ÒÑÖªÔÚº¬HNO3µÄÈÜÒºÖзÅÈëAl²»Éú³ÉH2£»Ä³ÎÞÉ«ÈÜÒºÀֻº¬ÓÐÏÂÁÐ11ÖÖÀë×ÓÖеļ¸ÖÖ£ºMg2£«¡¢Fe3+¡¢H£«¡¢Ag£«¡¢Ba2£«¡¢SO42£¡¢HCO3£¡¢OH£¡¢MnO4£¡¢NO3£¡¢CO32£¡£ÒÑÖª¸ÃÈÜÒºÄܸú½ðÊôÂÁ·´Ó¦£¬ÇҷųöµÄÆøÌåÖ»ÓÐÇâÆø¡£ÊԻشð£º
£¨1£©ÈôÈÜÒº¸úÂÁ·´Ó¦Ö»ÓÐAlO2£Éú³É£¬ÔòÔÈÜÒºÒ»¶¨º¬ÓеĴóÁ¿µÄÎïÖÊÊÇ________(Ìѧʽ)£¬·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_____________________________________________________£¬»¹¿ÉÄܺ¬ÓеĽ϶àµÄÎïÖÊÊÇ__________________(Ìѧʽ)¡£
£¨2£©ÈôÈÜÒº¸úÂÁ·´Ó¦ºóÓÐAl3£«Éú³É£¬ÔòÔÈÜÒºÖпÉÄÜ£¨°üÀ¨Ò»¶¨ÄÜ£©´óÁ¿´æÔÚµÄÀë×ÓÊÇ_____________¡£
£¨1£©Ba(OH)2 2Al£«2OH££«2H2O=2AlO2££«3H2¡ü Ba(NO3)2 £¨2£©H+ SO42- Mg2+
½âÎöÊÔÌâ·ÖÎö£º¸ÃÈÜÒºÄܸú½ðÊôÂÁ·´Ó¦£¬ÇҷųöµÄÆøÌåÖ»ÓÐÇâÆø£¬Õâ˵Ã÷¸ÃÈÜÒº¿ÉÄÜÏÔËáÐÔ£¬Ò²¿ÉÄÜÏÔ¼îÐÔ£¬ÓÖÒòΪ¸ÃÈÜÒºÊÇÎÞÉ«µÄ£¬ËùÒÔÒ»¶¨Ã»ÓÐFe3+¡¢MnO4££¬¾Ý´Ë¿ÉÒÔÅжϡ£
£¨1£©ÈôÈÜÒº¸úÂÁ·´Ó¦Ö»ÓÐAlO2£Éú³É£¬Õâ˵Ã÷¸ÃÒ»¶¨ÏÔ¼îÐÔ£¬ÔòMg2£«¡¢H£«¡¢Ag£«¡¢HCO3£Ò»¶¨²»ÄÜ´óÁ¿´æÔÚ¡£¸ù¾ÝÈÜÒºµÄµçÖÐÐÔ¿ÉÖª£¬Ò»¶¨º¬ÓÐBa2£«£¬ËùÒÔ¾ÍÒ»¶¨Ã»ÓÐSO42£¡¢CO32£¡£Òò´ËÔÈÜÒºÒ»¶¨º¬ÓеĴóÁ¿µÄÎïÖÊÊÇBa(OH)2¡£¶øNO3£²»ÄÜÈ·¶¨¡£ËùÒÔ»¹¿ÉÄܺ¬ÓеĽ϶àµÄÎïÖÊÊÇBa(NO3)2¡£Óйط´Ó¦µÄÀë×Ó·½³ÌʽΪ2Al£«2OH££«2H2O=2AlO2££«3H2¡ü¡£
£¨2£©ÈôÈÜÒº¸úÂÁ·´Ó¦ºóÓÐAl3£«Éú³É£¬Õâ˵Ã÷¸ÃÈÜÒºÒ»¶¨ÏÔËáÐÔ¡£ÔòHCO3£¡¢OH£¡¢CO32£Ò»¶¨²»ÄÜ´óÁ¿¹²´æ¡£ÓÖÒòΪÔÚº¬HNO3µÄÈÜÒºÖзÅÈëAl²»Éú³ÉH2£¬ËùÒÔ¸ÃÈÜÒºÖÐÒ²Ò»¶¨²»ÄÜ´óÁ¿´æÔÚNO3£¡£¸ù¾ÝÈÜÒºµÄµçÖÐÐÔ¿ÉÖª£¬ÈÜÒºÖÐÒ»¶¨´æÔÚSO42££¬ËùÒÔ¾ÍÒ»¶¨²»ÄÜ´óÁ¿´æÔÚAg£«¡¢Ba2£«£¬ËùÒÔÔÈÜÒºÖпÉÄÜ£¨°üÀ¨Ò»¶¨ÄÜ£©´óÁ¿´æÔÚµÄÀë×ÓÊÇ)H+¡¢SO42-¡¢Mg2+¡£
¿¼µã£º¿¼²éÀë×Ó¹²´æÒÔ¼°Àë×Ó¼ìÑéµÄÓйØÅжÏ
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£ºÌî¿ÕÌâ
³£¼ûµÄÎåÖÖÑÎA¡¢B¡¢C¡¢D¡¢E£¬ËüÃǵÄÑôÀë×Ó¿ÉÄÜÊÇNa+¡¢¡¢Cu2+¡¢Ba2+¡¢Al3+¡¢Ag+¡¢Fe3+£¬ÒõÀë×Ó¿ÉÄÜÊÇCl-¡¢¡£ÒÑÖª£º
¢ÙÎåÖÖÑξùÈÜÓÚË®£¬Ë®ÈÜÒº¾ùΪÎÞÉ«¡£
¢ÚDµÄÑæÉ«·´Ó¦³Ê»ÆÉ«¡£
¢ÛAµÄÈÜÒº³ÊÖÐÐÔ£¬B¡¢C¡¢EµÄÈÜÒº³ÊËáÐÔ£¬DµÄÈÜÒº³Ê¼îÐÔ¡£
¢ÜÈôÔÚÕâÎåÖÖÑεÄÈÜÒºÖзֱð¼ÓÈëBa£¨NO3£©2ÈÜÒº£¬Ö»ÓÐA¡¢CµÄÈÜÒº²»²úÉú³Áµí¡£
¢ÝÈôÔÚÕâÎåÖÖÑεÄÈÜÒºÖУ¬·Ö±ð¼ÓÈ백ˮ£¬EºÍCµÄÈÜÒºÖÐÉú³É³Áµí£¬¼ÌÐø¼Ó°±Ë®£¬CÖгÁµíÏûʧ¡£
¢Þ°ÑAµÄÈÜÒº·Ö±ð¼ÓÈëµ½B¡¢C¡¢EµÄÈÜÒºÖУ¬¾ùÄÜÉú³É²»ÈÜÓÚÏ¡ÏõËáµÄ³Áµí¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÎåÖÖÑÎÖУ¬Ò»¶¨Ã»ÓеÄÑôÀë×ÓÊÇ_________________________________________________£»
Ëùº¬µÄÒõÀë×ÓÏàͬµÄÁ½ÖÖÑεĻ¯Ñ§Ê½ÊÇ_____________________________________________¡£
£¨2£©DµÄ»¯Ñ§Ê½Îª_________£¬DÈÜÒºÏÔ¼îÐÔµÄÔÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©________________¡£
£¨3£©AºÍCµÄÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______________________________________________¡£
EºÍ°±Ë®·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______________________________________________________¡£
£¨4£©ÈôÒª¼ìÑéBÖÐËùº¬µÄÑôÀë×Ó£¬ÕýÈ·µÄʵÑé·½·¨ÊÇ__________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
äå¼°Æ仯ºÏÎï¹ã·ºÓ¦ÓÃÔÚÓлúºÏ³É¡¢»¯Ñ§·ÖÎöµÈÁìÓò¡£
£¨1£©º£Ë®Ìáäå¹ý³ÌÖÐäåÔªËصı仯ÈçÏ£º
¢Ù¹ý³Ì¢ñ£¬º£Ë®ÏÔ¼îÐÔ£¬µ÷ÆäpH£¼3.5ºó£¬ÔÙͨÈëÂÈÆø¡£
¢¡.ͨÈëÂÈÆøºó£¬·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______¡£
¢¢.µ÷º£Ë®pH¿ÉÌá¸ßCl2µÄÀûÓÃÂÊ£¬ÓÃƽºâÔÀí½âÊÍÆäÔÒòÊÇ______¡£
¢Ú¹ý³Ì¢ò£¬ÓÃÈÈ¿ÕÆø½«äå¸Ï³ö£¬ÔÙÓÃŨ̼ËáÄÆÈÜÒºÎüÊÕ¡£Íê³É²¢ÅäƽÏÂÁз½³Ìʽ¡£
Br2£«Na2CO3£½NaBrO3£«CO2£«______
¢Û¹ý³Ì¢ó£¬ÓÃÁòËáËữ¿ÉµÃBr2ºÍNa2SO4µÄ»ìºÏÈÜÒº¡£
ÏàͬÌõ¼þÏ£¬ÈôÓÃÑÎËáËữ£¬ÔòËùµÃäåµÄÖÊÁ¿¼õÉÙ£¬ÔÒòÊÇ______¡£
£¨2£©NaBrO3ÊÇÒ»ÖÖ·ÖÎöÊÔ¼Á¡£ÏòÁòËáËữµÄNaIÈÜÒºÖÐÖðµÎ¼ÓÈëNaBrO3ÈÜÒº£¬µ±¼ÓÈë2.6 mol NaBrO3ʱ£¬²âµÃ·´Ó¦ºóÈÜÒºÖÐäåºÍµâµÄ´æÔÚÐÎʽ¼°ÎïÖʵÄÁ¿·Ö±ðΪ£º
Á£×Ó | I2 | Br2 | IO3- |
ÎïÖʵÄÁ¿/mol | 0.5 | 1.3 | |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÓÐÏÂÁÐÎïÖÊ£º¢Ù ÇâÑõ»¯ÄƹÌÌ壻¢Ú ÍË¿£»¢Û ÂÈ»¯ÇâÆøÌ壻¢Ü Ï¡ÁòË᣻¢Ý ¶þÑõ»¯Ì¼ÆøÌ壻 ¢ÞÇâÑõ»¯¼ØÈÜÒº£»¢ß ̼ËáÄÆ·ÛÄ©£»¢à ÕáÌǾ§Ì壻¢á ÈÛÈÚÂÈ»¯ÄÆ£¬ÇëÓÃÐòºÅÌî¿Õ£º
£¨1£©ÉÏÊö״̬Ï¿ɵ¼µçµÄÊÇ________£»£¨2£©ÊôÓÚµç½âÖʵÄÊÇ________£»£¨3£©ÊôÓڷǵç½âÖʵÄÊÇ________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÏÖÓÐÏÂÁÐÎïÖÊ£º¢ÙNaCl¾§Ì壻¢ÚҺ̬SO2£»¢Û´¿´×Ë᣻¢ÜÁòËá±µ¡¡£»¢ÝÍ£»¢Þ¾Æ¾«£¨C2H5OH£©£»¢ßÈÛ»¯µÄKCl£»¢àNaOHÈÜÒº¡£
ÇëÓÃÒÔÉÏÎïÖʻشðÏÂÁÐÎÊÌâ¡££¨ÌîÐòºÅ£©
£¨1£©ÔÚÉÏÊö״̬ÏÂÄܵ¼µçµÄÎïÖÊÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡£»
£¨2£©ÊôÓÚÈõµç½âÖʵÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡£»
£¨3£©ÊôÓڷǵç½âÖÊ£¬µ«ÈÜÓÚË®ºóµÄË®ÈÜÒºÄܵ¼µçµÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡£»
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
¢Å°´ÒªÇóд³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ
¢ÙÇâÑõ»¯ÍÈÜÓÚÑÎËá
¢Ú̼ËáÄÆÈÜÒºÖкÍ×ãÁ¿µÄÑÎËáÈÜÒº»ìºÏ
£¨2£©ÏÖÓÐÒÔÏÂÎïÖÊ£º¢ÙNaCl¾§Ìå ¢ÚҺ̬HCl ¢ÛCaCO3¹ÌÌå ¢ÜÈÛÈÚKCl ¢ÝÕáÌÇ¢ÞÍ¢ßCO2¢àH2SO4¢áKOH¹ÌÌå ¢âË®¢ÏÑÎËá¢ÐCH3COOH ¢Ñ¾Æ¾«£¨ÌîÐòºÅ£©
¢ÙÒÔÉÏÎïÖÊÖÐÄܵ¼µçµÄÊÇ_________________________________¡£
¢ÚÒÔÉÏÎïÖÊÖÐÊôÓÚµç½âÖʵÄÊÇ_____________________________¡£
¢ÛÒÔÉÏÎïÖÊÖÐÊôÓڷǵç½âÖʵÄÊÇ___________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ÏÖÓÐÏÂÁÐÆßÖÖÎïÖÊ:
¢ÙÂÁ ¢ÚÕáÌÇ ¢ÛCO2 ¢ÜH2SO4 ¢ÝBa(OH)2 ¢ÞºìºÖÉ«µÄÇâÑõ»¯Ìú½ºÌå ¢ßHCl
£¨1£©ÉÏÊöÎïÖÊÖÐÊôÓÚµç½âÖʵÄÓÐ ¡£(ÌîÐòºÅ)
£¨2£©Ïò¢ÞµÄÈÜÒºÖÐÖ𽥵μӢߵÄÈÜÒº,¿´µ½µÄÏÖÏóÊÇ ¡£
£¨3£©ÉÏÊöÎïÖÊÖÐÓÐÁ½ÖÖÎïÖÊÔÚË®ÈÜÒºÖз¢Éú·´Ó¦,ÆäÀë×Ó·½³ÌʽΪ:H++OH-=H2O,Ôò¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
°´ÒªÇóÍê³ÉÏÂÁÐÌî¿Õ
(l)д³öÏÂÁÐÎïÖʵĵçÀë·½³Ìʽ£º
Fe2(SO4)3__________________________________________________£¬
NaHCO3______________________________________________________¡£
(2)д³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ£º
Ï¡ÑÎËáÓë̼Ëá¸Æ·´Ó¦____________________________________________£¬
ÇâÑõ»¯±µÈÜÒºÓëÏ¡ÁòËá·´Ó¦_______________________________________¡£
(3)д³öÓëÏÂÁÐÀë×Ó·½³ÌʽÏà¶ÔÓ¦µÄ»¯Ñ§·½³Ìʽ£º
H£«£«OH£=H2O __________________________________________£¬
CO32££«2H£«=CO2¡ü£«H2O__________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
ijNa2CO3¡¢NaAlO2µÄ»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈë1 mol¡¤L-1µÄÑÎËá,²âµÃÈÜÒºÖеÄC¡¢HC¡¢Al¡¢Al3+µÄÎïÖʵÄÁ¿Óë¼ÓÈëÑÎËáµÄÌå»ý±ä»¯¹ØϵÈçͼËùʾ,ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©
A£®Ô»ìºÏÈÜÒºÖеÄCÓëAlµÄÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã2 |
B£®V1¡ÃV2=1¡Ã5 |
C£®MµãʱÉú³ÉµÄCO2Ϊ0.05 mol |
D£®aÏß±íʾµÄÀë×Ó·½³ÌʽΪ:Al+H++H2OAl(OH)3¡ý |
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