ÌâÄ¿ÄÚÈÝ

(6·Ö)ÓÐһƿ³ÎÇåÈÜÒº£¬¿ÉÄܺ¬ÓÐNH£«4¡¢K£«¡¢Mg2£«¡¢Ba2£«¡¢Al3£«¡¢Fe3£«¡¢SO2¨D 4¡¢CO2¨D 3¡¢NO¨D 3¡¢Cl¨D¡¢I¨DÏÖ½øÐÐÈçÏÂʵÑ飺

£¨1£©²âÖªÈÜÒºÏÔÇ¿ËáÐÔ£»

£¨2£©È¡Ñù¼ÓÉÙÁ¿CCl4ºÍÊýµÎÐÂÖÆÂÈË®£¬CCl4²ãΪ×ϺìÉ«£»

£¨3£©ÁíÈ¡ÑùµÎ¼ÓÏ¡NaOHÈÜÒº£¬Ê¹ÈÜÒº±äΪ¼îÐÔ£¬´Ë¹ý³ÌÖоùÎÞ³ÁµíÉú³É£»

£¨4£©È¡ÉÙÁ¿ÉÏÊö¼îÐÔÈÜÒº£¬¼ÓNa2CO3ÈÜÒº³öÏÖ°×É«³Áµí£»

£¨5£©½«ÊµÑé(3)ÖеļîÐÔÈÜÒº¼ÓÈÈ£¬ÓÐÆøÌå·Å³ö£¬¸ÃÆøÌåÄÜʹʪºìɫʯÈïÊÔÖ½±äÀ¶¡£

     ÎÊ£º¢ÙÔ­ÈÜÒºÖп϶¨´æÔÚµÄÀë×ÓÊÇ____________________£»

    ¢Ú¿Ï¶¨²»´æÔÚµÄÀë×ÓÊÇ______________________________£»

    ¢Û²»ÄÜÈ·¶¨ÊÇ·ñ´æÔÚµÄÀë×ÓÊÇ______________________________¡£

 

 

¢ÙNH£«4¡¢Ba2£«¡¢I¨D

¢ÚMg2£«¡¢Al3£«¡¢Fe3£«¡¢CO2¨D 3¡¢NO¨D 3¡¢SO2¨D 4

¢ÛK£«¡¢Cl¨D

½âÎö:

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
»¯Ñ§Ñ§Ï°ÖУ¬ÓйØÎïÖÊÐÔÖʵÄѧϰÀë²»¿ªÊµÑ飬ÇëÔĶÁÏÂÁжÔÓ¦µÄÄÚÈÝ£¬²¢°´ÒªÇóÍê³ÉÌî¿Õ
£¨1£©ÊµÑéÊÒ¾­³£ÓÃÉÕ±­½øÐÐÐÔÖÊʵÑéÑо¿£¬ÈçÓÃͼ1ËùʾװÖü°Ï±íÖÐÊÔ¼Á£¬¹ýÒ»¶Îʱ¼äʵÑé1¡¢2¡¢3¡¢4ÉÕ±­¢ÙÖеÄÏÖÏó·Ö±ðÊÇ
±äÀ¶
±äÀ¶
£¬
ÎÞÃ÷ÏÔÏÖÏó
ÎÞÃ÷ÏÔÏÖÏó
£¬
Óа×É«½º×´³Áµí
Óа×É«½º×´³Áµí
£¬
±ä°×
±ä°×
£¬
ʵÑé±àºÅ ¢ÙÖеÄÎïÖÊ ¢ÚÖеÄÎïÖÊ
1 µí·Ûµâ»¯¼ØÈÜÒº ŨÏõËá
2 ·Ó̪ÈÜÒº ŨÁòËá
3 ÂÈ»¯ÂÁÈÜÒº Ũ°±Ë®
4 ʪÈóµÄºìÖ½ ±¥ºÍÂÈË®
£¨2£©ÓÃͼ2×°Ö㺷ÏͭмÖÆÏõËáÍ­£¬·´Ó¦½áÊøºó£¬¹ã¿ÚÆ¿ÄÚµÄÈÜÒºÖУ¬³ýÁ˺¬ÓÐNaOHÍ⣬»¹ÓÐ
NaNO3¡¢NaNO2
NaNO3¡¢NaNO2
£¨Ìîд»¯Ñ§Ê½£©
£¨3£©Å¨°±Ë®Í¨³£¿ÉÒÔÓÃÓÚʵÑéÊÒ¿ìËÙÖÆÈ¡°±Æø¼°ÆäÏà¹ØʵÑéµÄ̽¾¿£¬»Ø´ðÏÂÁÐÎÊÌ⣮

¢ÙÈôÒª²â¶¨Éú³ÉµÄNH3µÄÌå»ý£¬Ôò±ØÐëÑ¡ÔñµÄ×°ÖÃÊÇ
¢Ù¢Û
¢Ù¢Û
£¨Ìî×°ÖÃÐòºÅ£©£¬×°ÖÃÖÐËùÊ¢ÊÔ¼ÁÓ¦¾ßÓеÄÐÔÖÊÊÇ
°±ÆøÄÑÈÜÓÚ¸ÃÈܼÁ£¬¸ÃÈܼÁ²»Ò×»Ó·¢¡¢²»Óë°±Æø·´Ó¦
°±ÆøÄÑÈÜÓÚ¸ÃÈܼÁ£¬¸ÃÈܼÁ²»Ò×»Ó·¢¡¢²»Óë°±Æø·´Ó¦
£®ÊÕ¼¯¸ÉÔïµÄNH3£¬ÊÕ¼¯×°ÖÃӦѡÔñ£¨Ìî×°ÖÃÐòºÅ£©
¢Ý
¢Ý
£¬ÀíÓÉÊÇ
µ¼¹Ü½Ï³¤£¬ÃÞ»¨¿ÉÒÔ·ÀÖ¹¿ÕÆø¶ÔÁ÷£¬ÓÐÀûÓÚÊÕ¼¯µ½½ÏΪ¸ÉÔï´¿¾»µÄ°±Æø
µ¼¹Ü½Ï³¤£¬ÃÞ»¨¿ÉÒÔ·ÀÖ¹¿ÕÆø¶ÔÁ÷£¬ÓÐÀûÓÚÊÕ¼¯µ½½ÏΪ¸ÉÔï´¿¾»µÄ°±Æø
£®
¢ÚÏòŨCaCl2ÈÜÒºÖÐÏÈͨÈëNH3ÔÙͨÈëCO2ÆøÌå¿ÉÖÆÄÉÃ×¼¶£¨Á£×ÓÖ±¾¶ÔÚ1-10nmÖ®¼ä£©Ì¼Ëá¸Æ£¬ÊÔд³öÖÆÄÉÃ×¼¶Ì¼Ëá¸ÆµÄÀë×Ó·½³Ìʽ
Ca2++2NH3+CO2+H2O¡úCaCO3+NH4+
Ca2++2NH3+CO2+H2O¡úCaCO3+NH4+
£®
£¨4£©Í¼3ÊDZ½Óëäå·¢Éú·´Ó¦²¢½øÐвúÎï¼ìÑéµÄ·´Ó¦×°ÖÃʵÑé×°ÖÃÖеÄÀäÄý¹Ü¡°×óµÍÓҸߡ±µÄ·ÅÖÃÄ¿µÄÊÇ
ʹÀäÄý»ØÁ÷µÄ·´Ó¦Îï±½ºÍÒºäå»Øµ½£¨¢ñ£©ÖмÌÐø·´Ó¦
ʹÀäÄý»ØÁ÷µÄ·´Ó¦Îï±½ºÍÒºäå»Øµ½£¨¢ñ£©ÖмÌÐø·´Ó¦
£¬ÕûÌ×ʵÑé×°ÖÃÖÐÄÜ·ÀÖ¹µ¹ÎüµÄ×°ÖÃÊÇ
£¨¢ò£©£¨¢ó£©
£¨¢ò£©£¨¢ó£©
£¨Ìî×°ÖÃÐòºÅ£©
¶ÔÓÚ¹ÌÌåÁò»¯ÄƶÖÃÔÚ¿ÕÆøÖеı仯£¬ÓÐÈçϼÙÉ裺
¼ÙÉè¢Ù£º¹ÌÌåÁò»¯ÄÆÒ×±»¿ÕÆøÖеÄÑõÆøÑõ»¯Îªµ¥ÖÊÁò£®
¼ÙÉè¢Ú£º¹ÌÌåÁò»¯ÄÆÒ×±»¿ÕÆøÖеÄÑõÆøÑõ»¯ÎªÑÇÁòËáÄÆ£®
¼ÙÉè¢Û£º¹ÌÌåÁò»¯ÄÆÒ×±»¿ÕÆøÖеÄÑõÆøÑõ»¯ÎªÁòËáÄÆ£®
ΪÁË̽¾¿¹ÌÌåÁò»¯ÄƶÖÃÔÚ¿ÕÆøÖо¿¾¹ÓÐÔõÑùµÄ±ä»¯£¬Ä³»¯Ñ§Ñ§Ï°Ð¡×é½øÐÐÁËÈçÏÂʵÑ飺
¢Ù´ÓÊÔ¼ÁÆ¿ÖÐÈ¡³ö¹ÌÌåÁò»¯ÄÆÑùÆ·£¬·ÅÔÚÑв§ÖÐÑÐË飮
¢Ú½«Ñв§ÖеÄÑùƷ¶ÖÃÔÚ¿ÕÆøÖÐÁ½Ì죮
¢Û´ÓÑв§ÖÐÈ¡³öÒ»Ò©³×ÑùÆ··ÅÔÚÊÔ¹ÜÖУ¬¼ÓÈëÑÎËᣬÊÔÑùÈ«²¿Èܽ⣬µÃµ½³ÎÇåÈÜÒº£¬²¢·Å³ö´óÁ¿ÆøÅÝ£®
¢ÜÁ¢¼´¼ÓÈû£¬ÓÃÁ¦Õñµ´£¬²úÉú»ë×Ç£¬ÇÒÆøÅݵÄÁ¿´ó´ó¼õÉÙ£®
£¨5£©½âÊͼÓÈûÕñµ´ºó²úÉú»ë×Ç£¬ÇÒÆøÅÝ´óÁ¿¼õÉÙµÄÔ­Òò£¨Óû¯Ñ§·½³Ìʽ±íʾ£©
2H2S+SO2=2H2O+3S¡ý
2H2S+SO2=2H2O+3S¡ý
£®
£¨6£©Èç¹ûÒªÑéÖ¤¢ÛÊÇ·ñ³ÉÁ¢µÄʵÑé·½·¨ÊÇ
È¡ÉÙÁ¿Î´ÖªÒºÓÚÊÔ¹ÜÖУ¬ÏȵμӹýÁ¿µÄÏ¡ÑÎËᣬÎÞÈκÎÏÖÏó£¨ÎÞ³ÁµíÉú³É¡¢ÎÞÆøÌå³öÏÖ£©£¬ÔٵμÓÂÈ»¯±µÈÜÒº£¬³öÏÖ°×É«³Áµí
È¡ÉÙÁ¿Î´ÖªÒºÓÚÊÔ¹ÜÖУ¬ÏȵμӹýÁ¿µÄÏ¡ÑÎËᣬÎÞÈκÎÏÖÏó£¨ÎÞ³ÁµíÉú³É¡¢ÎÞÆøÌå³öÏÖ£©£¬ÔٵμÓÂÈ»¯±µÈÜÒº£¬³öÏÖ°×É«³Áµí
£®

(12·Ö) Óɼ¸ÖÖÑÎÈÜÓÚË®ÖÐÐγɵÄһƿ³ÎÇåµÄÈÜÒº£¬ÆäÖпÉÄܺ¬ÓÐNH4+¡¢Na+¡¢Mg2+¡¢Ba2+¡¢Al3+¡¢Fe3+¡¢Cu2+¡¢Cl£­¡¢Br£­¡¢I£­¡¢NO3£­¡¢CO32£­¡¢SO32£­¡¢SO42£­Öеļ¸ÖÖ¡£È¡¸ÃÈÜÒº½øÐÐÒÔÏÂʵÑ飺

£¨1£©ÓÃpHÊÔÖ½¼ìÑ飬pHÊÔÖ½³ÊºìÉ«¡£Åųý           Àë×ӵĴæÔÚ¡£

£¨2£©È¡³ö²¿·ÖÈÜÒº£¬¼ÓÈëÉÙÁ¿CCl4¼°¼¸µÎÐÂÖÆÂÈË®£¬¾­Õñµ´CCl4²ã³Ê×ϺìÉ«¡£Åųý

         Àë×Ó´æÔÚ¡£

£¨3£©ÁíÈ¡²¿·ÖÈÜÒº£¬Öð½¥ÏòÆäÖмÓÈëNaOHÈÜÒº£¬Ê¹ÈÜÒº´ÓËáÐÔÖð½¥±äΪ¼îÐÔ£¬Ôڵμӹý³ÌÖк͵μÓÍê±Ïºó£¬¾ùÎÞ³Áµí²úÉú¡£Ôò¿ÉÅųý                    Àë×ӵĴæÔÚ¡£

£¨4£©ÁíÈ¡²¿·ÖÉÏÊö¼îÐÔÈÜÒº£¬ÏòÆäÖмÓÈëNa2CO3ÈÜÒº£¬Óа×É«³ÁµíÉú³É¡£Ö¤Ã÷       

Àë×Ó´æÔÚ£¬ÓÖÅųý           Àë×Ó´æÔÚ¡£

£¨5£©¸ù¾ÝÉÏÊöʵÑéÊÂʵÄÜ·ñÈ·¶¨NH4+ÊÇ·ñ´æÔÚ£¬ÈôÄÜ£¬Çë¼òÊöÀíÓÉ             £¬Èô²»ÄÜ£¬Çë¼òÊö¼ìÑéËüµÄ·½·¨¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

£¨6£©Ô­ÈÜÒºÒ»¶¨´æÔÚµÄÀë×ÓÓР            £¬²»ÄÜÈ·¶¨µÄÀë×ÓÓС¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

 

(12·Ö) Óɼ¸ÖÖÑÎÈÜÓÚË®ÖÐÐγɵÄһƿ³ÎÇåµÄÈÜÒº£¬ÆäÖпÉÄܺ¬ÓÐNH4+¡¢Na+¡¢Mg2+¡¢Ba2+¡¢Al3+¡¢Fe3+¡¢Cu2+¡¢Cl£­¡¢Br£­¡¢I£­¡¢NO3£­¡¢CO32£­¡¢SO32£­¡¢SO42£­Öеļ¸ÖÖ¡£È¡¸ÃÈÜÒº½øÐÐÒÔÏÂʵÑ飺

£¨1£©ÓÃpHÊÔÖ½¼ìÑ飬pHÊÔÖ½³ÊºìÉ«¡£Åųý          Àë×ӵĴæÔÚ¡£

£¨2£©È¡³ö²¿·ÖÈÜÒº£¬¼ÓÈëÉÙÁ¿CCl4¼°¼¸µÎÐÂÖÆÂÈË®£¬¾­Õñµ´CCl4²ã³Ê×ϺìÉ«¡£Åųý

        Àë×Ó´æÔÚ¡£

£¨3£©ÁíÈ¡²¿·ÖÈÜÒº£¬Öð½¥ÏòÆäÖмÓÈëNaOHÈÜÒº£¬Ê¹ÈÜÒº´ÓËáÐÔÖð½¥±äΪ¼îÐÔ£¬Ôڵμӹý³ÌÖк͵μÓÍê±Ïºó£¬¾ùÎÞ³Áµí²úÉú¡£Ôò¿ÉÅųý                   Àë×ӵĴæÔÚ¡£

£¨4£©ÁíÈ¡²¿·ÖÉÏÊö¼îÐÔÈÜÒº£¬ÏòÆäÖмÓÈëNa2CO3ÈÜÒº£¬Óа×É«³ÁµíÉú³É¡£Ö¤Ã÷       

Àë×Ó´æÔÚ£¬ÓÖÅųý           Àë×Ó´æÔÚ¡£

£¨5£©¸ù¾ÝÉÏÊöʵÑéÊÂʵÄÜ·ñÈ·¶¨NH4+ÊÇ·ñ´æÔÚ£¬ÈôÄÜ£¬Çë¼òÊöÀíÓÉ            £¬Èô²»ÄÜ£¬Çë¼òÊö¼ìÑéËüµÄ·½·¨¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

£¨6£©Ô­ÈÜÒºÒ»¶¨´æÔÚµÄÀë×ÓÓР           £¬²»ÄÜÈ·¶¨µÄÀë×ÓÓС¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

 

(12·Ö)Óɼ¸ÖÖÑÎÈÜÓÚË®ÖÐÐγɵÄһƿ³ÎÇåµÄÈÜÒº£¬ÆäÖпÉÄܺ¬ÓÐNH4+¡¢Na+¡¢Mg2+¡¢Ba2+¡¢Al3+¡¢Fe3+¡¢Cu2+¡¢Cl£­¡¢Br£­¡¢I£­¡¢NO3£­¡¢CO32£­¡¢SO32£­¡¢SO42£­Öеļ¸ÖÖ¡£È¡¸ÃÈÜÒº½øÐÐÒÔÏÂʵÑ飺
£¨1£©ÓÃpHÊÔÖ½¼ìÑ飬pHÊÔÖ½³ÊºìÉ«¡£Åųý          Àë×ӵĴæÔÚ¡£
£¨2£©È¡³ö²¿·ÖÈÜÒº£¬¼ÓÈëÉÙÁ¿CCl4¼°¼¸µÎÐÂÖÆÂÈË®£¬¾­Õñµ´CCl4²ã³Ê×ϺìÉ«¡£Åųý
        Àë×Ó´æÔÚ¡£
£¨3£©ÁíÈ¡²¿·ÖÈÜÒº£¬Öð½¥ÏòÆäÖмÓÈëNaOHÈÜÒº£¬Ê¹ÈÜÒº´ÓËáÐÔÖð½¥±äΪ¼îÐÔ£¬Ôڵμӹý³ÌÖк͵μÓÍê±Ïºó£¬¾ùÎÞ³Áµí²úÉú¡£Ôò¿ÉÅųý                   Àë×ӵĴæÔÚ¡£
£¨4£©ÁíÈ¡²¿·ÖÉÏÊö¼îÐÔÈÜÒº£¬ÏòÆäÖмÓÈëNa2CO3ÈÜÒº£¬Óа×É«³ÁµíÉú³É¡£Ö¤Ã÷       
Àë×Ó´æÔÚ£¬ÓÖÅųý          Àë×Ó´æÔÚ¡£
£¨5£©¸ù¾ÝÉÏÊöʵÑéÊÂʵÄÜ·ñÈ·¶¨NH4+ÊÇ·ñ´æÔÚ£¬ÈôÄÜ£¬Çë¼òÊöÀíÓÉ            £¬Èô²»ÄÜ£¬Çë¼òÊö¼ìÑéËüµÄ·½·¨¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
£¨6£©Ô­ÈÜÒºÒ»¶¨´æÔÚµÄÀë×ÓÓР           £¬²»ÄÜÈ·¶¨µÄÀë×ÓÓС¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

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

¾«Ó¢¼Ò½ÌÍø