ÌâÄ¿ÄÚÈÝ

(10·Ö) ijͬѧÓÃÈçͼËùʾװÖÃ̽¾¿SO2µÄÐÔÖʼ°ÆäÓйØÊµÑ飮

(1)ʵÑéÊÒÓÃÑÇÁòËáÄÆ¹ÌÌåºÍÒ»¶¨Å¨¶ÈµÄÁòËá·´Ó¦ÖÆ±¸SO2ÆøÌ壬д³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ______________________________________________________________
(2)·Ö±ð½«SO2ÆøÌåͨÈëÏÂÁÐCÈÜÒºÖУ¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÉÙÁ¿SO2ͨÈë×ÏɫʯÈïÊÔÒº£¬ÏÖÏóÊÇ______________£¬¼ÌÐøÍ¨Èë¹ýÁ¿SO2ÆøÌ壬ÏÖÏóÊÇ________________£®
¢ÚSO2ͨÈë×ÏÉ«KMnO4ÈÜÒº£¬ÏÖÏóÊÇ______________£¬Æä·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________________£®
¢Û¹ýÁ¿SO2ÂýÂýµØÍ¨Èë³ÎÇåʯ»ÒË®ÖУ¬ÏÖÏó___________________________________
¢ÜÈôCΪ˫ÑõË®£¬ÔòͨÈëSO2ºó£¬Çë´óµ¨ÅжÏËùµÃÈÜÒºÊÇ________(ÌîÈÜÖʵĻ¯Ñ§Ê½)£¬Èô¸ÃÍ¬Ñ§ÖÆ±¸µÄSO2ÆøÌåÖлìÓÐCO2ÆøÌ壬²úÉúÔÓÖʵÄÔ­Òò¿ÉÄÜÊÇÑÇÁòËáÄÆ¹ÌÌåÖлìÓÐ__________£®

(1)Na2SO3£«H2SO4===Na2SO4£«SO2¡ü£«H2O
(2)¢Ù±äºì£¨1·Ö£©¡¡²»ÍÊÉ«»ò²»Ã÷ÏÔ
¢ÚÑÕÉ«Ö𽥱䵭»ò×ÏÉ«Ïûʧ    5SO2£«2MnO£«2H2O===5SO£«2Mn2£«£«4H£«
¢ÛÏȱä»ë×Ç£¬ºóÓÖ±ä³ÎÇå
¢ÜH2SO4    ¡¡Ì¼ËáÑλò̼ËáÇâÑÎ

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÊµÑéÊÒÖÆ±¸SO2µÄ·½³ÌʽΪ£ºNa2SO3£«H2SO4===Na2SO4£«SO2¡ü£«H2O
£¨2£©¢ÙSO2ËäÈ»ÓÐÆ¯°×ÐÔ£¬µ«ÊDz»ÄÜÆ¯°×ʯÈֻÄÜʹʯÈï±äºì²»ÄÜÍÊÉ«¡£
¢ÚSO2¾ßÓÐÇ¿»¹Ô­ÐÔËùÒÔͨÈëµ½KMnO4ÈÜÒºÖУ¬KMnO4ÈÜÒº×ÏÉ«ÍÊÈ¥¡£·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º5SO2£«2MnO£«2H2O===5SO£«2Mn2£«£«4H£«¡£
¢ÛCaSO3ÄÑÈÜCa(HSO3)2¿ÉÈÜ£¬ËùÒÔSO2ͨÈë³ÎÇåʯ»ÒË®ÖвúÉú³Áµí£¬¼ÌÐøÍ¨È룬³ÁµíÈܽ⡣
¢ÜH2O2¾ßÓÐÑõ»¯ÐÔ£¬SO2¾ßÓл¹Ô­ÐÔ£¬ËùÒÔ½«SO2ͨÈëµ½H2O2ÖÐÄÜ·´Ó¦Éú³ÉH2SO4¡£Èç¹ûÖÆµÃµÄSO2»ìÓÐCO2ÔÓÖÊ£¬ËµÃ÷Na2SO3ÖлìÓÐÁËNa2CO3»òNaHCO3¡£
¿¼µã£ºSO2µÄÐÔÖÊ
µãÆÀ£º±¾Ìâ·Ç³£»ù´¡£¬Ö÷Òª¿¼²éѧÉú¶ÔSO2ÐÔÖʵÄÕÆÎÕ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2011?ɽ¶«£©ÊµÑéÊÒÒÔº¬ÓÐCa2+¡¢Mg2+¡¢Cl-¡¢SO42-¡¢Br-µÈÀë×ӵıˮΪÖ÷ÒªÔ­ÁÏÖÆ±¸ÎÞË®CaCl2ºÍBr2£¬Á÷³ÌÈçÏ£º

£¨1£©²Ù×÷¢ñʹÓõÄÊÔ¼ÁÊÇ
ËÄÂÈ»¯Ì¼
ËÄÂÈ»¯Ì¼
£¬ËùÓõÄÖ÷ÒªÒÇÆ÷Ãû³ÆÊÇ
·ÖҺ©¶·
·ÖҺ©¶·
£®
£¨2£©¼ÓÈëÈÜÒºWµÄÄ¿µÄÊÇ
³ýÈ¥ÈÜÒºÖÐSO42-
³ýÈ¥ÈÜÒºÖÐSO42-
£®ÓÃCaOµ÷½ÚÈÜÒºYµÄpH£¬¿ÉÒÔ³ýÈ¥Mg2+£®ÓɱíÖÐÊý¾Ý¿ÉÖª£¬ÀíÂÛÉÏ¿ÉÑ¡ÔñµÄpH×î´ó·¶Î§ÊÇ
11.0¡ÜpH¡Ü12.2
11.0¡ÜpH¡Ü12.2
£®ËữÈÜÒºZʱ£¬Ê¹ÓõÄÊÔ¼ÁΪ
ÑÎËá
ÑÎËá
£®
¿ªÊ¼³ÁµíʱµÄpH ³ÁµíÍêȫʱµÄpH
Mg2+ 9.6 11.0
Ca2+ 12.2 c£¨OH-£©=1.8mol?L-1
£¨3£©ÊµÑéÊÒÓñ´¿ÇÓëÏ¡ÑÎËá·´Ó¦ÖÆ±¸²¢ÊÕ¼¯ÆøÌ壬ÏÂÁÐ×°ÖÃÖкÏÀíµÄÊÇ
b¡¢d
b¡¢d
£®

£¨4£©³£ÎÂÏ£¬H2SO3µÄµçÀë³£ÊýKa1=1.2¡Á10-2£¬Ka2=6.3¡Á10-8£»H2CO3µÄµçÀë³£ÊýKa1=4.5¡Á10-7£¬Ka2=4.7¡Á10-11£®Ä³Í¬Ñ§Éè¼ÆÊµÑéÑéÖ¤H2SO3ËáÐÔÇ¿ÓÚH2CO3£º½«SO2ºÍCO2ÆøÌå·Ö±ðͨÈëË®ÖÐÖÁ±¥ºÍ£¬Á¢¼´ÓÃËá¶È¼Æ²âÁ¿ÈÜÒºµÄpH£¬ÈôǰÕßµÄpHСÓÚºóÕߣ¬ÔòH2SO3ËáÐÔÇ¿ÓÚH2CO3£®¸ÃʵÑéÉè¼Æ²»ÕýÈ·£¬´íÎóÔÚÓÚ
ÓÃÓڱȽÏpHµÄÁ½ÖÖËáµÄÎïÖʵÄÁ¿Å¨¶È²»ÏàµÈ
ÓÃÓڱȽÏpHµÄÁ½ÖÖËáµÄÎïÖʵÄÁ¿Å¨¶È²»ÏàµÈ
£®
Éè¼ÆºÏÀíʵÑéÑéÖ¤H2SO3ËáÐÔÇ¿ÓÚH2CO3£¨¼òҪ˵Ã÷ʵÑé²½Öè¡¢ÏÖÏóºÍ½áÂÛ£©£®
ÈýÖֲο¼·½°¸ÈçÏ£º
·½°¸Ò»£ºÅäÖÆÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄNaHSO3ºÍNaHCO3ÈÜÒº£®ÓÃËá¶È¼Æ£¨»òpHÊÔÖ½£©²âÁ½ÈÜÒºµÄpH£®Ç°ÕßµÄpHСÓÚºóÕߣ¬Ö¤Ã÷H2SO3ËáÐÔÇ¿ÓÚH2CO3£®
·½°¸¶þ£º½«SO2ÆøÌåÒÀ´Îͨ¹ýNaHCO3£¨»òNa2CO3£©ÈÜÒº¡¢ËáÐÔKMnO4ÈÜÒº¡¢Æ·ºìÈÜÒº¡¢³ÎÇåʯ»ÒË®£®Æ·ºìÈÜÒº²»ÍÊÉ«£¬ÇÒ³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Ö¤Ã÷H2SO3ËáÐÔÇ¿ÓÚH2CO3£®
·½°¸Èý£º½«CO2ÆøÌåÒÀ´Îͨ¹ýNaHSO3£¨»òNa2SO3£©ÈÜÒº¡¢Æ·ºìÈÜÒº£®Æ·ºìÈÜÒº²»ÍÊÉ«£¬Ö¤Ã÷H2SO3ËáÐÔÇ¿ÓÚH2CO3
ÈýÖֲο¼·½°¸ÈçÏ£º
·½°¸Ò»£ºÅäÖÆÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄNaHSO3ºÍNaHCO3ÈÜÒº£®ÓÃËá¶È¼Æ£¨»òpHÊÔÖ½£©²âÁ½ÈÜÒºµÄpH£®Ç°ÕßµÄpHСÓÚºóÕߣ¬Ö¤Ã÷H2SO3ËáÐÔÇ¿ÓÚH2CO3£®
·½°¸¶þ£º½«SO2ÆøÌåÒÀ´Îͨ¹ýNaHCO3£¨»òNa2CO3£©ÈÜÒº¡¢ËáÐÔKMnO4ÈÜÒº¡¢Æ·ºìÈÜÒº¡¢³ÎÇåʯ»ÒË®£®Æ·ºìÈÜÒº²»ÍÊÉ«£¬ÇÒ³ÎÇåʯ»ÒË®±ä»ë×Ç£¬Ö¤Ã÷H2SO3ËáÐÔÇ¿ÓÚH2CO3£®
·½°¸Èý£º½«CO2ÆøÌåÒÀ´Îͨ¹ýNaHSO3£¨»òNa2SO3£©ÈÜÒº¡¢Æ·ºìÈÜÒº£®Æ·ºìÈÜÒº²»ÍÊÉ«£¬Ö¤Ã÷H2SO3ËáÐÔÇ¿ÓÚH2CO3
£®
ÒÇÆ÷×ÔÑ¡£®
¹©Ñ¡ÔñµÄÊÔ¼Á£ºCO2¡¢SO2¡¢Na2CO3¡¢NaHCO3¡¢Na2SO3¡¢NaHSO3¡¢ÕôÁóË®¡¢±¥ºÍʯ»ÒË®¡¢ËáÐÔKMnO4ÈÜÒº¡¢Æ·ºìÈÜÒº¡¢pHÊÔÖ½£®
µª»¯ÂÁ£¨AIN£©ÊÇÒ»ÖÖÐÂÐÍÎÞ»ú²ÄÁÏ£¬¹ã·ºÓ¦ÓÃÓÚ¼¯³Éµç·Éú²úÁìÓò£®Ä³»¯Ñ§Ñо¿Ð¡×éÀûÓÃ ÖÆÈ¡µª»¯ÂÁ£¬Éè¼ÆÈçͼ1ËùʾµÄʵÑé×°Öã®
¾«Ó¢¼Ò½ÌÍø
ÊԻشð£º
£¨1£©ÊµÑéÖÐÓñ¥ºÍNaNO2Óë NH4CÈÜÒºÖÆÈ¡ÑõÆøµÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨2£©×°ÖÃÖзÖҺ©¶·ÓëÕôÁóÉÕÆ¿Ö®¼äµÄµ¼¹ÜAµÄ×÷ÓÃÊÇ
 
£¨ÌîдÐòºÅ£©£®
a£®·ÀÖ¹NaNO2 ±¥ºÍÈÜÒºÕô·¢   b£®±£Ö¤ÊµÑé×°Öò»Â©Æø  c£®Ê¹NaNO2 ±¥ºÍÈÜÒºÈÝÒ×µÎÏÂ
£¨3£©°´Í¼Á¬½ÓºÃʵÑé×°Ö㬼ì²é×°ÖÃÆøÃÜÐԵķ½·¨ÊÇ
 
£®
£¨4£©»¯Ñ§Ñо¿Ð¡×éµÄ×°ÖôæÔÚÑÏÖØÎÊÌ⣬Çë˵Ã÷¸Ä½øµÄ°ì·¨
 
£®
£¨5£©·´Ó¦½áÊøºó£¬Ä³Í¬Ñ§ÓÃͼ2ËùʾװÖýøÐÐʵÑéÀ´²â¶¨µª»¯ÂÁÑù Æ·µÄÖÊÁ¿·ÖÊý£¨ÊµÑéÖе¼¹ÜÌå»ýºöÂÔ²»¼Æ£©£®ÒÑÖª£ºµª»¯ÂÁºÍNaOHÈÜÒº·´Ó¦Éú³ÉNa[Al£¨OH£©4]ºÍ°±Æø£®
¢Ù¹ã¿ÚÆ¿ÖеÄÊÔ¼ÁX×îºÃÑ¡ÓÃ
 
£¨ÌîдÐòºÅ£©£®a£®ÆûÓÍ        b£®¾Æ¾«       c£®Ö²ÎïÓÍ     d£®CCl4
¢Ú¹ã¿ÚÆ¿ÖеÄÒºÌåûÓÐ×°Âú£¨ÉÏ·½ÁôÓпռ䣩£¬ÔòʵÑé²âµÃNH3µÄÌå»ý½«
 
£¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢¡°²»±ä¡±£©£®
¢ÛÈôʵÑéÖгÆÈ¡µª»¯ÂÁÑùÆ·µÄÖÊÁ¿Îª10.0g£¬²âµÃ°±ÆøµÄÌå»ýΪ3.36L£¨±ê×¼×´¿ö£©£¬ÔòÑùÆ·ÖÐAlNµÄÖÊÁ¿·ÖÊýΪ
 
£®

¡¾10·Ö¡¿Ä³Ñ§ÉúÓõ¨·¯ÖÆÈ¡Ñõ»¯Í­¹ÌÌ壬²¢Ñо¿Ñõ»¯Í­ÄÜ·ñÔÚÂÈËá¼ØÊÜÈÈ·Ö½âʵÑéÖÐÆð´ß»¯×÷Óá£ÊµÑé²½ÖèÈçÏ£º

¢Ù³ÆÁ¿a gµ¨·¯¹ÌÌå·ÅÈëÉÕ±­ÖУ¬¼ÓË®ÖÆ³ÉÈÜÒº£¬ÏòÆäÖеμÓÇâÑõ»¯ÄÆÈÜÒºÖÁ³ÁµíÍêÈ«£»

¢Ú°Ñ²½Öè¢ÙÖеÄÈÜÒººÍ³Áµí×ªÒÆÖÁÕô·¢ÃóÖУ¬¼ÓÈÈÖÁÈÜÒºÖеijÁµíÈ«²¿±ä³ÉºÚÉ«Ñõ»¯Í­ÎªÖ¹£»

¢Û¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔ³ÆÁ¿ËùµÃ¹ÌÌåÖÊÁ¿Îªb g£»

¢Üȡһ¶¨ÖÊÁ¿µÄÉÏÊöÑõ»¯Í­¹ÌÌåºÍÒ»¶¨ÖÊÁ¿µÄÂÈËá¼Ø¹ÌÌ壬»ìºÏ¾ùÔȺó¼ÓÈÈ£¬ÊÕ¼¯·´Ó¦Éú³ÉµÄÑõÆø£¬ÈçÏÂͼËùʾ¡£

                      

Çë»Ø´ðÏÂÁÐÎÊÌâ¡£

(1)ÉÏÊö¸÷²½²Ù×÷ÖУ¬ÐèÒªÓõ½²£Á§°ôµÄÊÇ_________ (ÌîÐ´Ç°ÃæËùÊöʵÑé²½ÖèµÄÐòºÅ)¡£

(2)Óɵ¨·¯ÖƱ¸Ñõ»¯Í­µÄ²úÂÊ(ʵ¼Ê²úÁ¿ÓëÀíÂÛ²úÁ¿µÄ°Ù·Ö±È)Ϊ_________¡Á100%¡£

(3)Ϊ±£Ö¤Cu2+³ÁµíÍêÈ«£¬²½Öè¢ÙÖÐÈÜÒºµÄpHÓ¦´óÓÚ10¡£¼òÊöÓÃpHÊÔÖ½²â¶¨ÈÜÒºpHµÄ²Ù×÷£º____________________________________________________________________¡£

(4)Ϊ֤Ã÷Ñõ»¯Í­ÔÚÂÈËá¼ØµÄ·Ö½â·´Ó¦ÖÐÆð´ß»¯×÷Óã¬ÔÚÉÏÊöʵÑé¢Ù¡ª¢Üºó»¹Ó¦¸Ã½øÐеÄʵÑé²Ù×÷ÊÇ(°´ÊµÑéÏȺó˳ÐòÌîд×ÖĸÐòºÅ) _________¡£

a.¹ýÂË  b.ºæ¸É  c.ÈÜ½â  d.Ï´µÓ  e.³ÆÁ¿

(5)µ«ÓеÄͬѧÈÏΪ£¬»¹±ØÐëÁíÍâÔÙÉè¼ÆÒ»¸öʵÑé²ÅÄÜÖ¤Ã÷Ñõ»¯Í­ÔÚÂÈËá¼ØÊÜÈÈ·Ö½âµÄʵÑéÖÐÆð´ß»¯×÷Óá£ÄãÈÏΪ»¹Ó¦¸Ã½øÐеÄÁíÒ»¸öʵÑéÊÇ_________________________________¡£

£¨12·Öÿ¿ÕÁ½·Ö£©Ä³Í¬Ñ§ÓÃ0.10 mol/LµÄHClÈÜÒº²â¶¨Î´ÖªÅ¨¶ÈµÄNaOHÈÜÒº£¬ÆäʵÑé²Ù×÷ÈçÏ£º

A£®ÓÃËáʽµÎ¶¨¹ÜÁ¿È¡20.00mLHClÈÜҺעÈë×¶ÐÎÆ¿£¬Í¬Ê±µÎ¼Ó2-3µÎ·Ó̪ÊÔÒº£»
B£®ÓÃ0.10 mol/LµÄHClÈÜÒºÈóÏ´ËáʽµÎ¶¨¹Ü£»
C£®°ÑµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»£»
D£®È¡Ï¼îʽµÎ¶¨¹Ü£¬Óôý²âNaOHÈÜÒºÈóÏ´ºó£¬½«´ý²âNaOHÈÜҺעÈë¼îʽµÎ¶¨¹ÜÖÁ¾àÀë¿Ì¶È¡°0¡±ÒÔÉÏ2¡ª3cm´¦£¬ÔٰѼîʽµÎ¶¨¹Ü¹Ì¶¨ºÃ£¬µ÷½ÚÒºÃæ£»
E£®¼ì²éµÎ¶¨¹ÜÊÇ·ñ©ˮ£»
F£®Áíȡ׶ÐÎÆ¿£¬ÔÙÖØ¸´ÒÔÉϲÙ×÷1¡ª2 ´Î£»
G£®°Ñ×¶ÐÎÆ¿·ÅÔÚ¼îʽµÎ¶¨¹Üϱߣ¬Æ¿ÏµæÒ»ÕŰ×Ö½£¬±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿£¬Ö±µ½¼ÓÈëlµÎ¼îÒººóÈÜÒºÑÕɫͻ±ä²¢ÔÚ°ë·ÖÖÓÄÚ²»ÔÙ±äɫΪֹ£¬¼ÇÏµζ¨¹ÜÒºÃæËùÔڵĿ̶ȡ£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)£¨2·Ö£©µÎ¶¨²Ù×÷µÄÕýȷ˳ÐòÊÇ£º(Ìî×Öĸ)     ¡úC¡ú     ¡úB¡ú     ¡ú    ¡ú   __       ¡£
(1) G²½²Ù×÷ÖÐÔÚ×¶ÐÎÆ¿ÏµæÒ»ÕŰ×Ö½ÊǵÄ×÷ÓÃÊÇ                 ¡£
(3)D²½²Ù×÷ÖÐÒºÃæÓ¦µ÷½Úµ½           £¬¼â×첿·ÖÓ¦           ¡£
(4)µÎ¶¨ÖÕµã¶ÁÊýʱ£¬Èç¹ûÑöÊÓÒºÃæ£¬¶Á³öµÄÊýÖµ                  £¬ÈôµÎ¶¨Ç°Æ½ÊÓ¶ÁÊýÔòÓɴ˼ÆËãµÃµ½µÄNaOHÈÜҺŨ¶È       ¡£(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢¡°ÎÞÓ°Ï족)

£¨12·Ö ÿ¿ÕÁ½·Ö£©Ä³Í¬Ñ§ÓÃ0.10 mol/LµÄHClÈÜÒº²â¶¨Î´ÖªÅ¨¶ÈµÄNaOHÈÜÒº£¬ÆäʵÑé²Ù×÷ÈçÏ£º

A£®ÓÃËáʽµÎ¶¨¹ÜÁ¿È¡20.00mLHClÈÜҺעÈë×¶ÐÎÆ¿£¬Í¬Ê±µÎ¼Ó2-3µÎ·Ó̪ÊÔÒº£»

B£®ÓÃ0.10 mol/LµÄHClÈÜÒºÈóÏ´ËáʽµÎ¶¨¹Ü£»

C£®°ÑµÎ¶¨¹ÜÓÃÕôÁóˮϴ¾»£»

D£®È¡Ï¼îʽµÎ¶¨¹Ü£¬Óôý²âNaOHÈÜÒºÈóÏ´ºó£¬½«´ý²âNaOHÈÜҺעÈë¼îʽµÎ¶¨¹ÜÖÁ¾àÀë¿Ì¶È¡°0¡±ÒÔÉÏ2¡ª3cm´¦£¬ÔٰѼîʽµÎ¶¨¹Ü¹Ì¶¨ºÃ£¬µ÷½ÚÒºÃæ£»

E£®¼ì²éµÎ¶¨¹ÜÊÇ·ñ©ˮ£»

F£®Áíȡ׶ÐÎÆ¿£¬ÔÙÖØ¸´ÒÔÉϲÙ×÷1¡ª2 ´Î£»

G£®°Ñ×¶ÐÎÆ¿·ÅÔÚ¼îʽµÎ¶¨¹Üϱߣ¬Æ¿ÏµæÒ»ÕŰ×Ö½£¬±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿£¬Ö±µ½¼ÓÈëlµÎ¼îÒººóÈÜÒºÑÕɫͻ±ä²¢ÔÚ°ë·ÖÖÓÄÚ²»ÔÙ±äɫΪֹ£¬¼ÇÏµζ¨¹ÜÒºÃæËùÔڵĿ̶ȡ£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)£¨2·Ö£©µÎ¶¨²Ù×÷µÄÕýȷ˳ÐòÊÇ£º(Ìî×Öĸ)      ¡úC¡ú      ¡úB¡ú      ¡ú     ¡ú   __        ¡£

(1) G²½²Ù×÷ÖÐÔÚ×¶ÐÎÆ¿ÏµæÒ»ÕŰ×Ö½ÊǵÄ×÷ÓÃÊÇ                  ¡£

(3)D²½²Ù×÷ÖÐÒºÃæÓ¦µ÷½Úµ½            £¬¼â×첿·ÖÓ¦            ¡£

(4)µÎ¶¨ÖÕµã¶ÁÊýʱ£¬Èç¹ûÑöÊÓÒºÃæ£¬¶Á³öµÄÊýÖµ                   £¬ÈôµÎ¶¨Ç°Æ½ÊÓ¶ÁÊýÔòÓɴ˼ÆËãµÃµ½µÄNaOHÈÜҺŨ¶È        ¡£(Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±¡¢¡°ÎÞÓ°Ï족)

 

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

¾«Ó¢¼Ò½ÌÍø