ÌâÄ¿ÄÚÈÝ

A¡¢B¡¢C¡¢DËÄÖÖ¿ÉÈÜÐÔÑΣ¬ÖªÆäÑôÀë×Ó·Ö±ðÊÇNa+¡¢Ba2+¡¢Cu2+¡¢Ag+ ÖеÄijһÖÖ£¬
ÒõÀë×Ó·Ö±ðÊÇCl-¡¢SO42-¡¢CO32-¡¢NO3- ÖеÄijһÖÖ¡£ÏÖ×öÒÔÏÂʵÑ飺
¢Ù ½«ËÄÖÖÑθ÷È¡ÉÙÁ¿£¬·Ö±ðÈÜÓÚÊ¢ÓÐ5 mLÕôÁóË®µÄËÄÖ§ÊÔ¹ÜÖУ¬Ö»ÓÐBÑÎÈÜÒº³ÊÀ¶É«¡£
¢Ú ·Ö±ðÏò4Ö§ÊÔ¹ÜÖмÓÈë2 mLÏ¡ÑÎËᣬ·¢ÏÖAÑÎÈÜÒºÖвúÉú°×É«³Áµí£¬CÑÎÈÜÒºÖÐÓн϶àÆøÅݲúÉú£¬¶øDÑÎÈÜÒºÎÞÃ÷ÏÔÏÖÏó¡£
£¨1£©¸ù¾ÝÉÏÊöÊÂʵ£¬ÍƶÏÕâËÄÖÖÑεĻ¯Ñ§Ê½·Ö±ðΪ£º
A             ¡¢B             ¡¢C            ¡¢ D            ¡£
(2)д³öʵÑé²½Öè¢ÚÖÐÉæ¼°µ½µÄËùÓз´Ó¦µÄÀë×Ó·½³Ìʽ£º                                      ¡£

£¨1£©A£ºAgNO3£»B£ºCuSO4£»C£ºNa2CO3£¬ D£ºBaCl2£¨¸÷1·Ö£©
£¨2£©Ag++Cl-£½AgCl¡ý¡¢CO32-+2H+£½CO2¡ü+H2O£¨¸÷2·Ö£©

½âÎöÊÔÌâ·ÖÎö£º£¨1£©BÑεÄÈÜÒº³ÊÀ¶É«£¬ËµÃ÷BÑÎÖк¬ÓÐCu2+¡£·Ö±ðÏò4Ö§ÊÔ¹ÜÖмÓÈë2mLÏ¡ÑÎËᣬ·¢ÏÖAÑÎÈÜÒºÖвúÉú°×É«³Áµí£¬ËµÃ÷AÑÎÖк¬ÓÐAg+£»CÑÎÈÜÒºÖÐÓн϶àÆøÅݲúÉú£¬ËµÃ÷CÑÎÖк¬ÓÐCO32-£»ÓÖÒòΪA¡¢B¡¢C¡¢DËÄÖÖÑξùΪ¿ÉÈÜÐÔÑΣ¬¹ÊAÑÎΪAgNO3£»CÑÎΪNa2CO3£¬BÑÎΪCuSO4£»DÑÎΪBaCl2¡£
£¨2£©ÊµÑé²½Öè¢ÚÖÐÉæ¼°µ½µÄËùÓз´Ó¦µÄÀë×Ó·½³ÌʽÓÐAg++Cl-£½AgCl¡ý£¬CO32-+2H+£½CO2¡ü+H2O¡£
¿¼µã£º¿¼²éÀë×Ó¹²´æ¡¢Àë×Ó¼ìÑéÒÔ¼°Àë×Ó·½³ÌʽµÄÊéдµÈ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ij»ìºÏÈÜÒºÖпÉÄܺ¬ÓеÄÀë×ÓÈçϱíËùʾ£º

¿ÉÄÜ´óÁ¿º¬ÓеÄÑôÀë×Ó
H+¡¢Ag+¡¢Mg2+¡¢Al3+¡¢NH¡¢Fe3+
¿ÉÄÜ´óÁ¿º¬ÓеÄÒõÀë×Ó
Cl-¡¢Br-¡¢I-¡¢CO¡¢AlO
 
Ϊ̽¾¿Æä³É·Ö£¬½øÐÐÁËÒÔÏÂ̽¾¿ÊµÑé¡£
£¨1£©Ì½¾¿Ò»£º
¼×ͬѧȡһ¶¨Á¿µÄ»ìºÏÈÜÒº£¬ÏòÆäÖÐÖðµÎ¼ÓÈëÇâÑõ»¯ÄÆÈÜÒº£¬²úÉú³ÁµíµÄÎïÖʵÄÁ¿(n)Óë¼ÓÈëÇâÑõ »¯ÄÆÈÜÒºµÄÌå»ý£¨V£©µÄ¹ØϵÈçͼËùʾ¡£

¢Ù¸ÃÈÜÒºÖÐÒ»¶¨º¬ÓеÄÑôÀë×ÓÊÇ______________£¬Æä¶ÔÓ¦ÎïÖʵÄÁ¿Å¨¶ÈÖ®±ÈΪ________£¬Ò»¶¨²»´æÔÚµÄÒõÀë×ÓÊÇ_____________£»
¢ÚÇëд³ö³Áµí¼õÉÙ¹ý³ÌÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ_____________________________¡£
£¨2£©Ì½¾¿¶þ£º
ÒÒͬѧ¼ì²âµ½¸ÃÈÜÒºÖк¬ÓдóÁ¿µÄCl-¡¢Br-¡¢I-£¬ÈôÏò1 L¸Ã»ìºÏÈÜÒºÖÐͨÈëÒ»¶¨Á¿µÄCl2£¬ÈÜÒºÖÐCl-¡¢Br-¡¢I-µÄÎïÖʵÄÁ¿ÓëͨÈëCl2µÄÌå»ý£¨±ê×¼×´¿ö£©µÄ¹ØϵÈçϱíËùʾ£¬ ·ÖÎöºó»Ø´ðÏÂÁÐÎÊÌ⣺
Cl2µÄÌå»ý£¨±ê×¼×´¿ö£©
5.6 L
11.2 L
22.4 L
n (Cl-)
2.5 mol
3.0 mol
4.0 mol
n (Br-)
3.0 mol
2.8 mol
1.8 mol
n (I-)
x mol
0
0
 
¢Ùµ±Í¨ÈëCl2µÄÌå»ýΪ5.6 Lʱ£¬ÈÜÒºÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________£»
¢ÚÔ­ÈÜÒºÖÐCl-¡¢Br-¡¢I-µÄÎïÖʵÄÁ¿Å¨¶ÈÖ®±ÈΪ______________________¡£

¶þÑõ»¯ÂÈ£¨ClO2£©ÔÚ³£ÎÂÏÂÊÇÒ»ÖÖ»ÆÂÌÉ«Óд̼¤ÐÔÆøζµÄÆøÌ壬ÆäÈÛµãΪ£­59¡æ£¬·ÐµãΪ11£®0¡æ£¬Ò×ÈÜÓÚË®¡£¹¤ÒµÉÏÓÃÉÔ³±ÊªµÄKClO3ºÍ²ÝËᣨH2C2O4£©ÔÚ60¡æʱ·´Ó¦ÖƵá£Ä³Ñ§ÉúÄâÓÃÏÂͼËùʾװÖÃÄ£Ä⹤ҵÖÆÈ¡²¢ÊÕ¼¯ClO2¡£ÓÉÓڸ÷´Ó¦ÊÇÎüÈÈ·´Ó¦£¬ËùÒÔÔÚÇé¿öÏÂÓÐÀûÓڸ÷´Ó¦×Ô·¢½øÐÐ

£¨1£©AÖз´Ó¦²úÎïÓÐK2CO3¡¢ClO2ºÍCO2µÈ£¬Çëд³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º              ¡£
£¨2£©A±ØÐëÌí¼ÓζȿØÖÆ×°Ö㬳ý¾Æ¾«µÆÍ⣬»¹ÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢          £»B×°ÖñØÐë·ÅÔÚ±ùˮԡÖУ¬ÆäÔ­ÒòÊÇ                         ¡£
£¨3£©·´Ó¦ºóÔÚ×°ÖÃCÖпɵÃNaClO2ÈÜÒº¡£ÒÑÖªNaClO2±¥ºÍÈÜÒºÔÚζȵÍÓÚ38¡æʱÎö³ö¾§ÌåÊÇNaClO2¡¤3H2O£¬ÔÚζȸßÓÚ38¡æʱÎö³ö¾§ÌåÊÇNaClO2¡£Çë²¹³ä´ÓNaClO2ÈÜÒºÖÐÖƵÃNaClO2¾§ÌåµÄ²Ù×÷²½Ö裺¢Ù               £»¢Ú             £»¢ÛÏ´µÓ£»¢Ü¸ÉÔï¡£
£¨4£©ClO2ºÜ²»Îȶ¨£¬ÐèËæÓÃËæÖÆ£¬²úÎïÓÃË®ÎüÊյõ½ClO2ÈÜÒº¡£Îª²â¶¨ËùµÃÈÜÒºÖÐClO2µÄº¬Á¿£¬½øÐÐÁËÏÂÁÐʵÑ飺
²½Öè1£º×¼È·Á¿È¡ClO2ÈÜÒº10.00 mL£¬Ï¡ÊͳÉ100.00 mLÊÔÑù£»Á¿È¡V1 mLÊÔÑù¼ÓÈ뵽׶ÐÎÆ¿ÖУ»
²½Öè2£ºÓÃÏ¡ÁòËáµ÷½ÚÊÔÑùµÄpH¡Ü2.0£¬¼ÓÈë×ãÁ¿µÄKI¾§Ì壬¾²ÖÃƬ¿Ì£»
²½Öè3£º¼ÓÈëµí·Ûָʾ¼Á£¬ÓÃc mol/L Na2S2O3ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄNa2S2O3ÈÜÒºV2 mL¡££¨ÒÑÖª2 Na2S2O3 + I2£½Na2S4O6 + 2NaI£©
¢ÙÅäÖÆ100 mL c mol/LNa2S2O3±ê×¼ÈÜҺʱ£¬Óõ½µÄ²£Á§ÒÇÆ÷³ýÉÕ±­¡¢Á¿Í²¡¢²£Á§°ôÍ⻹ÓУº                ¡£
¢Úд³ö²½Öè2Öз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ                          ¡£
¢ÛÔ­ClO2ÈÜÒºµÄŨ¶ÈΪ          g / L£¨Óò½ÖèÖеÄ×Öĸ´úÊýʽ±íʾ£©¡£
¢ÜÈôµÎ¶¨Ç°µÎ¶¨¹Ü¼â×ìÖÐÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ£¬Ôò²â¶¨½á¹û        ¡£
ÈôµÎ¶¨¿ªÊ¼ÑöÊÓ¶ÁÊý£¬µÎ¶¨ÖÕµãʱÕýÈ·¶ÁÊý£¬Ôò²â¶¨½á¹û              ¡£
£¨Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±»ò¡°²»±ä¡± £©

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø