(12·Ö)ÖÐѧ»¯Ñ§Öг£¼ûµÄ¼¸ÖÖÎïÖÊ´æÔÚÈçϹØϵ£¬ÆäÖм×ÊǺÚÉ«·Ç½ðÊôµ¥ÖÊ£¬ÒÒÊÇÉú»îÖг£¼ûµÄ½ðÊôµ¥ÖÊ£¬CÔÚ³£ÎÂÏÂΪÎÞÉ«µÄÒºÌ壬DÊǺì×ØÉ«ÆøÌå¡££¨Í¼Öв¿·Ö²úÎïºÍ·´Ó¦Ìõ¼þÒÑÂÔÈ¥£©
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)д³ö¼×ÓëAµÄŨÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ_________________________________¡£
(2)CµÄµç×ÓʽÊÇ___________________________¡£
(3)³ýÈ¥G¹ÌÌåÖк¬ÓÐHÔÓÖʲÉÓõķ½·¨ÊÇ_____________________¡£
(4)AÈÜÒºÓëÒ»ÖÖÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌå·´Ó¦£¬Éú³ÉÒ»ÖÖÑΣ¬¸ÃÑεÄÈÜÒº³ÊËáÐÔ£¬ÆäÔÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©_____________________________¡£
(5)д³öÓÉFת»¯ÎªEµÄÀë×Ó·½³Ìʽ____________________________________£»
(6)д³öÓÉGת»¯ÎªHµÄÀë×Ó·½³Ìʽ____________________________________¡£
£¨12·Ö£©£¨1£©C+4HNO3£¨Å¨£©CO2¡ü+4NO2¡ü+2H2O £¨2·Ö£©
£¨2£©£¨2·Ö£© £¨3£©¼ÓÈÈ£¨2·Ö£©
£¨4£©NH+ H2ONH3¡¤H2O + H+ £¨2·Ö£©
£¨5£©2Fe3+ £«Fe3Fe2+£¨2·Ö£©
£¨6£©CO32- +H2O +CO2=2HCO3-£¨2·Ö£©
¡¾½âÎö¡¿×ۺϸ÷ÎïÖʵÄÑÕɫ״̬¼°¸÷ÎïÖʵÄת»¯¹Øϵ£¬¿ÉÅж¨¼×Ϊ̼¡¢AΪÏõË᣺
C+4HNO3£¨Å¨£©CO2¡ü+4NO2¡ü+2H2O £¨2·Ö£©ÆäÖеÄDΪNO2¡¢CΪ£È2£Ï¡¢BΪCO2£»
ÏõËáºÍ½ðÊôÒҵIJúÎïÓëÒÒµÄÄËÊǶàÉÙÓйأ¬¿ÉÖªÒÒΪ±ä¼Û½ðÊô£¬¿É²Â¶¨ÎªÌú¡£
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
£¨12 ·Ö£©£¨Ã¿¿Õ2·Ö£©CH3COOHÊÇÖÐѧ»¯Ñ§Öг£ÓõÄÒ»ÔªÈõËᣬÇë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Èô·Ö±ð½«pH=2µÄÑÎËáºÍ´×ËáÏ¡ÊÍ100±¶£¬ÔòÏ¡ÊͺóÈÜÒºµÄpH£ºÑÎËá ´×ËᣨÌî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©¡£
£¨2£©½«100mL0.1mol¡¤L-1µÄCH3COOHÈÜÒºÓë50mL 0.2mol¡¤L-1µÄNaOHÈÜÒº»ìºÏ£¬ËùµÃÈÜÒº³Ê ÐÔ£¬ÔÒò £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£
£¨3£©ÒÑ֪ij»ìºÏÈÜÒºÖÐÖ»º¬ÓÐCH3COO-¡¢H+¡¢Na+¡¢OH-ËÄÖÖÀë×Ó£¬ÇÒÀë×ÓŨ¶È´óС¹ØϵΪ£ºc(CH3COO-)>c(H+)> c(Na+)> c(OH-)£¬Ôò¸ÃÈÜÒºÖк¬ÓеÄÈÜÖÊΪ ¡£
£¨4£©ÒÑÖªKa(CH3COOH)= 1.76¡Á10-5£¬Ka(HNO2)= 4.6¡Á10-4£¬ÈôÓÃͬŨ¶ÈµÄNaOHÈÜÒº·Ö±ðÖк͵ÈÌå»ýÇÒpHÏàµÈµÄCH3COOHºÍHNO2£¬ÔòÏûºÄNaOHÈÜÒºµÄÌå»ý¹ØϵΪ£º
Ç°Õß ºóÕߣ¨Ìî¡°>£¬<»ò=¡±£©
£¨5£©ÒÑÖª25¡æʱ£¬0.1mol¡¤L-1´×ËáÈÜÒºµÄpHԼΪ3£¬ÏòÆäÖмÓÈëÉÙÁ¿´×ËáÄƾ§Ì壬·¢ÏÖÈÜÒºµÄpHÔö´ó¡£¶ÔÉÏÊöÏÖÏóÓÐÁ½ÖÖ²»Í¬µÄ½âÊÍ£º¼×ͬѧÈÏΪ´×ËáÄƳʼîÐÔ£¬ËùÒÔÈÜÒºµÄpHÔö´ó£»ÒÒͬѧ¸ø³öÁíÍâÒ»ÖÖ²»Í¬ÓÚ¼×ͬѧµÄ½âÊÍ£¬ÇëÄãд³öÒÒͬѧ¿ÉÄܵÄÀíÓÉ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìÄÚÃɹŰÍÑåÄ׶ûÊÐÒ»ÖиßÈý9ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ
(12·Ö)A¡¢B¡¢C¡¢D¾ùΪÖÐѧ»¯Ñ§Öг£¼ûµÄÎïÖÊ£¬ËüÃÇÖ®¼äµÄת»¯¹ØϵÈçÏÂͼ(²¿·Ö²úÎïÒÑÂÔÈ¥)£º
ÊԻشð£º
£¨1£©ÈôDÊǾßÓÐÑõ»¯ÐԵĵ¥ÖÊ£¬AÔªËØÊôÓÚÖ÷×å½ðÊôÔªËØ£¬ÔòAΪ________(ÌîÔªËØ·ûºÅ)¡£
£¨2£©ÈôDÊǽðÊô£¬CÈÜÒºÔÚ´¢´æʱӦ¼ÓÈëÉÙÁ¿D£¬ÆäÀíÓÉÊÇ(ÓñØÒªµÄÎÄ×ÖºÍÀë×Ó·½³Ìʽ±íʾ)____________________________________________________£»DÔÚ³±ÊªµÄ¿ÕÆøÖÐÒ×·¢ÉúÎüÑõ¸¯Ê´£¬Ð´³ö¸¯Ê´Ê±Ôµç³ØÕý¼«µÄµç¼«·´Ó¦Ê½£º ¡£
£¨3£©ÈôA¡¢B¡¢CΪº¬Í¬Ò»ÖÖ½ðÊôÔªËصÄÎÞ»ú»¯ºÏÎÔÚÈÜÒºÖÐAºÍC·´Ó¦Éú³ÉB¡£Çëд³öBת»¯ÎªCµÄËùÓпÉÄܵÄÀë×Ó·½³Ìʽ£º
______________________________¡¢_________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì¹ãÎ÷¹ðÁÖÖÐѧ¸ßÈý8ÔÂÔ¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ
(12·Ö)ÖÐѧ»¯Ñ§Öг£¼ûµÄ¼¸ÖÖÎïÖÊ´æÔÚÈçϹØϵ£¬ÆäÖм×ÊǺÚÉ«·Ç½ðÊôµ¥ÖÊ£¬ÒÒÊÇÉú»îÖг£¼ûµÄ½ðÊôµ¥ÖÊ£¬CÔÚ³£ÎÂÏÂΪÎÞÉ«µÄÒºÌ壬DÊǺì×ØÉ«ÆøÌå¡££¨Í¼Öв¿·Ö²úÎïºÍ·´Ó¦Ìõ¼þÒÑÂÔÈ¥£©
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)д³ö¼×ÓëAµÄŨÈÜÒº·´Ó¦µÄ»¯Ñ§·½³Ìʽ_________________________________¡£
(2)CµÄµç×ÓʽÊÇ___________________________¡£
(3)³ýÈ¥G¹ÌÌåÖк¬ÓÐHÔÓÖʲÉÓõķ½·¨ÊÇ_____________________¡£
(4)AÈÜÒºÓëÒ»ÖÖÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶µÄÆøÌå·´Ó¦£¬Éú³ÉÒ»ÖÖÑΣ¬¸ÃÑεÄÈÜÒº³ÊËáÐÔ£¬ÆäÔÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©_____________________________¡£
(5)д³öÓÉFת»¯ÎªEµÄÀë×Ó·½³Ìʽ____________________________________£»
(6)д³öÓÉGת»¯ÎªHµÄÀë×Ó·½³Ìʽ____________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìÕã½Ê¡º¼ÖÝÊиßÈýÉÏѧÆÚÆÚÖÐÆßУÁª¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ
(12·Ö)ÏÂͼÖУ¬Aµ½LΪ³£¼ûÎïÖÊ»ò¸ÃÎïÖʵÄË®ÈÜÒº£¬BÔÚAÆøÌåÖÐȼÉÕ²úÉú×ØÉ«ÑÌ£¬B¡¢GΪÖÐѧ»¯Ñ§Öг£¼û½ðÊôµ¥ÖÊ£¬IµÄÑæÉ«·´Ó¦Îª»ÆÉ«£¬×é³ÉJµÄÔªËØÔ×ÓºËÄÚÖ»ÓÐÒ»¸öÖÊ×Ó£¬FΪÎÞÉ«¡¢Óд̼¤ÐÔÆøζÆøÌ壬ÇÒÄÜʹƷºìÈÜÒºÍÊÉ«¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)KËùº¬µÄ»¯Ñ§¼üÓÐ ¡£
(2)DµÄË®ÈÜÒºÓëG·´Ó¦µÄ×ÜÀë×Ó·½³ÌʽΪ_______________________________________________¡£
(3)д³ö¹¤ÒµÉÏÖÆÈ¡µ¥ÖÊGµÄ»¯Ñ§·½³Ìʽ ¡£
(4)½«ÆøÌåFͨÈëÏÂÁÐ×°ÖÃÖÐ
д³öA¡¢BÖеÄÀë×Ó·½³Ìʽ£º
¡¢ ¡£
(5)Èô½«FͨÈëÒ»¶¨Á¿KµÄË®ÈÜÒºÖУ¬ÔòËùµÃÈÜÒºÖи÷Àë×ÓŨ¶ÈÒ»¶¨Âú×ãµÄ¹ØϵʽΪ
¡£
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