÒ©Æ·ºó£¬ÈôijÈ˵ÄÊÖÊÜÉË£¬Èç¹ûËûµÄÊÖÄܶ¯£¬µ«Ã»Óиоõ£¬ÄÇô±»×è¶ÏµÄλµã¿ÉÄÜÊÇ__________£»Èç¹ûËûÄܸоõµ½ÉË¿ÚÍ´£¬²¢ÇÒÊÖÒ²Äܶ¯£¬ÄÇô±»×è¶ÏµÄλµã¿ÉÄÜÊÇ__________¡£
£¨3£©·´À¡ÔÚÉñ¾ÍøÂçÖй㷺´æÔÚ¡£Í¼2±íʾ¼¹Ëèǰ½ÇÔ˶¯Éñ¾Ôª¡¢ÈòÉÜϸ°û£¨ÐË·ÜʱÄÜÊÍ·ÅÒÖÖÆÐÔÉñ
¾µÝÖÊ£©¹²Í¬Ö§Å伡ÈâÊÕËõµÄ;¾¶¡£ÈôÔÚaµã¸øÓèÊÊÒ˴̼¤£¬ÔòÔÚͼÖÐ__________£¨Ìî×Öĸ£©µãÄܲâµÃĤµçλ±ä»¯¡£
31£®£¨8·Ö£©Í¼1±íʾºôÂ×±´¶û´ó²ÝÔÖв¿·ÖʳÎïÍø£¬Í¼2±íʾ¸ÃʳÎïÍøÖÐÁ½ÉúÎïÖ®¼äµÄÊýÁ¿¹ØÏµ¡£Òò¿ª²É
Àí¿Æ×ÛºÏ µÚ12Ò³£¨¹²20Ò³£©