¡¾ÌâÄ¿¡¿ÃºÈ¼ÉÕÅŷŵÄÑÌÆøº¬ÓÐSO2ºÍNOx£¬²ÉÓÃNaClO2ÈÜÒº×÷ΪÎüÊÕ¼Á¿Éͬʱ¶ÔÑÌÆø½øÐÐÍÑÁò¡¢ÍÑÏõ¡£Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©ÉÏÊöÑÌÆø´¦Àí¹ý³ÌÖÐÉæ¼°µ½µÄ»¯Ñ§ÎïÖÊ×é³ÉÔªËØÖУ¬ÊôÓÚµÚÈýÖÜÆÚÔªËصÄÊÇ___£»Ð´³öNµÄºËÍâµç×ÓÅŲ¼Ê½___¡£
£¨2£©ÒÑÖªSO2·Ö×ӵĿռ乹ÐÍΪÕÛÏßÐΣ¬ÔòSO2Ϊ___£¨Ñ¡Ìî¡°¼«ÐÔ¡±¡¢¡°·Ç¼«ÐÔ¡±£©·Ö×Ó¡£
£¨3£©½«º¬ÓÐSO2ºÍNOxµÄÑÌÆøͨÈëÊ¢ÓÐNaClO2ÈÜÒºµÄ·´Ó¦Æ÷ÖУ¬·´Ó¦Ò»¶ÎʱÎʺ󣬲âµÃÈÜÒºÖÐÀë×ÓŨ¶ÈµÄÓйØÊý¾ÝÈçÏ£¨ÆäËûÀë×ÓºöÂÔ²»¼Æ£©£º
Àë×Ó | Na+ | SO42- | NO3- | OH- | Cl- |
Ũ¶È/£¨mol¡¤L-1£© | 5.5¡Á10-3 | 8.5¡Á10-4 | y | 2.0¡Á10-4 | 3.4¡Á10-3 |
¢Ù·´Ó¦ºóÈÜÒºpH___7£¬±íÖÐy=___mol¡¤L-1¡£
¢Úд³öNaClO2ÈÜÒºÎüÊÕSO2µÄÀë×Ó·½³Ìʽ___¡£
£¨4£©ÑÌÆøÖеÄSO2»¹¿É²ÉÓð±·¨ÍÑÁò³ýÈ¥£¬Æä·´Ó¦ÔÀí¿ÉÓÃÈçͼ±íʾ¡£
¢Ùд³öSO2¸ú°±Ë®·´Ó¦Éú³ÉNH4HSO3µÄ»¯Ñ§·½³Ìʽ___¡£
¢Ú(NH4)2SO4ÈÜÒºÖÐŨ¶È×î´óµÄÀë×ÓÊÇ___¡£
¡¾´ð°¸¡¿Na¡¢S¡¢Cl 1s22s22p3 ¼«ÐÔ £¾ 2.0¡Á10-4mol¡¤L-1 ClO2-+2SO2+4OH-¡ú2SO42-+Cl-+2H2O SO2+NH3¡¤H2O¡úNH4HSO3 NH4+
¡¾½âÎö¡¿
£¨1£©NÔ×ÓºËÍâÓÐ7¸öµç×Ó£¬¸ù¾ÝÄÜÁ¿×îµÍÔÀíÅŲ¼£»
£¨2£©SO2·Ö×ӵĿռ乹ÐÍΪÕÛÏßÐΣ¬ËùÒÔÕý¸ºµçºÉµÄÖØÐIJ»Öغϣ»
£¨3£©¢Ù¸ù¾Ý·´Ó¦ºóÈÜÒºÖÐÇâÀë×ÓµÄŨ¶ÈÅжÏpH£»¸ù¾ÝµçºÉÊؼÆËãyÖµ£»
¢ÚSO2±»NaClO2ÈÜÒºÑõ»¯ÎªSO42-£¬NaClO2±»»¹ÔΪCl-£»
£¨4£©¢ÙSO2¸ú°±Ë®1:1·´Ó¦Éú³ÉNH4HSO3£»
¢Ú(NH4)2SO4ÊÇÇ¿µç½âÖÊ£¬ÍêÈ«µçÀ룬 c(NH4+)½Ó½ü c(SO42-)µÄ¶þ±¶¡£
£¨1£©¸ù¾ÝÔªËØÖÜÆÚ±í£¬ÊôÓÚµÚÈýÖÜÆÚÔªËصÄÊÇNa¡¢S¡¢Cl£»NÔ×ÓºËÍâÓÐ7¸öµç×Ó£¬ºËÍâµç×ÓÅŲ¼Ê½ÊÇ1s22s22p3£»
£¨2£©SO2·Ö×ӵĿռ乹ÐÍΪÕÛÏßÐΣ¬ËùÒÔÕý¸ºµçºÉµÄÖØÐIJ»Öغϣ¬SO2Ϊ¼«ÐÔ·Ö×Ó£»
£¨3£©¢Ù·´Ó¦ºóÈÜÒºÖÐÇâÑõ¸ùÀë×ÓµÄŨ¶ÈÊÇ2.0¡Á10-4£¬ÇâÀë×ÓŨ¶ÈÊÇ5.0¡Á10-11£¬ÇâÀë×ÓŨ¶ÈСÓÚÇâÑõ¸ùÀë×ÓµÄŨ¶È£¬ËùÒÔÈÜÒº³Ê¼îÐÔ£¬pH£¾7£»¸ù¾ÝµçºÉÊغã5.5¡Á10-3+5.0¡Á10-11=(8.5¡Á10-4)¡Á2+y+2.0¡Á10-4+3.4¡Á10-3£¬½âµÃy=2.0¡Á10-4mol¡¤L-1£»
¢ÚSO2±»NaClO2ÈÜÒºÑõ»¯ÎªSO42-£¬NaClO2±»»¹ÔΪCl-£¬·´Ó¦µÄÀë×Ó·½³ÌʽÊÇClO2-+2SO2+4OH-¡ú2SO42-+Cl-+2H2O£»
£¨4£©¢ÙSO2¸ú°±Ë®1:1·´Ó¦Éú³ÉNH4HSO3£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇSO2+NH3¡¤H2O¡úNH4HSO3£»
¢Ú(NH4)2SO4ÊÇÇ¿µç½âÖÊ£¬ÍêÈ«µçÀ룬 c(NH4+)½Ó½ü c(SO42-)µÄ¶þ±¶£¬ËùÒÔŨ¶È×î´óµÄÀë×ÓÊÇNH4+¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿¶¬¼¾µÄ½µÑ©¸ø½»Í¨´øÀ´ÁËÖî¶à²»±ã£¬ÆäÖд×Ëá¼Ø£¨CH3COOK£©ÊÇÈÚѩЧ¹û×îºÃµÄÈÚÑ©¼Á£¬ÏÂÁйØÓÚ´×Ëá¼Ø»ò´×ËáµÄ˵·¨ÕýÈ·µÄÊÇ£¨ £©
A.1molCH3COOKµÄÖÊÁ¿Îª98g/mol
B.CH3COOHĦ¶ûÖÊÁ¿¾ÍÊÇËüµÄÏà¶Ô·Ö×ÓÖÊÁ¿
C.Ò»¸öCH3COOHµÄÖÊÁ¿Ô¼Îªg
D.º¬ÓÐ6.02¡Á1023¸ö̼Ô×ÓµÄCH3COOKµÄÎïÖʵÄÁ¿ÊÇ1mol
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿I£®Ä³»¯Ñ§ÊµÑéС×éΪÁË̽¾¿ÔÚʵÑéÊÒÖƱ¸Cl2µÄ¹ý³ÌÖÐÓÐË®ÕôÆøºÍHCl»Ó·¢³öÀ´£¬Í¬Ê±Ö¤Ã÷ÂÈÆøµÄijЩÐÔÖÊ£¬Ä³Í¬Ñ§Éè¼ÆÁËÈçͼËùʾµÄʵÑé×°Öã¨ÂÈÆøÒ×ÈÜÓÚCCl4£¬HCl²»ÈÜÓÚCCl4)¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)×°ÖÃAÖÐÁ¬½ÓÏðƤ¹ÜµÄÄ¿µÄÊÇ_____________________________________¡£
(2)×°ÖÃAÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________________________£¬×°ÖÃBÖÐÊ¢·ÅµÄÊÔ¼ÁÊÇ___________¡£
(3)×°ÖÃDÓëEÖгöÏֵIJ»Í¬ÏÖÏó˵Ã÷µÄÎÊÌâÊÇ_________________________¡£
(4)ÓÐͬѧ»ùÓÚʵÑéµÄÑϽ÷ÐÔ¿¼ÂÇ£¬ÈÏΪ¿ÉÔÚF¡¢GÁ½¸ö×°ÖÃÖ®¼äÔÙ¼ÓÒ»¸ö×°ÓÐʪÈóµÄµí·ÛKIÊÔÖ½µÄ×°Öã¬ÆäÄ¿µÄÊÇ___________________________________¡£
¢ò£®Ä³Ñо¿ÐÔѧϰС×éÓûÖƱ¸Æ¯°×·Û£¬Éè¼ÆÔÚÉÏÊö×°ÖÃAºóÁ¬½ÓÈçÓÒͼËùʾµÄ×éºÏ×°Öãº
(5)×°ÖâÚÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌΪ___________________________________¡£
(6)ÖÆƯ°×·ÛµÄ·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬·´Ó¦Î¶ȽϸßʱÓи±·´Ó¦·¢Éú£¬¸Ä½ø¸ÃʵÑé×°ÖÃÒÔ¼õÉÙ¸±·´Ó¦·¢ÉúµÄ·½·¨ÊÇ___________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿(1)µÚÈýÖÜÆÚ8ÖÖÔªËØ°´µ¥ÖÊÈÛµã¸ßµÍµÄ˳ÐòÈçͼ£¬ÆäÖÐÐòºÅ¡°8¡±´ú±í___(ÌîÔªËØ·ûºÅ)£»ÆäÖе縺ÐÔ×î´óµÄÊÇ___£¨ÌîͼÖеÄÐòºÅ£©¡£
(2)Çëд³ö±ÈÔªËØ1µÄÔ×ÓÐòÊý´ó8µÄÔªËصĻù̬Ô×Óµç×ÓÅŲ¼Ê½___¡£
(3)ÔªËØ7µÄµ¥Öʾ§ÌåÖÐÔ×ӵĶѻý·½Ê½Èçͼ¼×Ëùʾ£¬Æ侧°ûÌØÕ÷ÈçͼÒÒËùʾ£¬Ô×ÓÖ®¼äÏ໥λÖùØϵµÄƽÃæͼÈçͼ±ûËùʾ¡£
ÈôÒÑÖª7µÄÔ×Ӱ뾶Ϊd cm£¬NA´ú±í°¢·ü¼ÓµÂÂÞ³£Êý£¬7µÄÏà¶ÔÔ×ÓÖÊÁ¿ÎªM£¬Çë»Ø´ð£º
¢Ù¾§°ûÖÐ7Ô×ÓµÄÅäλÊýΪ___£¬Ò»¸ö¾§°ûÖÐ7Ô×ÓµÄÊýĿΪ___£»
¢Ú¸Ã¾§ÌåµÄÃܶÈΪ___ g/cm3(ÓÃ×Öĸ±íʾ)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÒÑÖªX¡¢YºÍZÈýÖÖÔªËصÄÔ×ÓÐòÊýÖ®ºÍµÈÓÚ42¡£XÔªËØÔ×ÓµÄ4p¹ìµÀÉÏÓÐ3¸öδ³É¶Ôµç×Ó£¬YÔªËØÔ×ÓµÄ×îÍâ²ã2p¹ìµÀÉÏÓÐ2¸öδ³É¶Ôµç×Ó¡£X¸úY¿ÉÐγɻ¯ºÏÎïX2Y3£¬ZÔªËØ¿ÉÒÔÐγɸºÒ»¼ÛÀë×Ó¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)XÔªËØÔ×Ó»ù̬ʱµÄµç×ÓÅŲ¼Ê½Îª_______________£»
(2)YÔªËØÔ×ӵļ۲ãµç×ӵĹìµÀ±íʾʽΪ_____________£»
(3)XÓëZ¿ÉÐγɻ¯ºÏÎïXZ3£¬¸Ã»¯ºÏÎïµÄ¿Õ¼ä¹¹ÐÍΪ________£»
(4)ÒÑÖª»¯ºÏÎïX2Y3ÔÚÏ¡ÁòËáÈÜÒºÖпɱ»½ðÊôп»¹ÔΪXZ3£¬²úÎﻹÓÐZnSO4ºÍH2O£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ___________£»
(5)±È½ÏXµÄÇ⻯ÎïÓëͬ×åµÚ¶þ¡¢µÚÈýÖÜÆÚÔªËØËùÐγɵÄÇ⻯Îï·Ðµã¸ßµÍ______¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿»¯ºÏÎïE£¨¶þÒÒËá-1£¬4-»·¼º¶þ´¼õ¥£©ÊÇÒ»ÖÖÖÆ×÷½¨Öþ²ÄÁϵÄÔÁÏ¡£ÆäºÏ³É·ÏßÈçÏ£º
Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©AÖк¬ÓеĹÙÄÜÍÅÊÇ______£»EµÄ·Ö×ÓʽÊÇ______£»ÊÔ¼ÁaÊÇ______¡£
£¨2£©Ð´³ö·´Ó¦ÀàÐÍ£ºB¡úC______¡£
£¨3£©CÓëD·´Ó¦Éú³ÉEµÄ»¯Ñ§·½³Ìʽ£º______¡£
£¨4£©CµÄͬ·ÖÒì¹¹Ì壬ÄÜʹʯÈïÊÔ¼ÁÏÔºìÉ«£¬Ð´³ö¸ÃÎïÖʵÄÒ»Öֽṹ¼òʽ______¡£
£¨5£©Éè¼ÆÒ»ÌõÒÔ»·¼º´¼£¨£©ÎªÔÁÏ£¨ÆäËûÎÞ»úÊÔ¼ÁÈÎÈ¡£©ºÏ³ÉAµÄºÏ³É·Ïß¡£
£¨ºÏ³É·Ïß³£Óõıíʾ·½Ê½Îª£ºAB¡¡Ä¿±ê²úÎ_____________________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿ÎªÌ½¾¿Ä³³ÈÉ«º¬½á¾§Ë®µÄÑÎXµÄ×é³ÉºÍÐÔÖÊ£¬Éè¼Æ²¢Íê³ÉÈçͼʵÑé¡£Çë»Ø´ð£º
£¨1£©XµÄ»¯Ñ§Ê½ÊÇ___¡£
£¨2£©ÒÒ»¯Ñ§ÐÔÖÊÓëÆä×é³ÉÔªËصĵ¥ÖÊÏàËÆ£¬ÏÂÁÐÎïÖÊÖÐÄÜÓëÒÒ·¢Éú·´Ó¦µÄÊÇ__£¨ÌîдÏàÓ¦µÄ×Öĸ£©¡£
A.Mg B.CaCl2 C.NaOH D.K2SO4
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿(1)ÔÚ¢ÙLi¡¡¢Úʯī¡¡¢ÛC60¡¡¢ÜMg ¢Ý CH3CH2OH ¢ÞC ¢ßLi ¢à CH3OCH3 ÖУº____»¥ÎªÍ¬Î»ËØ£» ____»¥ÎªÍ¬·ÖÒì¹¹Ì壻___»¥ÎªÍ¬ËØÒìÐÎÌå(ÌîÐòºÅ)
(2)ÏÖÓТÙCaCl2 ¢Ú½ð¸Õʯ ¢ÛNH4Cl ¢ÜNa2SO4 ¢Ý±ù µÈÎåÖÖÎïÖÊ£¬°´ÏÂÁÐÒªÇó»Ø´ð£º
¢ÙÈÛ»¯Ê±²»ÐèÒªÆÆ»µ»¯Ñ§¼üµÄÊÇ___________£¬ÈÛµã×î¸ßµÄÊÇ_______¡£(ÌîÐòºÅ)
¢ÚÖ»º¬ÓÐÀë×Ó¼üµÄÎïÖÊÊÇ______£¬¾§ÌåÒÔ·Ö×Ó¼ä×÷ÓÃÁ¦½áºÏµÄÊÇ______¡£(ÌîÐòºÅ)
(3)д³öÏÂÁÐÎïÖʵĵç×Óʽ
¢ÙH2O______
¢ÚNaOH______
¢ÛNH3______
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¡¾ÌâÄ¿¡¿Óá°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±Ìî¿Õ£º
£¨1£©Í¬ÎÂͬѹÏ£¬H2(g)+Cl2(g)£½2HCl(g)£¬¹âÕպ͵ãȼÌõ¼þµÄ¦¤H£¨»¯Ñ§¼ÆÁ¿ÊýÏàͬ£©·Ö±ðΪ¦¤H1¡¢¦¤H2£¬¦¤H1______¦¤H2£»
£¨2£©ÏàͬÌõ¼þÏ£¬2molÇâÔ×ÓËù¾ßÓеÄÄÜÁ¿____1molÇâ·Ö×ÓËù¾ßÓеÄÄÜÁ¿£»
£¨3£©ÒÑÖª³£ÎÂʱºìÁױȰ×Á×Îȶ¨£¬±È½ÏÏÂÁз´Ó¦ÖЦ¤HµÄ´óС£º¦¤H1_____¦¤H2¡£
¢Ù4P(°×Á×£¬s) +5O2(g)£½2P2O5(s) ¦¤H1£¬¢Ú4P(ºìÁ×£¬s)+5O2(g)£½2P2O5(s) ¦¤H2£»
£¨4£©ÒÑÖª£º101 kPaʱ£¬2C(s) +O2(g)£½2CO(g) ¦¤H£½£221kJ¡¤mol£1£¬Ôò̼µÄȼÉÕÈÈÊýÖµ____110.5 kJ¡¤mol£1£»
£¨5£©ÒÑÖª£ºÏ¡ÈÜÒºÖУ¬H£«(aq)+OH£(aq)£½H2O(l) ¦¤H£½£57.3kJ/mol£¬ÔòŨÁòËáÓëÏ¡NaOHÈÜÒº·´Ó¦Éú³É1 molË®£¬·Å³öµÄÈÈÁ¿________57.3 kJ£»
£¨6£©¿ÉÄæ·´Ó¦£ºaA(Æø)+bB(Æø)cC(Æø)+dD(Æø)£»¦¤H£½Q£¬¸ù¾Ýͼ»Ø´ð£º
P1______ P2£»¢Ú£¨a+b£©______£¨c+d£©£»¢Ût1¡æ______ t2¡æ¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com