ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿²ÝËᾧÌåµÄ×é³É¿É±íʾΪH2C2O4¡¤xH2O£¬ÎªÁ˲ⶨxÖµ£¬½øÐÐÏÂÊöʵÑ飺
¢Ù³ÆÈ¡n g²ÝËᾧÌåÅä³É100.00 mLË®ÈÜÒº£»
¢ÚÈ¡25.00 mLËùÅäÖƵIJÝËáÈÜÒºÖÃÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÏ¡ÁòËᣬÓÃŨ¶ÈΪa mol¡¤L£1µÄKMnO4ÈÜÒºµÎ¶¨£¬ÒÑÖª2KMnO4+5H2C2O4+3H2SO4=K2SO4+10CO2¡ü+2MnSO4+8H2O
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©ÊµÑéÖв»¿¼ÂÇÌú¼Ų̈µÈ¼Ð³ÖÒÇÆ÷Í⣬ÐèÒªµÄÒÇÆ÷ÓУ¨ÌîÐòºÅ£©___________£¬»¹È±ÉÙµÄÒÇÆ÷ÓУ¨ÌîÃû³Æ£©____________¡£
A.ÍÐÅÌÌìƽ£¨´øíÀÂ룬Ä÷×Ó£© B.µÎ¶¨¹Ü C. 100mLÈÝÁ¿Æ¿ D.ÉÕ±E.©¶· F.׶ÐÎÆ¿ G.²£Á§°ô H.ÉÕÆ¿
£¨2£©ÊµÑéÖÐKMnO4ÈÜҺӦװÔÚ____ʽµÎ¶¨¹ÜÖУ¬µÎ¶¨ÖÕµãµÄÅжÏÒÀ¾ÝÊÇ_________________
£¨3£©ÈôÔڵζ¨Ç°Ã»ÓÐÓÃamol¡¤L-1µÄKMnO4ÈÜÒº¶ÔµÎ¶¨¹Ü½øÐÐÈóÏ´£¬ÔòËù²âµÃµÄxÖµ»á___________£¨Æ«´ó¡¢Æ«Ð¡¡¢ÎÞÓ°Ï죩¡£
£¨4£©ÈôµÎ¶¨ÖÕµã¶ÁÊýʱĿ¹âÑöÊÓ£¬Ôò¼ÆËã³öµÄxÖµ¿ÉÄÜ_______________£¨ÌîÆ«´ó¡¢Æ«Ð¡¡¢ÎÞÓ°Ï죩¡£
£¨5£©µÎ¶¨¹ý³ÌÖÐÓÃÈ¥V mL a mol¡¤L£1µÄKMnO4ÈÜÒº£¬ÔòËùÅäÖƵIJÝËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_______mol¡¤L£1 ¡£
¡¾´ð°¸¡¿ABCDFG ½ºÍ·µÎ¹Ü Ëá µÎÈë×îºóÒ»µÎKMnO4ÈÜÒº£¬ÈÜÒº´ÓÎÞÉ«±äΪ×ÏÉ«ÇÒ°ë·ÖÖÓ²»ÍÊÈ¥£¬ËµÃ÷µÎ¶¨µ½ÖÕµã ƫС ƫС 0.1aV
¡¾½âÎö¡¿
£¨1£©ÊµÑéÓÐÁ½¸ö¹ý³Ì£ºÅäÖÆ׼ȷŨ¶ÈµÄ²ÝËáÈÜÒº£¬ËùÐèÒªµÄʵÑéÒÇÆ÷Ö÷ÒªÓÐÌìƽ£¨º¬íÀÂ룩¡¢ÉÕ±¡¢Ò©³×¡¢100mLÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü¡¢²£Á§°ôµÈ£»
ÓøßÃÌËá¼ØÈÜÒº½øÐе樲â²ÝËáµÄÎïÖʵÄÁ¿£ºËùÐèÒªµÄʵÑéÒÇÆ÷Ö÷ÒªÓÐÉÕ±¡¢ËáʽµÎ¶¨¹Ü¡¢Ìú¼Ų̈£¨´øµÎ¶¨¹Ü¼Ð£©¡¢×¶ÐÎÆ¿µÈ£»¸ù¾ÝÒÔÉϲÙ×÷ÖÐʹÓõÄÒÇÆ÷½øÐнâ´ð£»
£¨2£©KMnO4ÈÜÒº¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬Äܹ»¸¯Ê´ÏðƤ¹Ü£¬¹ÊKMnO4ÈÜҺӦװÔÚËáʽµÎ¶¨¹ÜÖУ»¸ù¾ÝËáÐÔ¸ßÃÌËá¼Ø±¾ÉíÏÔ×ÏÉ«¼°²Ù×÷¹æ·¶·ÖÎö×÷´ð£»
£¨3£©£¨4£©Ïȸù¾ÝʵÑéµÄ¹æ·¶²Ù×÷¶Ô¹«Ê½c£¨´ý²â£©=µÄÓ°Ïì·ÖÎöÎó²î£¬ÔÙ½áºÏʵÑé¼ÆËã¹ý³Ì·ÖÎö³ö²ÝËẬÁ¿£¬²ÝËẬÁ¿Ô½´ó£¬Æä½á¾§Ë®µÄÖÊÁ¿Ô½Ð¡£¬¼´ËùµÃxֵԽС£¬¾Ý´Ë·ÖÎö×÷´ð£»
£¨5£©¸ù¾ÝÒÑÖªµÄ»¯Ñ§·½³ÌʽÖÐÎïÖʵÄÁ¿ÓëÆ仯ѧ¼ÆÁ¿ÊýÖ®¼äµÄ¹Øϵ×÷´ð¡£
£¨1£©ÎªÁË׼ȷÅäÖÆÒ»¶¨Å¨¶ÈµÄ²ÝËáÈÜÒº£¬ËùÐèÒªµÄʵÑéÒÇÆ÷Ö÷ҪΪ£ºÍÐÅÌÌìƽ£¨´øíÀÂ룬Ä÷×Ó£©¡¢100mLÈÝÁ¿Æ¿¡¢ÉÕ±¡¢²£Á§°ô¡¢½ºÍ·µÎ¹Ü£»ÓøßÃÌËá¼ØÈÜÒº½øÐе樲â²ÝËáµÄÎïÖʵÄÁ¿£¬ËùÐèÒªµÄʵÑéÒÇÆ÷Ö÷ÒªÓУºÉÕ±¡¢ËáʽµÎ¶¨¹Ü¡¢Ìú¼Ų̈£¨´øµÎ¶¨¹Ü¼Ð£©¡¢×¶ÐÎÆ¿µÈ£¬¹ÊÑ¡ABCDFG£¬»¹È±ÉÙµÄÒÇÆ÷Ϊ£º½ºÍ·µÎ¹Ü£¬
¹Ê´ð°¸Îª£ºABCDFG£»½ºÍ·µÎ¹Ü£»
£¨2£©KMnO4ÈÜÒº¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬¿ÉÒÔ¸¯Ê´¼îʽµÎ¶¨¹ÜÖеÄÏðƤ¹Ü£¬¹ÊKMnO4ÈÜҺӦװÔÚËáʽµÎ¶¨¹ÜÖУ¬µ±µÎÈë×îºóÒ»µÎKMnO4ÈÜÒº£¬ÈÜÒº´ÓÎÞÉ«±äΪ×ÏÉ«ÇÒ°ë·ÖÖÓ²»ÍÊÈ¥£¬ËµÃ÷µÎ¶¨µ½Öյ㣬
¹Ê´ð°¸Îª£ºË᣻µÎÈë×îºóÒ»µÎKMnO4ÈÜÒº£¬ÈÜÒº´ÓÎÞÉ«±äΪ×ÏÉ«ÇÒ°ë·ÖÖÓ²»ÍÊÈ¥£¬ËµÃ÷µÎ¶¨µ½Öյ㣻
£¨3£©Ôڵζ¨Ç°Ã»ÓÐÓÃa mol¡¤L-1µÄKMnO4ÈÜÒº¶ÔµÎ¶¨¹Ü½øÐÐÈóÏ´£¬Ï൱ÓÚÏ¡ÊÍÁËKMnO4±ê×¼ÈÜÒº£¬Ôò»áʹV(±ê×¼)Æ«´ó£¬¸ù¾Ýc£¨´ý²â£©=¿ÉÖª£¬c£¨´ý²â£©Æ«´ó£¬µ¼ÖÂn(H2C2O4)»áÆ«´ó£¬Ôò¼ÆËã³öµÄx»áƫС£¬
¹Ê´ð°¸Îª£ºÆ«Ð¡£»
£¨4£©ÈôµÎ¶¨ÖÕµã¶ÁÊýʱĿ¹âÑöÊÓ£¬»áʹV(±ê×¼)Æ«´ó£¬¸ù¾Ýc£¨´ý²â£©=¿ÉÖª£¬c£¨´ý²â£©Æ«´ó£¬µ¼ÖÂn(H2C2O4)»áÆ«´ó£¬Ôò¼ÆËã³öµÄx»áƫС£¬
¹Ê´ð°¸Îª£ºÆ«Ð¡£»
£¨5£©¸ù¾Ý¹Øϵʽ2KMnO4 5H2C2O4¿ÉµÃn (H2C2O4) = = mol£¬Ôò
c(H2C2O4) = = 0.1 aV molL-1£¬
¹Ê´ð°¸Îª£º0.1 aV¡£
¡¾ÌâÄ¿¡¿Ä³Î¶Èʱ£¬ÔÚ2LµÄÃܱÕÈÝÆ÷ÖУ¬X¡¢Y¡¢Z£¨¾ùΪÆøÌ壩ÈýÖÖÎïÖʵÄÁ¿Ëæʱ¼äµÄ±ä»¯ÇúÏßÈçͼËùʾ¡£
£¨1£©ÓÉͼÖÐËù¸øÊý¾Ý½øÐзÖÎö£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________¡£
£¨2£©ÈôÉÏÊö·´Ó¦ÖÐX¡¢Y¡¢Z·Ö±ðΪH2¡¢N2¡¢NH3,ijζÈÏ£¬ÔÚÈÝ»ýºã¶¨Îª2.0LµÄÃܱÕÈÝÆ÷ÖгäÈë2.0molN2ºÍ2.0molH2£¬Ò»¶Îʱ¼äºó·´Ó¦´ïƽºâ״̬£¬ÊµÑéÊý¾ÝÈçϱíËùʾ£º
t/s | 0 | 50 | 150 | 250 | 350 |
n(NH3) | 0 | 0.24 | 0.36 | 0.40 | 0.40 |
0~50sÄÚµÄƽ¾ù·´Ó¦ËÙÂÊ v(N2) = __________ mol¡¤L£1¡¤min£1£¬250sʱ£¬H2µÄת»¯ÂÊΪ____________£¥¡£
£¨3£©ÒÑÖª£º¼üÄÜÖ¸ÔÚ±ê×¼×´¿öÏ£¬½«1molÆø̬·Ö×ÓAB£¨g£©½âÀëΪÆø̬Ô×ÓA£¨g£©£¬B£¨g£©ËùÐèµÄÄÜÁ¿£¬Ó÷ûºÅE±íʾ£¬µ¥Î»ÎªkJ/mol¡£µÄ¼üÄÜΪ946kJ/mol£¬H-HµÄ¼üÄÜΪ436kJ/mol£¬N-HµÄ¼üÄÜΪ391kJ/mol£¬ÔòÉú³É1molNH3¹ý³ÌÖÐ___£¨Ìî¡°ÎüÊÕ¡±»ò¡°·Å³ö¡±£©µÄÄÜÁ¿Îª____ kJ£¬ ·´Ó¦´ïµ½(2)ÖеÄƽºâ״̬ʱ,¶ÔÓ¦µÄÄÜÁ¿±ä»¯µÄÊýֵΪ____kJ¡£
£¨4£©·´Ó¦´ïƽºâʱÈÝÆ÷ÄÚ»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿±ÈÆðʼʱ____£¨ÌîÔö´ó¡¢¼õС»ò²»±ä£©£¬»ìºÏÆøÌåÃܶȱÈÆðʼʱ______£¨ÌîÔö´ó¡¢¼õС»ò²»±ä£©¡£
£¨5£©Îª¼Ó¿ì·´Ó¦ËÙÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ_______£¨Ìî·ûºÅ£©
a£®½µµÍÎÂ¶È b£®Ôö´óѹǿ c£®ºãÈÝʱ³äÈëHeÆø
d£®ºãѹʱ³äÈëHeÆø e£®¼°Ê±·ÖÀëNH3 f£®¼ÓÈë´ß»¯¼Á