¡¾ÌâÄ¿¡¿ÊµÑéÊÒÖƱ¸ËÄË®¼×ËáÍ­[Cu(HCOO)24H2O]¾§ÌåʵÑé²½ÖèÈçÏ¡£

(1)¼îʽ̼ËáÍ­µÄÖƱ¸£º

a.²½ÖèiÊǽ«Ò»¶¨Á¿µ¨·¯ºÍ̼ËáÇâÄƹÌÌåÒ»Æð·Åµ½Ñв§ÖÐÑÐÄ¥£¬ÆäÄ¿µÄÊÇ______¡£

b.²½ÖèiiÊÇÔÚ½Á°èϽ«¹ÌÌå»ìºÏÎï·Ö¶à´Î»ºÂý¼ÓÈëÈÈË®ÖУ¬·´Ó¦Î¶ȿØÖÆÔÚ70¡æ¡ª80¡æ£¬Èç¹û¿´µ½_____(дʵÑéÏÖÏó)£¬ËµÃ÷ζȹý¸ß¡£

c.Ïà¹ØµÄ»¯Ñ§·½³Ìʽ______¡£

(2)ËÄË®¼×ËáÍ­¾§ÌåµÄÖƱ¸£º½«¼îʽ̼ËáÍ­¹ÌÌå·ÅÈëÉÕ±­ÖУ¬¼ÓÈëÒ»¶¨Á¿ÈȵÄÕôÁóË®£¬ÔÙÖðµÎ¼ÓÈë¼×ËáÖÁ¼îʽ̼ËáÍ­Ç¡ºÃÈ«²¿Èܽ⣬³ËÈȹýÂ˳ýÈ¥ÉÙÁ¿²»ÈÜÐÔÔÓÖÊ£¬È»ºóÕô·¢ÀäÈ´¹ýÂË£¬ÔÚÓÃÉÙÁ¿ÎÞË®ÒÒ´¼Ï´µÓ¾§Ìå2¡ª3´ÎÁÀ¸É£¬µÃµ½²úÆ·¡£

a.Ïà¹ØµÄ»¯Ñ§·½³Ìʽ______¡£

b.³ÃÈȹýÂËÖУ¬±ØÐë³ÃÈȵÄÔ­ÒòÊÇ______¡£

c.ÓÃÒÒ´¼Ï´µÓ¾§ÌåµÄÄ¿µÄ______¡£

¡¾´ð°¸¡¿ÑÐϸ²¢»ìºÏ¾ùÔÈ ÓкÚÉ«¹ÌÌåÉú³É ·ÀÖ¹¼×ËáÍ­¾§ÌåÎö³ö Ï´È¥¾§Ìå±íÃæµÄË®ºÍÆäËüÔÓÖÊ£¬¼õÉÙ¼×ËáÍ­µÄËðʧ

¡¾½âÎö¡¿

¢Åµ¨·¯ºÍ̼ËáÇâÄÆÁ½ÖÖ¹ÌÌåÎïÖÊÒªÒª³ä·Ö½Ó´¥²ÅÄܳä·Ö·´Ó¦£¬¼îʽ̼ËáÍ­ÊÜÈÈÒ׷ֽ⣬Éú³ÉºÚÉ«µÄÑõ»¯Í­£¬ÁòËáÍ­Óë̼ËáÇâÄÆ·´Ó¦Éú³É¼îʽ̼ËáÍ­£¬¸ù¾ÝÔªËØÊغãÊéд»¯Ñ§·½³Ìʽ¡£

¢Æ¼îʽ̼ËáÍ­Óë¼×Ëá·´Ó¦Éú³É¼×ËáÍ­£¬¸ù¾ÝÔªËØÊغãÊéд»¯Ñ§·½³Ìʽ£¬¼×ËáÍ­µÄÈܽâ¶ÈËæζȵÄÉý¸ß¶ø±ä´ó£¬Ðè³ÃÈȹýÂË£¬¸ù¾Ý¡°ÏàËÆÏàÈÜÔ­Àí¡±¿ÉÖª¼×ËáÍ­Ò×ÈÜÓÚË®£¬ÄÑÈÜÓÚÓлúÈܼÁ¡£

¢Åa.µ¨·¯ºÍ̼ËáÇâÄÆÁ½ÖÖ¹ÌÌåÎïÖÊÒªÒª³ä·Ö½Ó´¥²ÅÄܳä·Ö·´Ó¦£¬ËùÒÔÑÐÄ¥µÄ×÷ÓÃÊÇ°ÑÒ©Æ·ÑÐϸ²¢»ìºÏ¾ùÔÈ£»¹Ê´ð°¸Îª£ºÑÐϸ²¢»ìºÏ¾ùÔÈ¡£

b.ζȹý¸ß£¬Cu(OH)2CuCO3»á·Ö½âÉú³ÉºÚÉ«µÄÑõ»¯Í­£»¹Ê´ð°¸Îª£ºÓкÚÉ«¹ÌÌåÉú³É¡£

c.ÁòËáÍ­Óë̼ËáÇâÄÆ·´Ó¦Éú³É¼îʽ̼ËáÍ­£¬¸ù¾ÝÔªËØÊØ¿ÉÖª»¯Ñ§·½³ÌʽΪ2CuSO4 + 4NaHCO3 = Cu(OH)2CuCO3¡ý + 3CO2¡ü+ 2Na2SO4 + H2O£¬¹Ê´ð°¸Îª£º2CuSO4 + 4NaHCO3 = Cu(OH)2CuCO3¡ý + 3CO2¡ü+ 2Na2SO4 + H2O¡£

¢Æa.¼îʽ̼ËáÍ­Óë¼×Ëá·´Ó¦ÖƵÃËÄË®¼×ËáÍ­[Cu(HCOO)24H2O]¾§Ì壬¸ù¾ÝÔªËØÊØ¿ÉÖª»¯Ñ§·½³ÌʽΪCu(OH)2CuCO3 +4HCOOH + 5H2O =2 Cu(HCOO)24H2O + CO2¡ü£»¹Ê´ð°¸Îª£ºCu(OH)2CuCO3 +4HCOOH + 5H2O =2 Cu(HCOO)24H2O + CO2¡ü¡£

b.¼×ËáÍ­µÄÈܽâ¶ÈËæζȵÄÉý¸ß¶ø±ä´ó£¬Èç¹ûÀäÈ´£¬»áÓо§ÌåÎö³ö£¬½µµÍ²úÂÊ£¬Òò´ËÐè³ÃÈȹýÂË£»¹Ê´ð°¸Îª£º·ÀÖ¹¼×ËáÍ­¾§ÌåÎö³ö¡£

c.¼×ËáÍ­Ò×ÈÜÓÚË®£¬ÄÑÈÜÓھƾ«£¬²»ÄÜÓÃÕôÁóˮϴµÓ£¬¹ÊÐèÓÃÒÒ´¼½øÐÐÏ´µÓ£¬Ï´È¥¾§Ìå±íÃæµÄÒºÌåÔÓÖÊ£»¹Ê´ð°¸Îª£ºÏ´È¥¾§Ìå±íÃæµÄË®ºÍÆäËüÔÓÖÊ£¬¼õÉÙ¼×ËáÍ­µÄËðʧ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³³ÎÇå͸Ã÷ÈÜÒºÖÐÖ»¿ÉÄܺ¬ÓТÙAl3+£¬¢ÚMg2+£¬¢ÛFe3+£¬¢ÜFe2+£¬¢ÝH+£¬¢ÞCO32-£¬¢ßNO3-Öеļ¸ÖÖ£¬Ïò¸ÃÈÜÒºÖÐÖðµÎ¼ÓÈëNaOHÈÜÒºÖÁ¹ýÁ¿£¬Éú³É³ÁµíµÄÖÊÁ¿ÓëNaOHµÄÎïÖʵÄÁ¿µÄ¹ØϵÈçͼËùʾ£®Ôò¸ÃÈÜÒºÖÐÒ»¶¨º¬ÓеÄÀë×ÓÊÇ

A.¢Ý¢Þ¢ßB.¢Ú¢Û¢Ý¢ßC.¢Ù¢Ú¢Û¢Þ¢ßD.¢Ù¢Ú¢Û¢Ü¢Ý¢ß

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁл¯Ñ§·´Ó¦ÏȺó˳ÐòÅжÏÕýÈ·µÄÊÇ

A.Ïòº¬ÓеÈÎïÖʵÄÁ¿µÄ¡¢KOHµÄ»ìºÏÈÜÒºÖÐͨÈ룬Óë·´Ó¦µÄÎïÖÊÒÀ´ÎÊÇ£ºKOH¡¢ ¡¢

B.Ïòº¬ÓеÈÎïÖʵÄÁ¿µÄ¡¢¡¢µÄ»ìºÏÈÜÒºÖмÓÈëZn£¬ÓëZn·´Ó¦µÄÀë×ÓÒÀ´ÎÊÇ£º¡¢¡¢

C.Ïòº¬ÓеÈÎïÖʵÄÁ¿µÄ¡¢¡¢µÄ»ìºÏÈÜÒºÖеμÓÑÎËᣬÓëÑÎËá·´Ó¦µÄÎïÖÊÒÀ´ÎÊÇ£º¡¢¡¢¡¢

D.ÔÚº¬ÓеÈÎïÖʵÄÁ¿µÄ¡¢µÄÈÜÒºÖУ¬»ºÂýͨÈëÂÈÆø£¬ÓëÂÈÆø·´Ó¦µÄÀë×ÓÒÀ´ÎÊÇ£º¡¢¡¢

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¸ù¾ÝÌâÒâÌî¿Õ¡£

¢ñ.³£ÎÂÏ£¬Ä³Ë®ÈÜÒºMÖдæÔÚµÄÀë×ÓÓÐNa£«¡¢A2£­¡¢HA£­¡¢H£«ºÍOH£­£¬´æÔڵķÖ×ÓÓÐH2O¡¢H2A¡£¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺

(1)д³öËáH2AµÄµçÀë·½³Ìʽ£º

________________________________________________________________________¡£

(2)ʵÑé²âµÃNaHAÈÜÒºµÄpH>7£¬Çë·ÖÎöNaHAÈÜÒºÏÔ¼îÐÔµÄÔ­Òò£º

________________________________________________________________________¡£

Çëд³öNaHAÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳Ðò£º

________________________________________________________________________¡£

¢ò.(1)Èçͼ¼×±íʾÓÃÏàͬŨ¶ÈµÄNaOHÈÜÒº·Ö±ðµÎ¶¨Å¨¶ÈÏàͬµÄÈýÖÖÒ»ÔªËᣬÓÉͼ¿ÉÈ·¶¨ËáÐÔ×îÇ¿µÄÊÇ________(Ìî¡°¢Ù¡±¡°¢Ú¡±»ò¡°¢Û¡±)¡£ÈçͼÒÒ±íʾÓÃÏàͬŨ¶ÈµÄAgNO3±ê×¼ÈÜÒº·Ö±ðµÎ¶¨Å¨¶ÈÏàͬµÄº¬Cl£­¡¢Br£­¼°I£­µÄ»ìºÏÈÜÒº£¬ÓÉͼ¿ÉÈ·¶¨Ê×ÏȳÁµíµÄÀë×ÓÊÇ________¡£

(2)25 ¡æʱ£¬Ksp(AgCl)£½1.8¡Á10£­10¡£ÔÚ1 L 0.1 mol¡¤L£­1 NaClÈÜÒºÖмÓÈë1 L 0.2 mol¡¤L£­1AgNO3ÈÜÒº£¬³ä·Ö·´Ó¦ºóÈÜÒºÖÐc(Cl£­)£½________ mol¡¤L£­1(¼ÙÉè»ìºÏºóÈÜÒºµÄÌå»ý±ä»¯ºöÂÔ²»¼Æ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿»¯Ñ§Éϳ£ÓÃȼÉÕ·¨²â¶¨ÓлúÎïµÄ·Ö×Óʽ£¬ÕâÖÖ·½·¨ÊÇÔڵ篼ÓÈÈʱÓô¿ÑõÆøÑõ»¯¹ÜÄÚÑùÆ·£¬¸ù¾Ý²úÎïµÄÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É¡£ÏÂͼËùʾµÄÊÇÓÃȼÉÕ·¨È·¶¨ÓлúÎï·Ö×ÓʽµÄ³£ÓÃ×°Öá£

ʵÑé̽¾¿Ð¡×é³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄÑùÆ·£¨Ö»º¬C¡¢H¡¢OÈýÖÖÔªËØÖеÄÁ½ÖÖ»òÈýÖÖ£©£¬½øÐÐʵÑé,ͨ¹ý²â¶¨²úÉúµÄCO2ºÍË®µÄÖÊÁ¿È·¶¨ÓлúÎïµÄ×é³É£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©CÖÐŨÁòËáµÄ×÷ÓÃÊdzýÈ¥ÑõÆøÖеÄË®ÕôÆø£¬ÊµÑé×°ÖõÄÁ¬½Ó˳ÐòÓ¦ÊÇ£º___________£¨Ã¿ÖÖ×°ÖÃÖ»ÓÃÒ»´Î£©£»

£¨2£©ÊµÑéÊý¾Ý¼Ç¼ºÍ´¦Àí

ÎïÀíÁ¿ÊµÑéÐòºÅ

ȼÉÕÓлúÎïµÄÖÊÁ¿

¢Ù

¢Ú

ʵÑéÇ°ÖÊÁ¿

ʵÑéºóÖÊÁ¿

ʵÑéÇ°ÖÊÁ¿

ʵÑéºóÖÊÁ¿

1

m1

m2

m3

m4

m5

ÉϱíÖТ١¢¢Ú·Ö±ðÖ¸Äĸö×°Öã¿____________ ¡¢ _____________¡£

£¨3£©ÈôʵÑé׼ȷ³ÆÈ¡4.4 gÑùÆ·£¬¾­È¼ÉÕºó²âµÃ²úÉúCO28.8 g£¬Ë®ÕôÆø3.6g¡£ÒªÈ·¶¨¸ÃÓлúÎïµÄ·Ö×Óʽ£¬»¹±ØÐëÖªµÀµÄÊý¾ÝÊÇ________£»

£¨4£©ÏàͬÌõ¼þÏ£¬Èô¸ÃÓлúÎïÕôÆø¶ÔÇâÆøµÄÏà¶Ô·Ö×ÓÖÊÁ¿Îª22£¬ÇÒËüµÄºË´Å¹²ÕñÇâÆ×ÉÏÓÐÁ½¸ö·å£¬ÆäÇ¿¶È±ÈΪ3:1£¬ÊÔͨ¹ý¼ÆËãÈ·¶¨¸ÃÓлúÎïµÄ½á¹¹¼òʽ___________£¬Óë¸ÃÓлúÎïÏà¶Ô·Ö×ÓÖÊÁ¿ÏàͬµÄÌþµÄÒ»ÂÈ´úÎïÓÐ_____ÖÖ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³pH =1µÄÈÜÒºX£¬ÆäÖпÉÄܺ¬ÓÐAl3+¡¢Fe2+¡¢Fe3+¡¢Ba2+¡¢NH4+¡¢CO32£­¡¢SO42£­¡¢SiO32£­¡¢NO3£­ÖеÄÒ»ÖÖ»ò¼¸ÖÖ£¬È¡500 mL¸ÃÈÜÒº½øÐÐʵÑ飬ÆäÏÖÏó¼°×ª»¯ÈçͼËùʾ¡£

ÒÑÖª£º·´Ó¦¹ý³ÌÖÐÓÐÒ»ÖÖÆøÌåÊǺì×ØÉ«¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)½öÓÉÇ¿ËáÐÔÌõ¼þ±ã¿ÉÅжÏÈÜÒºXÖÐÒ»¶¨²»´æÔÚµÄÀë×ÓÓÐ______¡£

(2)ÈÜÒºXÖУ¬¹ØÓÚNO3£­µÄÅжÏÒ»¶¨ÕýÈ·µÄÊÇ______Ìî×Öĸ¡£

a.Ò»¶¨ÓÐ

b.Ò»¶¨Ã»ÓÐ

c.¿ÉÄÜÓÐ

(3)·´Ó¦¢ÙÖвúÉúÆøÌåAµÄÀë×Ó·½³ÌʽΪ______¡£

(4)·´Ó¦¢ßÖÐÉú³É³ÁµíKµÄÀë×Ó·½³ÌʽΪ______¡£

(5)ÈÜÒºXÖв»ÄÜÈ·¶¨µÄÀë×ÓÊÇ______¡£

(6)ÈôʵÑé²â¶¨A¡¢F¡¢K¾ùΪ0.01 mol£¬ÊÔÈ·¶¨³ÁµíCÊÇ______£¬ÆäÎïÖʵÄÁ¿·¶Î§ÊÇ______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐʵÑéÖУ¬·½°¸¢ñºÍ·½°¸¢ò¾ù¿ÉÐУ¬ÇÒ·½°¸¢ò¸üºÏÀíµÄÊÇ

ʵÑéÄ¿µÄ

·½°¸¢ñ

·½°¸¢ò

A

³ýÈ¥ÒÒËáÒÒõ¥ÖеÄÉÙÁ¿ÒÒËá

¼Ó±¥ºÍÈÜÒº³ä·Ö·´Ó¦ºó£¬¾²ÖᢷÖÒº

ÕôÁó

B

¼ø±ð̼ËáÄƺÍ̼ËáÇâÄÆÈÜÒº

·Ö±ðµÎ¼Ó³ÎÇåʯ»ÒË®

·Ö±ð¼ÓÈÈÁ½ÈÜÒº

C

¼ìÑéÑÇÁòËáÄÆÊÇ·ñ±»Ñõ»¯

µÎÈëÈÜÒº

µÎÈëÑÎËáËữµÄÈÜÒº

D

±È½ÏÂÈÔªËØ¡¢µâÔªËصķǽðÊôÐÔÇ¿Èõ

·Ö±ð¼ÓÈÈÂÈ»¯Çâ¡¢µâ»¯Ç⣬±È½ÏÈÈÎȶ¨ÐÔ

ÔÚµí·Ûµâ»¯¼ØÊÔÖ½ÉϵμÓÂÈË®

A.AB.BC.CD.D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ïã²ÝÈ©ÊÇÒ»ÖÖʳƷÌí¼Ó¼Á£¬¿ÉÓÉÓú´´Ä¾·Ó×÷Ô­ÁϺϳɣ¬ºÏ³É·ÏßÈçÏ£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

A. ·´Ó¦1¡ú2ÊôÓڼӳɷ´Ó¦£¬ÇÒÉú³ÉµÄ»¯ºÏÎï2¾ßÓÐÒ»¸öÊÖÐÔ̼ԭ×Ó

B. »¯ºÏÎï2ÔÚÒ»¶¨Ìõ¼þÏ¿ɷ¢ÉúÏûÈ¥·´Ó¦

C. ¼ìÑéÖƵõÄÏã²ÝÈ©ÖÐÊÇ·ñ»ìÓл¯ºÏÎï3£¬¿ÉÓÃÂÈ»¯ÌúÈÜÒº

D. µÈÎïÖʵÄÁ¿ËÄÖÖ»¯ºÏÎï·Ö±ðÓë×ãÁ¿NaOH·´Ó¦£¬ÏûºÄNaOHÎïÖʵÄÁ¿Ö®±ÈΪ1£º3£º2£º4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¢ñ.NH4Al(SO4)2ÊÇʳƷ¼Ó¹¤ÖÐ×îΪ¿ì½ÝµÄʳƷÌí¼Ó¼Á£¬ÓÃÓÚ±º¿¾Ê³Æ·£»NH4HSO4ÔÚ·ÖÎöÊÔ¼Á¡¢Ò½Ò©¡¢µç×Ó¹¤ÒµÖÐÓÃ;¹ã·º¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÏàͬÌõ¼þÏ£¬0.1 mol¡¤L£­1NH4Al(SO4)2ÖÐc(NH4+)________(Ìî¡°µÈÓÚ¡±¡¢¡°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±)0.1 mol¡¤L£­1NH4HSO4ÖÐc(NH4+)¡£

(2)Èçͼ1ÊÇ0.1 mol¡¤L£­1µç½âÖÊÈÜÒºµÄpHËæζȱ仯µÄͼÏñ¡£

¢ÙÆäÖзûºÏ0.1 mol¡¤L£­1NH4Al(SO4)2µÄpHËæζȱ仯µÄÇúÏßÊÇ________(Ìî×Öĸ)¡£

¢ÚÊÒÎÂʱ£¬0.1 mol¡¤L£­1NH4Al(SO4)2ÖÐ2c(SO42-)£­c(NH4+)£­3c(Al3£«)£½________mol¡¤L£­1(ÌîÊýÖµ±í´ïʽ)¡£

(3)ÊÒÎÂʱ£¬Ïò100 mL 0.1 mol¡¤L£­1NH4HSO4ÈÜÒºÖеμÓ0.1 mol¡¤L£­1NaOHÈÜÒº£¬µÃµ½µÄÈÜÒºpHÓëNaOHÈÜÒºÌå»ýµÄ¹ØϵÇúÏßÈçͼ2Ëùʾ¡£ÊÔ·ÖÎöͼÖÐa¡¢b¡¢c¡¢dËĸöµã£¬Ë®µÄµçÀë³Ì¶È×î´óµÄÊÇ____________£»ÔÚbµã£¬ÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄÅÅÁÐ˳ÐòÊÇ______________¡£

¢ò.pCÊÇÖ¸¼«Ï¡ÈÜÒºÖÐÈÜÖÊÎïÖʵÄÁ¿Å¨¶ÈµÄ³£ÓöÔÊý¸ºÖµ£¬ÀàËÆpH¡£ÈçijÈÜÒºÈÜÖʵÄŨ¶ÈΪ1¡Á10£­3mol¡¤L£­1£¬Ôò¸ÃÈÜÒºÖиÃÈÜÖʵÄpC£½£­lg10£­3£½3¡£ÒÑÖªH2CO3ÈÜÒºÖдæÔÚÏÂÁÐƽºâ£ºCO2£«H2OH2CO3¡¢H2CO3H£«£«HCO3-¡¢HCO3-H£«£«CO32-ͼ3ΪH2CO3¡¢HCO3-¡¢CO32-ÔÚ¼ÓÈëÇ¿Ëá»òÇ¿¼îÈÜÒººó£¬´ïµ½Æ½ºâʱÈÜÒºÖÐÈýÖֳɷֵÄpCpHͼ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÔÚpH£½9ʱ£¬H2CO3ÈÜÒºÖÐŨ¶È×î´óµÄº¬Ì¼ÔªËصÄÀë×ÓΪ______¡£

(2)pH<4ʱ£¬ÈÜÒºÖÐH2CO3µÄpC×ÜÊÇÔ¼µÈÓÚ3µÄÔ­ÒòÊÇ____________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸