ÌâÄ¿ÄÚÈÝ

£¨1£©ÏÂͼΪ³£¼ûÒÇÆ÷µÄ²¿·Ö½á¹¹(ÓеÄÒÇÆ÷±»·Å´ó)AͼÖÐÒºÃæËùʾÈÜÒºµÄÌå»ýΪ     mL£¬ÓÃÉÏÊöËÄÖÖÒÇÆ÷ÖеÄijÖÖ²âÁ¿Ò»ÒºÌåµÄÌå»ý£¬Æ½ÊÓʱ¶ÁÊýΪN mL£¬ÑöÊÓʱ¶ÁÊýΪM mL£¬ÈôM>N£¬ÔòËùʹÓõÄÒÇÆ÷ÊÇ____        (Ìî×Öĸ±êºÅ)¡£
£¨2£©ÔÚ»¯Ñ§·ÖÎöÖУ¬³£ÐèÓÃKMnO4±ê×¼ÈÜÒº£¬ÓÉÓÚKMnO4¾§ÌåÔÚÊÒÎÂϲ»Ì«Îȶ¨£¬Òò¶øºÜÄÑÖ±½ÓÅäÖÆ׼ȷÎïÖʵÄÁ¿Å¨¶ÈµÄKMnO4ÈÜÒº¡£ÊµÑéÊÒÒ»°ãÏȳÆÈ¡Ò»¶¨ÖÊÁ¿µÄKMnO4¾§Ì壬´ÖÅä³É´óÖÂŨ¶ÈµÄKMnO4ÈÜÒº£¬ÔÙÓÃÐÔÖÊÎȶ¨¡¢Ïà¶Ô·Ö×ÓÖÊÁ¿½Ï´óµÄ»ù×¼ÎïÖʲÝËáÄÆ[Mr(Na2C2O4)£½134£®0]¶Ô´ÖÅäµÄKMnO4ÈÜÒº½øÐб궨£¬²â³öËùÅäÖƵÄKMnO4ÈÜÒºµÄ׼ȷŨ¶È£¬·´Ó¦Ô­ÀíΪ£º5C2O42£­£«2MnO4£­£«16H£«¡ú10CO2¡ü£«2Mn2+£«8H2O

ÒÔÏÂÊDZ궨KMnO4ÈÜÒºµÄʵÑé²½Ö裺
²½ÖèÒ»£ºÏÈ´ÖÅäŨ¶ÈԼΪ0£®15mol¡¤L-1µÄ¸ßÃÌËá¼ØÈÜÒº500 mL¡£
²½Öè¶þ£º×¼È·³ÆÈ¡Na2C2O4¹ÌÌåm g·ÅÈë׶ÐÎÆ¿ÖУ¬ÓÃÕôÁóË®ÈܽⲢ¼ÓÏ¡ÁòËáËữ£¬¼ÓÈÈÖÁ70¡«80¡æ£¬Óò½ÖèÒ»ËùÅä¸ßÃÌËá¼ØÈÜÒº½øÐеζ¨¡£¼Ç¼Ïà¹ØÊý¾Ý¡£
²½ÖèÈý£º                                                       ¡£
²½ÖèËÄ£º¼ÆËãµÃ¸ßÃÌËá¼ØµÄÎïÖʵÄÁ¿Å¨¶È¡£ÊԻشðÏÂÁÐÎÊÌ⣺
¢Ù¸ÃµÎ¶¨ÊµÑé             £¨¡°ÐèÒª¡±»ò¡°²»ÐèÒª¡±£©¼Óָʾ¼Á¡£
¢Ú²½Öè¶þÖе樲Ù×÷ͼʾÕýÈ·µÄÊÇ__________£¨Ìî±àºÅ£©¡£

¢Û²½Öè¶þµÄµÎ¶¨¹ý³Ìζȱ仯²¢²»Ã÷ÏÔ£¬µ«²Ù×÷¹ý³ÌÖз¢ÏÖÇ°Ò»½×¶ÎÈÜÒºÍÊÉ«½ÏÂý£¬Öмä½×¶ÎÍÊÉ«Ã÷ÏÔ±ä¿ì£¬×îºó½×¶ÎÍÊÉ«ÓÖ±äÂý¡£ÊÔ¸ù¾ÝÓ°Ï컯ѧ·´Ó¦ËÙÂʵÄÌõ¼þ·ÖÎö£¬ÈÜÒºÍÊÉ«Ã÷ÏÔ±ä¿ìµÄÔ­Òò¿ÉÄÜÊÇ_______             £¬×îºóÓÖ±äÂýµÄÔ­ÒòÊÇ         ¡£¢ÜÇëд³ö²½ÖèÈýµÄ²Ù×÷ÄÚÈÝ                           ¡£¢ÝÈômµÄƽ¾ùÊýֵΪ1.340g£¬µÎ¶¨µÄKMnO4ÈÜҺƽ¾ùÓÃÁ¿Îª25.00mL£¬ÔòKMnO4ÈÜÒºµÄŨ¶ÈΪ                 mol¡¤L-1¡£

£¨1£©28.0£»C
£¨2£©¢Ù²»ÐèÒª ¢ÚA
¢ÛÉú³ÉµÄMn2+Ϊ´ß»¯¼Á£¬Ê¹·´Ó¦ËÙÂʱä´ó¡£·´Ó¦ÎïŨ¶È¼õС£¬ËùÒÔËÙÂʱäС
¢ÜÖظ´²½Öè¶þ£¨Á½µ½Èý´Î£©
¢Ý0.1600£¨Ð´0.16»ò 0.160¾ùËãÕýÈ·£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨Ò»£©£¨1£©ÏÂͼΪ³£¼ûÒÇÆ÷µÄ²¿·Ö½á¹¹£¨ÓеÄÒÇÆ÷±»·Å´ó£©

AͼÖÐÒºÃæËùʾÈÜÒºµÄÌå»ýΪ
28.0
28.0
mL£¬ÓÃÉÏÊöËÄÖÖÒÇÆ÷ÖеÄijÖÖ²âÁ¿Ò»ÒºÌåµÄÌå»ý£¬Æ½ÊÓʱ¶ÁÊýΪN mL£¬ÑöÊÓʱ¶ÁÊýΪM mL£¬ÈôM£¾N£¬ÔòËùʹÓõÄÒÇÆ÷ÊÇ
C
C
£¨Ìî×Öĸ±êºÅ£©£®
£¨2£©´ÖÑξ­Ìá´¿ºóµÃµ½NaClÈÜÒº£¬ÔÙ¾­Õô·¢¡¢½á¾§¡¢ºæ¸ÉµÃ¾«ÑΣ®
¢ÙÕô·¢²Ù×÷ÖÐʹÓõ½µÄ´ÉÖÊÒÇÆ÷µÄÃû³ÆΪ
Õô·¢Ãó
Õô·¢Ãó
£»
¢Ú¸Ãͬѧ½«ËùµÃ¾«ÑÎÅä³ÉÈÜÒº£¬ÓÃÓÚÁíһʵÑ飮ʵÑéÖÐÐèÒªÓÃ80mL l mol/LµÄNaClÈÜÒº£¬ÅäÖƹý³ÌÖÐÓÃÍÐÅÌÌìƽ³ÆÈ¡µÄ¾«ÑÎÖÊÁ¿Îª
5.9
5.9
g£¬ÓÃÓÚ¶¨ÈݵIJ£Á§ÒÇÆ÷µÄ¹æ¸ñºÍÃû³ÆΪ
100mLÈÝÁ¿Æ¿
100mLÈÝÁ¿Æ¿
£®
£¨¶þ£©Áù¸öδÌù±êÇ©µÄÊÔ¼ÁÆ¿Öзֱð×°ÓÐÒÔÏÂÏ¡ÈÜÒº¢ÙFeSO4¡¢¢ÚH2SO4¡¢¢ÛBaCl2¡¢¢ÜH2O2¡¢¢ÝAl£¨NO3£©3¡¢¢ÞNaOH£®
£¨1£©Ä³Í¬Ñ§Ïë²»ÓÃÆäËûÊÔ¼Á£¬½öͨ¹ýÓÃÊÔ¹ÜÈ¡ÉÙÁ¿ÉÏÊöÈÜÒº½øÐÐÁ½Á½»ìºÏʵÑé¶ø¸øÊÔ¼ÁÆ¿ÌùÉÏÕýÈ·µÄ±êÇ©£¬ËûÄܳɹ¦Âð£¿
ÄÜ
ÄÜ
 £¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©£®
£¨2£©ÊµÑéÖз¢ÏÖ£¬ÓÐÒ»×éÔÚ»ìºÏʱ£¬Ëæ×ÅÊÔ¼ÁµÎ¼ÓµÄ˳Ðò²»Í¬¶ø³öÏÖÃ÷ÏÔ²»Í¬µÄÏÖÏ󣮸Ã×éÊÇ
¢ÝºÍ¢Þ
¢ÝºÍ¢Þ
£»ÓÐÒ»×éÔÚ»ìºÏʱ£¬Ëæ×ÅÊÔ¼ÁµÎ¼ÓºóµÄʱ¼ä²»Í¬¶ø³öÏÖÃ÷ÏÔ²»Í¬µÄÏÖÏ󣬸Ã×éÊÇ
¢ÙºÍ¢Þ
¢ÙºÍ¢Þ
£¬ÆäÖÐÉæ¼°Ñõ»¯»¹Ô­·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
4Fe£¨OH£©2+O2+2H2O¨T4Fe£¨OH£©3
4Fe£¨OH£©2+O2+2H2O¨T4Fe£¨OH£©3
£®
£¨3£©¼ø±ðºó£¬¸ÃͬѧÓÖÓÃFeSO4×öÁËÈý¸öÌåÏÖFe2+»¹Ô­ÐÔµÄʵÑ飬ÿ´Î¼ÓÈëÉÏÊöÁ½ÖÖÒѼø±ðµÄÎïÖʵÄÈÜÒºÓëÆä»ìºÏ£®ÊÔд³öÆäÖÐÁ½¸ö·´Ó¦µÄÀë×Ó·½³Ìʽ
3Fe2++4H++NO3-¨T3Fe3++NO¡ü+2H2O
3Fe2++4H++NO3-¨T3Fe3++NO¡ü+2H2O
¡¢
2Fe2++2H++H2O2¨T2Fe3++2H2O
2Fe2++2H++H2O2¨T2Fe3++2H2O
£®
£¨4£©Êµ¼ÊÉÏʵÑéÊÒÅäÖƵÄFeSO4ÈÜÒº²»¿É³¤ÆÚ±£´æ£¬¶ÌÆÚ±£´æʱ¿ÉÔÚFeSO4ÈÜÒºÅäÖÆʱ¼ÓÈëÉÙÁ¿µÄÉÏÊö
¢Ú
¢Ú
£¨Ìî±àºÅ£©·Àֹˮ½â£¬ÈôÄÜÔÙ¼ÓÉÏ
¼¸Ã¶Ìú¶¤
¼¸Ã¶Ìú¶¤
£¨ÌîÎïÖÊÃû³Æ£©Ð§¹û»á¸üºÃ£®

£¨1£©ÏÂͼΪ³£¼ûÒÇÆ÷µÄ²¿·Ö½á¹¹(ÓеÄÒÇÆ÷±»·Å´ó)AͼÖÐÒºÃæËùʾÈÜÒºµÄÌå»ýΪ      mL£¬ÓÃÉÏÊöËÄÖÖÒÇÆ÷ÖеÄijÖÖ²âÁ¿Ò»ÒºÌåµÄÌå»ý£¬Æ½ÊÓʱ¶ÁÊýΪN mL£¬ÑöÊÓʱ¶ÁÊýΪM mL£¬ÈôM>N£¬ÔòËùʹÓõÄÒÇÆ÷ÊÇ____        (Ìî×Öĸ±êºÅ)¡£

£¨2£©ÔÚ»¯Ñ§·ÖÎöÖУ¬³£ÐèÓÃKMnO4±ê×¼ÈÜÒº£¬ÓÉÓÚKMnO4¾§ÌåÔÚÊÒÎÂϲ»Ì«Îȶ¨£¬Òò¶øºÜÄÑÖ±½ÓÅäÖÆ׼ȷÎïÖʵÄÁ¿Å¨¶ÈµÄKMnO4ÈÜÒº¡£ÊµÑéÊÒÒ»°ãÏȳÆÈ¡Ò»¶¨ÖÊÁ¿µÄKMnO4¾§Ì壬´ÖÅä³É´óÖÂŨ¶ÈµÄKMnO4ÈÜÒº£¬ÔÙÓÃÐÔÖÊÎȶ¨¡¢Ïà¶Ô·Ö×ÓÖÊÁ¿½Ï´óµÄ»ù×¼ÎïÖʲÝËáÄÆ[Mr(Na2C2O4)£½134£®0]¶Ô´ÖÅäµÄKMnO4ÈÜÒº½øÐб궨£¬²â³öËùÅäÖƵÄKMnO4ÈÜÒºµÄ׼ȷŨ¶È£¬·´Ó¦Ô­ÀíΪ£º5C2O42£­£«2MnO4£­£«16H£«¡ú10CO2¡ü£«2Mn2+£«8H2O

ÒÔÏÂÊDZ궨KMnO4ÈÜÒºµÄʵÑé²½Ö裺

²½ÖèÒ»£ºÏÈ´ÖÅäŨ¶ÈԼΪ0£®15mol¡¤L-1µÄ¸ßÃÌËá¼ØÈÜÒº500 mL¡£

²½Öè¶þ£º×¼È·³ÆÈ¡Na2C2O4¹ÌÌåm g·ÅÈë׶ÐÎÆ¿ÖУ¬ÓÃÕôÁóË®ÈܽⲢ¼ÓÏ¡ÁòËáËữ£¬¼ÓÈÈÖÁ70¡«80¡æ£¬Óò½ÖèÒ»ËùÅä¸ßÃÌËá¼ØÈÜÒº½øÐеζ¨¡£¼Ç¼Ïà¹ØÊý¾Ý¡£

²½ÖèÈý£º                                                       ¡£

²½ÖèËÄ£º¼ÆËãµÃ¸ßÃÌËá¼ØµÄÎïÖʵÄÁ¿Å¨¶È¡£ÊԻشðÏÂÁÐÎÊÌ⣺

¢Ù¸ÃµÎ¶¨ÊµÑé             £¨¡°ÐèÒª¡±»ò¡°²»ÐèÒª¡±£©¼Óָʾ¼Á¡£

¢Ú²½Öè¶þÖе樲Ù×÷ͼʾÕýÈ·µÄÊÇ__________£¨Ìî±àºÅ£©¡£

¢Û²½Öè¶þµÄµÎ¶¨¹ý³Ìζȱ仯²¢²»Ã÷ÏÔ£¬µ«²Ù×÷¹ý³ÌÖз¢ÏÖÇ°Ò»½×¶ÎÈÜÒºÍÊÉ«½ÏÂý£¬Öмä½×¶ÎÍÊÉ«Ã÷ÏÔ±ä¿ì£¬×îºó½×¶ÎÍÊÉ«ÓÖ±äÂý¡£ÊÔ¸ù¾ÝÓ°Ï컯ѧ·´Ó¦ËÙÂʵÄÌõ¼þ·ÖÎö£¬ÈÜÒºÍÊÉ«Ã÷ÏÔ±ä¿ìµÄÔ­Òò¿ÉÄÜÊÇ_______             £¬×îºóÓÖ±äÂýµÄÔ­ÒòÊÇ         ¡£¢ÜÇëд³ö²½ÖèÈýµÄ²Ù×÷ÄÚÈÝ                           ¡£¢ÝÈômµÄƽ¾ùÊýֵΪ1.340g£¬µÎ¶¨µÄKMnO4ÈÜҺƽ¾ùÓÃÁ¿Îª25.00mL£¬ÔòKMnO4ÈÜÒºµÄŨ¶ÈΪ                 mol¡¤L-1¡£

 

£¨1£©ÏÂͼΪ³£¼ûÒÇÆ÷µÄ²¿·Ö½á¹¹(ÓеÄÒÇÆ÷±»·Å´ó)AͼÖÐÒºÃæËùʾÈÜÒºµÄÌå»ýΪ       mL£¬ÓÃÉÏÊöËÄÖÖÒÇÆ÷ÖеÄijÖÖ²âÁ¿Ò»ÒºÌåµÄÌå»ý£¬Æ½ÊÓʱ¶ÁÊýΪN mL£¬ÑöÊÓʱ¶ÁÊýΪM mL£¬ÈôM>N£¬ÔòËùʹÓõÄÒÇÆ÷ÊÇ____         (Ìî×Öĸ±êºÅ)¡£

£¨2£©ÔÚ»¯Ñ§·ÖÎöÖУ¬³£ÐèÓÃKMnO4±ê×¼ÈÜÒº£¬ÓÉÓÚKMnO4¾§ÌåÔÚÊÒÎÂϲ»Ì«Îȶ¨£¬Òò¶øºÜÄÑÖ±½ÓÅäÖÆ׼ȷÎïÖʵÄÁ¿Å¨¶ÈµÄKMnO4ÈÜÒº¡£ÊµÑéÊÒÒ»°ãÏȳÆÈ¡Ò»¶¨ÖÊÁ¿µÄKMnO4¾§Ì壬´ÖÅä³É´óÖÂŨ¶ÈµÄKMnO4ÈÜÒº£¬ÔÙÓÃÐÔÖÊÎȶ¨¡¢Ïà¶Ô·Ö×ÓÖÊÁ¿½Ï´óµÄ»ù×¼ÎïÖʲÝËáÄÆ[Mr(Na2C2O4)£½134£®0]¶Ô´ÖÅäµÄKMnO4ÈÜÒº½øÐб궨£¬²â³öËùÅäÖƵÄKMnO4ÈÜÒºµÄ׼ȷŨ¶È£¬·´Ó¦Ô­ÀíΪ£º5C2O42£­£«2MnO4£­£«16H£«¡ú10CO2¡ü£«2Mn2+£«8H2O

ÒÔÏÂÊDZ궨KMnO4ÈÜÒºµÄʵÑé²½Ö裺

²½ÖèÒ»£ºÏÈ´ÖÅäŨ¶ÈԼΪ0£®15mol¡¤L-1µÄ¸ßÃÌËá¼ØÈÜÒº500 mL¡£

²½Öè¶þ£º×¼È·³ÆÈ¡Na2C2O4¹ÌÌåm g·ÅÈë׶ÐÎÆ¿ÖУ¬ÓÃÕôÁóË®ÈܽⲢ¼ÓÏ¡ÁòËáËữ£¬¼ÓÈÈÖÁ70¡«80¡æ£¬Óò½ÖèÒ»ËùÅä¸ßÃÌËá¼ØÈÜÒº½øÐеζ¨¡£¼Ç¼Ïà¹ØÊý¾Ý¡£

²½ÖèÈý£º                                                        ¡£

²½ÖèËÄ£º¼ÆËãµÃ¸ßÃÌËá¼ØµÄÎïÖʵÄÁ¿Å¨¶È¡£ÊԻشðÏÂÁÐÎÊÌ⣺

¢Ù¸ÃµÎ¶¨ÊµÑé              £¨¡°ÐèÒª¡±»ò¡°²»ÐèÒª¡±£©¼Óָʾ¼Á¡£

¢Ú²½Öè¶þÖе樲Ù×÷ͼʾÕýÈ·µÄÊÇ__________£¨Ìî±àºÅ£©¡£

¢Û²½Öè¶þµÄµÎ¶¨¹ý³Ìζȱ仯²¢²»Ã÷ÏÔ£¬µ«²Ù×÷¹ý³ÌÖз¢ÏÖÇ°Ò»½×¶ÎÈÜÒºÍÊÉ«½ÏÂý£¬Öмä½×¶ÎÍÊÉ«Ã÷ÏÔ±ä¿ì£¬×îºó½×¶ÎÍÊÉ«ÓÖ±äÂý¡£ÊÔ¸ù¾ÝÓ°Ï컯ѧ·´Ó¦ËÙÂʵÄÌõ¼þ·ÖÎö£¬ÈÜÒºÍÊÉ«Ã÷ÏÔ±ä¿ìµÄÔ­Òò¿ÉÄÜÊÇ_______              £¬×îºóÓÖ±äÂýµÄÔ­ÒòÊÇ          ¡£¢ÜÇëд³ö²½ÖèÈýµÄ²Ù×÷ÄÚÈÝ                            ¡£¢ÝÈômµÄƽ¾ùÊýֵΪ1.340g£¬µÎ¶¨µÄKMnO4ÈÜҺƽ¾ùÓÃÁ¿Îª25.00mL£¬ÔòKMnO4ÈÜÒºµÄŨ¶ÈΪ                  mol¡¤L-1¡£

 

£¨Ò»£©£¨1£©ÏÂͼΪ³£¼ûÒÇÆ÷µÄ²¿·Ö½á¹¹£¨ÓеÄÒÇÆ÷±»·Å´ó£©

AͼÖÐÒºÃæËùʾÈÜÒºµÄÌå»ýΪ______mL£¬ÓÃÉÏÊöËÄÖÖÒÇÆ÷ÖеÄijÖÖ²âÁ¿Ò»ÒºÌåµÄÌå»ý£¬Æ½ÊÓʱ¶ÁÊýΪN mL£¬ÑöÊÓʱ¶ÁÊýΪM mL£¬ÈôM£¾N£¬ÔòËùʹÓõÄÒÇÆ÷ÊÇ______£¨Ìî×Öĸ±êºÅ£©£®
£¨2£©´ÖÑξ­Ìá´¿ºóµÃµ½NaClÈÜÒº£¬ÔÙ¾­Õô·¢¡¢½á¾§¡¢ºæ¸ÉµÃ¾«ÑΣ®
¢ÙÕô·¢²Ù×÷ÖÐʹÓõ½µÄ´ÉÖÊÒÇÆ÷µÄÃû³ÆΪ______£»
¢Ú¸Ãͬѧ½«ËùµÃ¾«ÑÎÅä³ÉÈÜÒº£¬ÓÃÓÚÁíһʵÑ飮ʵÑéÖÐÐèÒªÓÃ80mL l mol/LµÄNaClÈÜÒº£¬ÅäÖƹý³ÌÖÐÓÃÍÐÅÌÌìƽ³ÆÈ¡µÄ¾«ÑÎÖÊÁ¿Îª______g£¬ÓÃÓÚ¶¨ÈݵIJ£Á§ÒÇÆ÷µÄ¹æ¸ñºÍÃû³ÆΪ______£®
£¨¶þ£©Áù¸öδÌù±êÇ©µÄÊÔ¼ÁÆ¿Öзֱð×°ÓÐÒÔÏÂÏ¡ÈÜÒº¢ÙFeSO4¡¢¢ÚH2SO4¡¢¢ÛBaCl2¡¢¢ÜH2O2¡¢¢ÝAl£¨NO3£©3¡¢¢ÞNaOH£®
£¨1£©Ä³Í¬Ñ§Ïë²»ÓÃÆäËûÊÔ¼Á£¬½öͨ¹ýÓÃÊÔ¹ÜÈ¡ÉÙÁ¿ÉÏÊöÈÜÒº½øÐÐÁ½Á½»ìºÏʵÑé¶ø¸øÊÔ¼ÁÆ¿ÌùÉÏÕýÈ·µÄ±êÇ©£¬ËûÄܳɹ¦Âð£¿______ £¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©£®
£¨2£©ÊµÑéÖз¢ÏÖ£¬ÓÐÒ»×éÔÚ»ìºÏʱ£¬Ëæ×ÅÊÔ¼ÁµÎ¼ÓµÄ˳Ðò²»Í¬¶ø³öÏÖÃ÷ÏÔ²»Í¬µÄÏÖÏ󣮸Ã×éÊÇ______£»ÓÐÒ»×éÔÚ»ìºÏʱ£¬Ëæ×ÅÊÔ¼ÁµÎ¼ÓºóµÄʱ¼ä²»Í¬¶ø³öÏÖÃ÷ÏÔ²»Í¬µÄÏÖÏ󣬸Ã×éÊÇ______£¬ÆäÖÐÉæ¼°Ñõ»¯»¹Ô­·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______£®
£¨3£©¼ø±ðºó£¬¸ÃͬѧÓÖÓÃFeSO4×öÁËÈý¸öÌåÏÖFe2+»¹Ô­ÐÔµÄʵÑ飬ÿ´Î¼ÓÈëÉÏÊöÁ½ÖÖÒѼø±ðµÄÎïÖʵÄÈÜÒºÓëÆä»ìºÏ£®ÊÔд³öÆäÖÐÁ½¸ö·´Ó¦µÄÀë×Ó·½³Ìʽ______¡¢______£®
£¨4£©Êµ¼ÊÉÏʵÑéÊÒÅäÖƵÄFeSO4ÈÜÒº²»¿É³¤ÆÚ±£´æ£¬¶ÌÆÚ±£´æʱ¿ÉÔÚFeSO4ÈÜÒºÅäÖÆʱ¼ÓÈëÉÙÁ¿µÄÉÏÊö______£¨Ìî±àºÅ£©·Àֹˮ½â£¬ÈôÄÜÔÙ¼ÓÉÏ______£¨ÌîÎïÖÊÃû³Æ£©Ð§¹û»á¸üºÃ£®

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

¾«Ó¢¼Ò½ÌÍø