ÁòËṤҵβÆøÖжþÑõ»¯ÁòµÄº¬Á¿³¬¹ý 0.05%(Ìå»ý·ÖÊý)ʱÐè¾´¦Àíºó²ÅÄÜÅÅ·Å¡£Ä³Ð£ÐËȤС×éÓû²â¶¨ÁòËṤҵβÆøÖжþÑõ»¯ÁòµÄº¬Á¿£¬²ÉÓÃÒÔÏ·½°¸£º
¼×·½°¸£ºÈçÏÂͼËùʾ£¬Í¼ÖÐÆøÌåÁ÷Á¿¼ÆBÓÃÓÚ׼ȷ²âÁ¿Í¨¹ýβÆøµÄÌå»ý¡£½«Î²ÆøͨÈëÒ»¶¨Ìå»ýÒÑ֪Ũ¶ÈµÄµâË®ÖвⶨSO2µÄº¬Á¿¡£µ±Ï´ÆøÆ¿CÖÐÈÜÒºÀ¶É«Ïûʧʱ£¬Á¢¼´¹Ø±Õ»îÈûA¡£
(1)Ï´ÆøÆ¿CÖе¼¹ÜÄ©¶ËÁ¬½ÓÒ»¸ö¶à¿×ÇòÅÝD£¬¿ÉÒÔÌá¸ßʵÑéµÄ׼ȷ¶È£¬ÆäÀíÓÉÊÇ________________________________________________________________________
________________________________________________________________________¡£
(2)Ï´ÆøÆ¿CÖеÄÈÜÒº¿ÉÒÔÓÃÆäËûÊÔ¼ÁÌæ´ú£¬ÇëÄã¾Ù³öÒ»ÖÖ£º__________________¡£
(3)Ï´ÆøÆ¿CÖÐÈÜÒºÀ¶É«Ïûʧºó£¬Ã»Óм°Ê±¹Ø±Õ»îÈûA£¬²âµÃµÄSO2º¬Á¿__________(Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족)¡£
ÒÒ·½°¸£ºÊµÑé²½ÖèÈçÏÂÃæÁ÷³ÌͼËùʾ£º
(4)д³ö²½Öè¢ÚÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ________________________________________¡£
(5)²½Öè¢ÛÖÐÏ´µÓ³ÁµíµÄ·½·¨ÊÇ____________________________________________¡£
(6)ͨ¹ýµÄβÆøÌå»ýΪV L(ÒÑ»»Ëã³É±ê×¼×´¿ö)ʱ£¬¸ÃβÆøÖжþÑõ»¯ÁòµÄº¬Á¿(Ìå»ý·ÖÊý)Ϊ______________(Óú¬ÓÐV¡¢mµÄ´úÊýʽ±íʾ)¡£
±û·½°¸£º½«ÒÒ·½°¸Öв½Öè¢ÙÊ¡ÂÔ£¬Ö±½Ó½«Î²ÆøͨÈë¹ýÁ¿Ba(OH)2ÈÜÒºÖУ¬ÆäÓಽÖèÓëÒÒ·½°¸Ïàͬ¡£
(7)ÄãÈÏΪ±û·½°¸ÊÇ·ñºÏÀí£¬ËµÃ÷ÀíÓÉ£º______________________________________
________________________________________________________________________¡£
´ð°¸¡¡(1)Ôö´óÆøÌåÓëÈÜÒºµÄ½Ó´¥Ãæ»ý£¬ÓÐÀûÓÚSO2ÓëµâË®³ä·Ö·´Ó¦
(2)ËáÐÔ¸ßÃÌËá¼ØÈÜÒº(»òÆäËûºÏÀí´ð°¸)
(3)Æ«µÍ
(4)H2SO4£«Ba(OH)2===BaSO4¡ý£«2H2O
(5)Ïò©¶·Àï×¢ÈëÕôÁóË®£¬Ê¹Ë®Ãæû¹ý³ÁµíÎ´ýË®Á÷¾¡ºó£¬Öظ´²Ù×÷2¡«3´Î
(6)(»ò%)
(7)²»ºÏÀí¡£ÑÇÁòËá±µÈÝÒ×±»(ÑõÆø)²¿·ÖÑõ»¯ÎªÁòËá±µ
½âÎö¡¡(1)¼×·½°¸ÀûÓõÄÔÀíΪSO2£«I2£«2H2O===H2SO4£«2HI¡£Ï´ÆøÆ¿CÖе¼¹ÜÄ©¶ËÁ¬½ÓÒ»¸ö¶à¿×ÇòÅÝD£¬¿ÉÒÔÔö´óSO2ÓëµâË®µÄ½Ó´¥Ãæ»ý£¬Ê¹SO2ºÍµâË®³ä·Ö·´Ó¦¡£(2)Ï´ÆøÆ¿CÖеÄÈÜÒº»¹¿ÉÒÔÓÃËáÐÔ¸ßÃÌËá¼ØÈÜÒº¡¢äåË®µÈ´úÌæ¡£(3)Ï´ÆøÆ¿CÖÐÈÜÒºÀ¶É«Ïûʧºó£¬Ã»Óм°Ê±¹Ø±Õ»îÈûA£¬ÔòͨÈëβÆøµÄÌå»ýÔö´ó£¬Òò´ËSO2º¬Á¿Æ«µÍ¡£(4)ÒÒ·½°¸ÀûÓõÄÔÀíΪSO2£«H2O2===H2SO4£¬H2SO4£«Ba(OH)2===BaSO4¡ý£«2H2O¡£²½Öè¢ÚÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪH2SO4£«Ba(OH)2===BaSO4¡ý£«2H2O¡£(5)²½Öè¢ÛÖÐÏ´µÓ³ÁµíʱÐèÒªÖظ´Ï´µÓ2¡«3´Î¡£(6)¸ù¾ÝSO2¡«BaSO4£¬ÔòβÆøÖжþÑõ»¯ÁòµÄÌå»ýΪ¡Á22.4 L£¬¹ÊβÆøÖжþÑõ»¯ÁòµÄº¬Á¿Îª¡Á100%£½%¡£(7)±û·½°¸ÀûÓõÄÔÀíÊÇSO2£«Ba(OH)2===BaSO3¡ý£«H2O£¬ÓÉÓÚBaSO3ÈÝÒ×±»¿ÕÆøÖеÄÑõÆøÑõ»¯ÎªBaSO4£¬Òò´ËBaSO3µÄÖÊÁ¿mµÄÊýÖµ²»×¼È·£¬¹Ê¸Ã·½°¸²»ºÏÀí¡£
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
ij»¯Ñ§ÐËȤС×éÓÃÂÁƬÓëÏ¡ÁòËá·´Ó¦ÖÆÈ¡ÇâÆø£¬ÒÔÏÂÄܹ»¼Ó¿ì¸Ã·´Ó¦ËÙÂʵÄÊÇ¢ÙÓÃ98%µÄŨÁòËá´úÌæÏ¡ÁòËá ¢Ú¼ÓÈÈ ¢Û¸ÄÓÃÂÁ·Û ¢ÜÔö´óÏ¡ÁòËáµÄÌå»ý ¢Ý¼ÓË® ¢Þ¼ÓÈëÉÙÁ¿ÁòËáÍ
A£®È«²¿ B£®¢Ú¢Û¢Þ C£®¢Ù¢Ú¢Û¢Þ D£®¢Ú¢Û¢Ü¢Þ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÒÒËá³È»¨õ¥ÊÇÒ»ÖÖʳÓÃÏãÁÏ£¬Æä½á¹¹¼òʽÈçÏÂͼËùʾ£¬
¹ØÓÚ¸ÃÓлúÎïµÄÏÂÁÐÐðÊöÖÐÕýÈ·µÄÊÇ
¢Ù·Ö×ÓʽΪC12H20O2
¢ÚÄÜʹËáÐÔKMnO4ÈÜÒºÍÊÉ«
¢ÛÄÜ·¢Éú¼Ó³É·´Ó¦£¬µ«²»ÄÜ·¢ÉúÈ¡´ú·´Ó¦
¢ÜËüµÄͬ·ÖÒì¹¹ÌåÖпÉÄÜÓз¼Ïã×廯ºÏÎÇÒÊôÓÚ·¼Ïã×廯ºÏÎïµÄͬ·ÖÒì¹¹ÌåÓÐ8ÖÖ
¢Ý1 mol¸ÃÓлúÎïË®½âʱֻÄÜÏûºÄ1 mol NaOH¡£
¢Þ1 mol¸ÃÓлúÎïÔÚÒ»¶¨Ìõ¼þϺÍH2·´Ó¦£¬¹²ÏûºÄH2 Ϊ3 mol
A£®¢Ù¢Ú¢Û B£®¢Ù¢Ú¢Ý C£®¢Ù¢Ú¢Ý¢Þ D£®¢Ù¢Ú¢Ü¢Ý¢Þ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ij»¯Ñ§ÐËȤС×éΪ̽¾¿SO2µÄÐÔÖÊ£¬°´ÈçͼËùʾװÖýøÐÐʵÑé¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)×°ÖÃAÖÐÊ¢·ÅÑÇÁòËáÄƵÄÒÇÆ÷Ãû³ÆÊÇ________£¬ÆäÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________________________________________________________________¡£
(2)ʵÑé¹ý³ÌÖУ¬×°ÖÃB¡¢CÖз¢ÉúµÄÏÖÏó·Ö±ðÊÇ_______________________________
______________¡¢_______________________________________________________£»
ÕâЩÏÖÏó·Ö±ð˵Ã÷SO2¾ßÓеÄÐÔÖÊÊÇ________________ºÍ________________£»×°ÖÃBÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ___________________________________________________
________________________________________________________________________¡£
(3)×°ÖÃDµÄÄ¿µÄÊÇ̽¾¿SO2ÓëÆ·ºì×÷ÓõĿÉÄæÐÔ£¬Çëд³öʵÑé²Ù×÷¼°ÏÖÏó________________________________________________________________________
________________________________________________________________________¡£
(4)βÆø¿É²ÉÓÃ________ÈÜÒºÎüÊÕ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ijÐËȤС×éÔÚʵÑéÊÒÓÃͺÍÁòËáΪÔÁÏ£¬²ÉÓöàÖÖ·½·¨ÖÆÈ¡ÁòËáÍ£¬ÖƱ¸·½·¨ÈçÏ£º
·½·¨Ò»
(1)ŨÁòËáÊÔ¼ÁÆ¿ÉÏÊʺÏÌùÉϵıêÇ©ÊÇ________(ÌîÐòºÅ)¡£
(2)¼×ͬѧȡ6.4 gÍƬºÍ10 mL 18 mol¡¤L£1ŨÁòËᣬ·ÅÔÚÊÔ¹ÜÖй²ÈÈʱ·¢ÏÖ£¬ÍÓëÈȵÄŨÁòËá·´Ó¦ºó²¢Ã»Óеõ½Ô¤ÆÚµÄÀ¶É«ÈÜÒº£¬¶øÊÇÔÚÊԹܵײ¿¿´µ½»Ò°×É«³Áµí¡£¼×ͬѧΪÁËÑéÖ¤ÆäÖлҰ×É«³ÁµíµÄÖ÷Òª³É·Ö£¬Éè¼ÆÏÂÁÐʵÑ飺
ʵÑé²½Ö裺Çãµ¹µôÉϲãÒºÌåºó£¬ÏòËùµÃ»Ò°×É«µÄ¹ÌÌåÖмÓÈëÊÊÁ¿ÕôÁóË®£¬±ß¼Ó±ß½Á°è¡£
ʵÑéÏÖÏó£º_____________________________________________________________¡£
ʵÑé½áÂÛ£ºËùµÃ»Ò°×É«¹ÌÌåµÄ»¯Ñ§Ê½Îª__________¡£
(3)ÒÒ»¹¹Û²ìµ½¼ÓÈȹý³ÌÖУ¬ÊÔ¹ÜÄÚ±ÚÉϲ¿Îö³öÉÙÁ¿µ»ÆÉ«¹ÌÌåÎïÖÊ£¬³ÖÐø¼ÓÈÈ£¬µ»ÆÉ«¹ÌÌåÎïÖÊÓÖÂýÂýµØÈÜÓÚŨÁòËá¶øÏûʧ¡£µ»ÆÉ«¹ÌÌåÏûʧµÄÔÒòÊÇ(Óû¯Ñ§·½³Ìʽ»Ø´ð)________________________________________________________________________¡£
Ö±µ½×îºó·´Ó¦Íê±Ï£¬·¢ÏÖÊÔ¹ÜÖл¹ÓÐÍƬʣÓ࣬ÒÒ¸ù¾Ý×Ô¼ºËùѧµÄ»¯Ñ§ÖªÊ¶£¬ÈÏΪÊÔ¹ÜÖл¹ÓÐÁòËáÊ£Óà¡£ËûÕâÑùÈÏΪµÄÀíÓÉÊÇ______________________________________¡£
·½·¨¶þ
(4)±ûͬѧÈÏΪ¼×Éè¼ÆµÄʵÑé·½°¸²»ºÃ£¬Ëû×Ô¼ºÉè¼ÆµÄ˼·ÊÇ2Cu£«O22CuO£¬CuO£«H2SO4===CuSO4£«H2O¡£
¶Ô±È¼×µÄ·½°¸£¬ÄãÈÏΪ±ûͬѧµÄÓŵãÊÇ¢Ù_____________________________________£»
¢Ú________________________________________________________________________¡£
·½·¨Èý
(5)¶¡Í¬Ñ§È¡Ò»ÍƬºÍÏ¡ÁòËá·ÅÔÚÊÔ¹ÜÖУ¬ÔÙÏòÆäÖеÎÈëË«ÑõË®·¢ÏÖÈÜÒºÖð½¥³ÊÀ¶É«£¬Ð´³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ_______________________________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
¶þÑõ»¯ÁòÊÇÁòµÄÖØÒª»¯ºÏÎÔÚÉú²ú¡¢Éú»îÖÐÓй㷺ӦÓ᣶þÑõ»¯ÁòÓж¾£¬²¢ÇÒÊÇÐγÉËáÓêµÄÖ÷ÒªÆøÌå¡£ÎÞÂÛÊÇʵÑéÊÒÖƱ¸»¹Êǹ¤ÒµÉú²ú£¬¶þÑõ»¯ÁòβÆøÎüÊÕ»òÑÌÆøÍÑÁò¶¼·Ç³£ÖØÒª¡£Íê³ÉÏÂÁÐÌî¿Õ£º
(1)ʵÑéÊÒ¿ÉÓÃͺÍŨÁòËá¼ÓÈÈ»òÁòËáºÍÑÇÁòËáÄÆ·´Ó¦ÖÆÈ¡¶þÑõ»¯Áò¡£
Èç¹ûÓÃÁòËáºÍÑÇÁòËáÄÆ·´Ó¦ÖÆÈ¡¶þÑõ»¯Áò£¬²¢Ï£ÍûÄÜ¿ØÖÆ·´Ó¦Ëٶȣ¬ÉÏͼÖпÉÑ¡Óõķ¢Éú×°ÖÃÊÇ________(Ìîд×Öĸ)¡£
(2)ÈôÓÃÁòËáºÍÑÇÁòËáÄÆ·´Ó¦ÖÆÈ¡3.36 L(±ê×¼×´¿ö)¶þÑõ»¯Áò£¬ÖÁÉÙÐèÒª³ÆÈ¡ÑÇÁòËáÄÆ________g(±£ÁôһλСÊý)£»Èç¹ûÒÑÓÐ40%ÑÇÁòËáÄÆ(ÖÊÁ¿·ÖÊý)±»Ñõ»¯³ÉÁòËáÄÆ£¬ÔòÖÁÉÙÐè³ÆÈ¡¸ÃÑÇÁòËáÄÆ________g (±£ÁôһλСÊý)¡£
(3)ʵÑéÊÒ¶þÑõ»¯ÁòβÆøÎüÊÕÓ빤ҵÑÌÆøÍÑÁòµÄ»¯Ñ§ÔÀíÏàͨ¡£Ê¯»Òʯ¸à·¨ºÍ¼î·¨Êdz£ÓõÄÑÌÆøÍÑÁò·¨¡£
ʯ»Òʯ¸à·¨µÄÎüÊÕ·´Ó¦ÎªSO2£«Ca(OH)2¨D¡úCaSO3¡ý£«H2O¡£ÎüÊÕ²úÎïÑÇÁòËá¸ÆÓɹܵÀÊäËÍÖÁÑõ»¯ËþÑõ»¯£¬·´Ó¦Îª2CaSO3£«O2£«4H2O¨D¡ú2CaSO4¡¤2H2O¡£ÆäÁ÷³ÌÈçÏÂͼ£º
¼î·¨µÄÎüÊÕ·´Ó¦ÎªSO2£«2NaOH¨D¡úNa2SO3£«H2O¡£¼î·¨µÄÌصãÊÇÇâÑõ»¯ÄƼîÐÔÇ¿¡¢ÎüÊտ졢ЧÂʸߡ£ÆäÁ÷³ÌÈçÏÂͼ£º
ÒÑÖª£º
ÊÔ¼Á | Ca(OH)2 | NaOH |
¼Û¸ñ(Ôª/kg) | 0.36 | 2.9 |
ÎüÊÕSO2µÄ³É±¾(Ôª/mol) | 0.027 | 0.232 |
ʯ»Òʯ¸à·¨ºÍ¼î·¨ÎüÊÕ¶þÑõ»¯ÁòµÄ»¯Ñ§ÔÀíÏà֮ͬ´¦ÊÇ
________________________________________________________________________¡£
ºÍ¼î·¨Ïà±È£¬Ê¯»Ò£Ê¯¸à·¨µÄÓŵãÊÇ_______________________________________£¬
ȱµãÊÇ__________________________________________________________________¡£
(4)ÔÚʯ»Òʯ¸à·¨ºÍ¼î·¨µÄ»ù´¡ÉÏ£¬Éè¼ÆÒ»¸ö¸Ä½øµÄ¡¢ÄÜʵÏÖÎïÁÏÑ»·µÄÑÌÆøÍÑÁò·½°¸(ÓÃÁ÷³Ìͼ±íʾ)¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁл¯Ñ§·´Ó¦µÄÀë×Ó·½³ÌʽÕýÈ·µÄÊÇ (¡¡¡¡)
A£®ÔÚÏ¡°±Ë®ÖÐͨÈë¹ýÁ¿CO2£ºNH3¡¤H2O£«CO2===NH£«HCO
B£®ÉÙÁ¿SO2ͨÈëƯ°×·ÛÈÜÒºÖУºSO2£«H2O£«Ca2£«£«2ClO£===CaSO3¡ý£«2HClO
C£®ÓÃÏ¡HNO3ÈܽâFeS¹ÌÌ壺FeS£«2H£«===Fe2£«£«H2S¡ü
D£®ÏòNH4HCO3ÈÜÒºÖмÓÈë×ãÁ¿Ba(OH)2ÈÜÒº£º2HCO£«Ba2£«£«2OH£===BaCO3¡ý£«CO£«2H2O
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁÐÏÖÏóÖУ¬ÄÜÓüüÄÜ´óС½âÊ͵ÄÊÇ
A. µªÆøµÄ»¯Ñ§ÐÔÖʱÈÑõÆøÎȶ¨ B. ³£Î³£Ñ¹Ï£¬äå³ÊҺ̬£¬µâ³Ê¹Ì̬
C. Ï¡ÓÐÆøÌåÒ»°ãºÜÄÑ·¢Éú»¯Ñ§·´Ó¦ D. ÏõËáÒ×»Ó·¢£¬¶øÁòËáÄѻӷ¢
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
ÏÂÁÐÓйØÔªËØÖÜÆÚ±íÖÐÔªËØÐÔÖʵÄÐðÊöÖÐÕýÈ·µÄÊÇ£¨ £©
A. ͬÖ÷×åÔªËØ´ÓÉϵ½Ï£¬·Ç½ðÊôÐÔÖð½¥ÔöÇ¿
B. ÔªËصÄ×î¸ßÕý»¯ºÏ¼Û¾ùµÈÓÚËüËùÔÚµÄ×åÐòÊý
C. ͬÖÜÆÚµÄÖ÷×åÔªËصÄÔ×Ӱ뾶ԽС£¬Ô½ÄÑʧȥµç×Ó
D. µÚ¢õA×åÔªËصÄÇ⻯ÎïµÄ·Ðµã£¬´ÓÉϵ½ÏÂÒÀ´ÎÖð½¥Éý¸ß
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