¡¾ÌâÄ¿¡¿ÄÜÔ´ÊÇÈËÀàÉú´æºÍ·¢Õ¹µÄÖØÒªÖ§Öù¡£¸ù¾ÝÒªÇó»Ø´ðÏÂÁÐÎÊÌâ¡£

(1)ÏÂÁÐ˵·¨ÖдíÎóµÄÊÇ______¡£

A£®¶¼ÊÇÇå½àÄÜÔ´

B£®µçÄÜ£¬H2ÊôÓÚ¶þ´ÎÄÜÔ´£¬Ë®Á¦ÊÇÒ»¼¶ÄÜÔ´

C£®ÃººÍʯÓÍÊôÓÚ»¯Ê¯È¼ÁÏ£¬È¼ÉÕ»á²úÉúÎÂÊÒÆøÌå

D£®ÇâÄÜÈÈÖµ¸ß£¬¶Ô»·¾³ÎÞÎÛȾ

(2)ÎÒ¹úĿǰʹÓõÄÖ÷ÒªÄÜÔ´ÊÇ»¯Ê¯È¼ÁÏ£¬½üÄêÀ´ÎÒ¹úÄÜÔ´×ÜÏû·ÑÁ¿ÓëÈ˾ùÄÜÔ´Ïû·ÑÁ¿Èçͼ¡£

¢ÙÔÚÎÒ¹ú£¬ÃºµÄÏû·ÑÁ¿¾Þ´ó£¬ÇÒúÊôÓÚ²»¿ÉÔÙÉúÄÜÔ´£¬Ñ°ÕÒÐÂÄÜÔ´ÊÇÄÜÔ´Ñо¿µÄÖ÷Òª¿ÎÌâÖ®Ò»¡£ÏÂÁв»ÊôÓÚÐÂÄÜÔ´µÄÊÇ______¡£

A£®Ì«ÑôÄÜ B£®ÇâÄÜ C£®µØÈÈÄÜ D£®º£ÑóÄÜ

E£®ÉúÎïÖÊÄÜ F£®Ê¯ÓÍ G£®ÌìÈ»Æø

¢ÚÒÑÖª£¬1 kgú´óÔ¼·ÅÈÈ2.9¡Á104 kJ¡£·çÄÜÄÜÁ¿¾Þ´ó£¬µØÇòÉÏÒ»Äê¿ÉÀûÓõķçÄÜÏ൱ÓÚ5¡Á1014 kw¡¤hµÄµçÄÜ¡£ÊÔ¼ÆË㣬һÄêµÄ·çÄÜÈôÄÜÈ«²¿ÀûÓ㬿ÉÏ൱ÓÚ½ÚÊ¡______¶Öȼú¡£(ÒÑÖª£¬1 kw¡¤h£½3.6¡Á103 kJ)

¡¾´ð°¸¡¿A FG 6.2¡Á1010

¡¾½âÎö¡¿

Çå½àÄÜÔ´£¬¼´ÂÌÉ«ÄÜÔ´£¬ÊÇÖ¸²»ÅÅ·ÅÎÛȾÎï¡¢Äܹ»Ö±½ÓÓÃÓÚÉú²úÉú»îµÄÄÜÔ´£¬Ëü°üÀ¨ºËÄܺ͡°¿ÉÔÙÉúÄÜÔ´¡±¡£ÐÂÄÜÔ´Ò»°ãÊÇÖ¸ÔÚм¼Êõ»ù´¡ÉϼÓÒÔ¿ª·¢ÀûÓõĿÉÔÙÉúÄÜÔ´£¬°üÀ¨Ì«ÑôÄÜ¡¢ÉúÎïÖÊÄÜ¡¢·çÄÜ¡¢µØÈÈÄÜ¡¢²¨ÀËÄÜ¡¢ÇâÄÜ¡¢º£ÑóÄܵȡ£

£¨1£©A£®Çå½àÄÜÔ´£¬¼´ÂÌÉ«ÄÜÔ´£¬ÊÇÖ¸²»ÅÅ·ÅÎÛȾÎï¡¢Äܹ»Ö±½ÓÓÃÓÚÉú²úÉú»îµÄÄÜÔ´£¬Ëü°üÀ¨ºËÄܺ͡°¿ÉÔÙÉúÄÜÔ´¡±¡£Òò´Ë»¯Ê¯È¼ÁϺͽ¹Â¯Æø²»ÄܳÆΪÇå½àÄÜÔ´£¬AÏî´íÎó£»

B£®µçÄÜ£¬H2ÊÇÓÉÒ»´ÎÄÜÔ´ÖƱ¸ËùµÃµÄ¶þ´ÎÄÜÔ´£¬Ë®Á¦ÊÇÒ»¼¶ÄÜÔ´£¬BÏîÕýÈ·£»

C£®Ãº¡¢Ê¯ÓÍ¡¢ÌìÈ»Æø¾ùÊôÓÚ»¯Ê¯È¼ÁÏ£¬È¼ÉÕʱ»á²úÉúÎÂÊÒÆøÌå¶þÑõ»¯Ì¼£¬CÏîÕýÈ·£»

D£®ÇâÄÜÈÈÖµ¸ß£¬¶Ô»·¾³ÎÞÎÛȾ£¬ÊÇÒ»ÖÖÀíÏëµÄÐÂÄÜÔ´£¬DÏîÕýÈ·£»

´ð°¸Ñ¡A¡£

(2)

¢ÙÐÂÄÜÔ´Ò»°ãÊÇÖ¸ÔÚм¼Êõ»ù´¡ÉϼÓÒÔ¿ª·¢ÀûÓõĿÉÔÙÉúÄÜÔ´£¬°üÀ¨Ì«ÑôÄÜ¡¢ÉúÎïÖÊÄÜ¡¢·çÄÜ¡¢µØÈÈÄÜ¡¢²¨ÀËÄÜ¡¢ÇâÄÜ¡¢º£ÑóÄܵȣ¬¶øú¡¢Ê¯ÓÍ¡¢ÌìÈ»Æø¾ùÊôÓÚ»¯Ê¯È¼Áϲ»ÊôÓÚÐÂÄÜÔ´¡£Òò´Ë²»ÊôÓÚÐÂÄÜÔ´µÄӦѡÔñFG£»

¢ÚÒÑÖª£¬1 kgú´óÔ¼·ÅÈÈ2.9¡Á104 kJ¡£Ò»Äê¿ÉÀûÓõķçÄÜÏ൱ÓÚ5¡Á1014 kw¡¤hµÄµçÄÜ£¬¸ù¾Ý1 kw¡¤h£½3.6¡Á103 kJ£¬¿É»»Ëã³öÒ»Äê¿ÉÀûÓõķçÄÜÌṩµÄÄÜÁ¿Îª5¡Á1014 kw¡¤h¡Á3.6¡Á103 kJ=1.8¡Á1018 kJ¡£Ï൱ÓÚȼúÓÃÁ¿£º= 6.2¡Á1013 kg=6.2¡Á1010t¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÎÒ¹ú¿Æѧ¼Ò³É¹¦ÊµÏÖ¼×ÍéÔÚ´ß»¯¼Á¼°ÎÞÑõÌõ¼þÏ£¬Ò»²½¸ßЧÉú²úÒÒÏ©¡¢·¼ÌþºÍÇâÆøµÈ»¯Ñ§Æ·£¬ÎªÌìÈ»Æø»¯¹¤¿ª·¢ÁËÒ»Ìõ¸ïÃüÐÔ¼¼Êõ¡£ÒÔ¼×ÍéΪԭÁϺϳɲ¿·Ö»¯¹¤²úÆ·Á÷³ÌÈçÏ£¨²¿·Ö·´Ó¦Ìõ¼þÒÑÂÔÈ¥£©£º

£¨1£©E µÄÃû³ÆΪ____£¬BÎïÖʵĽṹ¼òʽ£º______£»

£¨2£©ÉÏÊö¢Û¡«¢Þת»¯·´Ó¦ÖУ¬ÊôÓÚÈ¡´ú·´Ó¦µÄÓÐ______£¨Ó÷´Ó¦ÐòºÅÌîд£©£»

£¨3£©Ð´³ö·´Ó¦¢ßµÄ·´Ó¦·½³Ìʽ£º______£»

£¨4£©ÈçͼΪʵÑéÊÒÖÆÈ¡EµÄ×°ÖÃͼ£¬Í¼ÖÐaÊÔ¼ÁΪ_______£»

£¨5£©Ä³Í¬Ñ§ÔÚÊÔ¹ÜbÖмÓÈë6.0¿ËÒÒËáºÍ×ãÁ¿ÒÒ´¼²ÉÓÃÊʵ±Ìõ¼þʹ·´Ó¦³ä·Ö½øÐУ¬½áÊøºóÔÚÊÔ¹Üb»ØÊÕµ½3.0¿ËÒÒËᣬÔò¸ÃͬѧÔÚ±¾´ÎʵÑéÖÐÖƵÃÒÒËáÒÒõ¥µÄ×î´óÖÊÁ¿Îª_______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿½üÆÚ·¢ÏÖ£¬H2SÊǼÌNO¡¢COÖ®ºóµÄµÚÈý¸öÉúÃüÌåϵÆøÌåÐźŷÖ×Ó£¬Ëü¾ßÓвÎÓëµ÷½ÚÉñ¾­ÐźŴ«µÝ¡¢ÊæÕÅѪ¹Ü¼õÇá¸ßѪѹµÄ¹¦ÄÜ¡£»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÏÂÁÐÊÂʵÖУ¬²»ÄܱȽÏÇâÁòËáÓëÑÇÁòËáµÄËáÐÔÇ¿ÈõµÄÊÇ________(Ìî±êºÅ)¡£

A£®ÇâÁòËá²»ÄÜÓë̼ËáÇâÄÆÈÜÒº·´Ó¦£¬¶øÑÇÁòËá¿ÉÒÔ

B£®ÇâÁòËáµÄµ¼µçÄÜÁ¦µÍÓÚÏàͬŨ¶ÈµÄÑÇÁòËá

C£®0.10 mol¡¤L£­1µÄÇâÁòËáºÍÑÇÁòËáµÄpH·Ö±ðΪ4.5ºÍ2.1

D£®ÇâÁòËáµÄ»¹Ô­ÐÔÇ¿ÓÚÑÇÁòËá

£¨2£©ÏÂͼÊÇͨ¹ýÈÈ»¯Ñ§Ñ­»·ÔڽϵÍζÈÏÂÓÉË®»òÁò»¯Çâ·Ö½âÖƱ¸ÇâÆøµÄ·´Ó¦ÏµÍ³Ô­Àí¡£

ͨ¹ý¼ÆË㣬¿É֪ϵͳ(¢ñ)ºÍϵͳ(¢ò)ÖÆÇâµÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ_______£¬ÖƵõÈÁ¿H2ËùÐèÄÜÁ¿½ÏÉÙµÄÊÇ________¡£

£¨3£©H2SÓëCO2ÔÚ¸ßÎÂÏ·¢Éú·´Ó¦£ºH2S(g)£«CO2(g)COS(g)£«H2O(g)¡£ÔÚ610 Kʱ£¬½«0.10 mol CO2Óë0.40 mol H2S³äÈë2.5 LµÄ¿Õ¸ÖÆ¿ÖУ¬·´Ó¦Æ½ºâºóË®µÄÎïÖʵÄÁ¿·ÖÊýΪ0.02¡£

¢ÙH2SµÄƽºâת»¯ÂʦÁ1£½________%£¬·´Ó¦Æ½ºâ³£ÊýK£½________¡£

¢ÚÔÚ620 KÖظ´ÊµÑ飬ƽºâºóË®µÄÎïÖʵÄÁ¿·ÖÊýΪ0.03£¬H2SµÄת»¯ÂʦÁ2_______¦Á1£¬¸Ã·´Ó¦µÄ¦¤H________0¡£(Ìî¡°>¡±¡°<¡±»ò¡°£½¡±)

¢ÛÏò·´Ó¦Æ÷ÖÐÔÙ·Ö±ð³äÈëÏÂÁÐÆøÌ壬ÄÜʹH2Sת»¯ÂÊÔö´óµÄÊÇ________(Ìî±êºÅ)¡£

A£®H2S B£®CO2 C£®COS D£®N2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚijѧÉúµÄ»¯Ñ§ÊµÑ鱨¸æÖÐÓÐÈçϼǼÆäÖÐʵÑéÊý¾ÝºÏÀíµÄ£¨ £©

A.ÓÃ100 mLÁ¿Í²Á¿È¡5.26 mLÏ¡ÁòËáÈÜÒº

B.ÓÃÍÐÅÌÌìƽ³ÆÈ¡11.7 g CuO·ÛÄ©

C.Óù㷺pHÊÔÖ½²âµÃÈÜÒºµÄpHΪ3.5

D.ζȼÆÉÏÏÔʾµÄÊÒζÁÊýΪ25.68 ¡æ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿FeÊÇÈÕ³£Éú»îÖÐ×î³£ÓõĽðÊôÖ®Ò»¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÉîÂñÔÚ³±ÊªÍÁÈÀÖеÄÌú¹ÜµÀ£¬ÔÚÁòËáÑλ¹Ô­¾ú(¸Ã»¹Ô­¾ú×î¼ÑÉú´æ»·¾³ÔÚpHΪ7¡«8Ö®¼ä)×÷ÓÃÏ£¬Äܱ»SO42-¸¯Ê´£¬Æäµç»¯Ñ§¸¯Ê´Ô­ÀíÈçÏÂͼËùʾ£¬Ð´³öÕý¼«µÄµç¼«·´Ó¦Ê½___________________¡£

(2)ÔÚ1800Kʱ£¬2Fe(s)+3/2O2(g)=Fe2O3(s) H=-354.2kJ/mol£»3Fe(s)+2O2(g)=Fe3O4(s) H=-550.9kJ/molÔò·´Ó¦£º2Fe3O4(s)+1/2O2(g) ===3Fe2O3(s)µÄHΪ_____ kJ¡¤mol1£¬ËÄÑõ»¯ÈýÌúÔÚ³äÂúÑõÆøµÄ¼¯ÆøÆ¿Öз´Ó¦Éú³ÉFe2O3_____(Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±)×Ô·¢½øÐС£

(3)Fe3+ºÍI£­ÔÚË®ÈÜÒºÖеķ´Ó¦ÈçÏ£º2I£­+2Fe3+2Fe2+ +I2(ÔÚË®ÈÜÒºÖÐ)¡£

¢Ù298Kʱ£¬Ïò5mL 0.1molL1 µÄKIÈÜÒºÖеμÓ0.1molL1 FeCl3ÈÜÒº£¬µÃµ½c(Fe2+)Óë¼ÓÈëFeCl3 ÈÜÒºÌå»ý¹ØϵÈçÏÂͼËùʾ£º

¸ÃζÈϵμÓ5mL FeCl3ÈÜҺʱ£¬Fe3+µÄƽºâת»¯ÂÊ=_____%£¬Æ½ºâ³£ÊýK=_____£¬ÈôÒªÌá¸ßFe3+µÄת»¯ÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÊÇ________________________¡£

¢ÚÔÚÒѾ­´ïµ½Æ½ºâµÄÉÏÊö·´Ó¦ÌåϵÖУ¬¼ÓÈë±½¶ÔI2½øÐÐÝÍÈ¡£¬±£³ÖζȲ»±ä£¬·´Ó¦ËÙÂÊ_____ (Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)£¬´Ëʱ¦Ô(Õý)_____¦Ô(Äæ)(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)¡£

¢Û¦Ô(Õý)ÓëFe3+¡¢IµÄŨ¶È¹ØϵΪ¦Ô=kc(I£­)mc(Fe3+)n(kΪ³£Êý)

c(I)molL1

c(Fe3+)molL1

¦Ô (molL1s1 )

(1)

0.20

0.80

0.032k

(2)

0.60

0.40

0.144k

(3)

0.80

0.20

0.128k

ͨ¹ý·ÖÎöËù¸øÊý¾Ý¼ÆËã¿ÉÖª£ºÔÚ¦Ô=kc(I£­)mc(Fe3+)n ÖУ¬m£¬nµÄֵΪ_____(Ìî×Öĸ´úºÅ)¡£

A£®m=1£¬n=1¡¡ B£®m=2£¬n=1 ¡¡C£®m=2£¬n=2 ¡¡D£®m=1£¬n=2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿»¯Ñ§·´Ó¦¹ý³ÌÖз¢ÉúÎïÖʱ仯µÄͬʱ£¬³£³£°éÓÐÄÜÁ¿µÄ±ä»¯£®ÕâÖÖÄÜÁ¿µÄ±ä»¯³£ÒÔÈÈÁ¿µÄÐÎʽ±íÏÖ³öÀ´£¬½Ð×ö·´Ó¦ÈÈ£®ÓÉÓÚ·´Ó¦µÄÇé¿ö²»Í¬£¬·´Ó¦ÈÈ¿ÉÒÔ·ÖΪÐí¶àÖÖ£¬ÈçȼÉÕÈȺÍÖкÍÈȵȡ£

(1)ÏÂÁС÷H±íʾÎïÖÊȼÉÕÈȵÄÊÇ ______ £»±íʾÎïÖÊÖкÍÈȵÄÊÇ ______ (Ìî¡°¡÷H1¡±¡¢¡°¡÷H2¡±¡¢¡°¡÷H3¡±µÈ)£®

A£®2H2(g)+O2(g)¨T2H2O(l)¡÷H1

B£®C(s)+1/2O2(g)¨TCO(g)¡÷H2

C£®CH4(g)+2O2(g)¨TCO2(g)+2H2O(g)¡÷H3

D£®C(s)+O2(g)¨TCO2(g)¡÷H4

E£®C6H12O6(s)+6O2(g)¨T6CO2(g)+6H2O(l)¡÷H5¡¡

F£®NaOH(aq)+HCl(aq)¨TNaCl(aq)+H2O(l)¡÷H6

G£®2NaOH(aq)+H2SO4(aq)¨TNa2SO4(aq)+2H2O(l)¡÷H7

H£®CH3COOH(aq)+NaOH(aq)¨TCH3COONa(aq)+H2O(l)¡÷H8

(2)ÒÀ¾ÝÊÂʵ£¬Ð´³öÏÂÁз´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£®

¢ñ¡¢ÔÚ25¡æ¡¢101kPaÏ£¬1gCH3OHȼÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ22.68kJ£®Ôò±íʾ¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ _______________________£»

¢ò¡¢1.00L1.00mol/LH2SO4ÈÜÒºÓë2.00L1.00mol/LNaOHÈÜÒºÍêÈ«·´Ó¦£¬·Å³ö114.6kJµÄÈÈÁ¿£¬±íʾÆäÖкÍÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ ________________________£»

¢ó¡¢ÒÑÖª²ð¿ª1molH-H¼ü£¬1molN-H¼ü£¬1molN¡ÔN¼ü·Ö±ðÐèÒªµÄÄÜÁ¿ÊÇ436kJ¡¢391kJ¡¢ 946kJ£¬ÔòN2ÓëH2·´Ó¦Éú³ÉNH3µÄÈÈ»¯Ñ§·½³ÌʽΪ _______________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ïã²ÝËá¹ã·ºÓÃÓÚʳƷµ÷ζ¼Á£¬ ×÷Ï㾫¡¢ ÏãÁÏ£¬ ºÏ³ÉÏß·ÈçÏ£º

£¨1£©C µÄ½á¹¹¼òʽÊÇ_____£¬ B ת»¯Îª C µÄ·´Ó¦ÀàÐÍΪ_____¡£

£¨2£©A Öк¬ÓеĹÙÄÜÍÅÃû³ÆÊÇ_____¡£

£¨3£©B µÄ·Ö×ÓʽÊÇ_____¡£

£¨4£©ÔÚ B ÖеÎÈë FeCl3ÈÜÒº£¬ ÆäÏÖÏóÊÇ_____¡£

£¨5£©Óë F ¹ÙÄÜÍÅÏàͬµÄ·¼Ïã×廯ºÏÎïµÄͬ·ÖÒì¹¹ÌåÓÐ_____ÖÖ¡£

£¨6£©Ð´³ö F ÓëÒÒ´¼½øÐÐõ¥»¯·´Ó¦µÄ·½³Ìʽ_____¡£

£¨7£©Ð´³öÓɵĺϳÉÏß·ͼ_____¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÔÚ25¡æ¡¢101 kPaÏÂ,1g¼×´¼È¼ÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ22.68kJ,ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÊéдÕýÈ·µÄÊÇ()

A. CH3OH(l)+3/2O2(g)CO2(g)+2H2O(l) ¦¤H=+725.76 kJ¡¤mol-1

B. 2CH3OH(l)+3O2(g)2CO2(g)+4H2O(l) ¦¤H=-1 451.52 kJ¡¤mol-1

C. 2CH3OH(l)+3O2(g)2CO2(g)+4H2O(l) ¦¤H=-725.76 kJ¡¤mol-1

D. 2CH3OH(l)+3O2(g)2CO2(g)+4H2O(l)¦¤H=+1451.52 kJ¡¤mol-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¿×ȸʯµÄÖ÷Òª³É·ÖÊÇCu2(OH)2CO3£¨º¬Fe2O3¡¢FeCO3¡¢Al2O3¡¢SiO2ÔÓÖÊ£©£¬¹¤ÒµÉÏÓÿ×ȸʯÖƱ¸ÁòËáÍ­µÄµÚÒ»²½ÐèÓùýÁ¿µÄÁòËáÈܽⲢ¹ýÂË¡£³£ÎÂÏ£¬·Ö±ðÈ¡ÂËÒº²¢ÏòÆäÖмÓÈëÖ¸¶¨ÎïÖÊ£¬·´Ó¦ºóµÄÈÜÒºÖдóÁ¿´æÔÚµÄÀë×Ó×éÕýÈ·µÄÊÇ

A. ¼ÓÈë¹ýÁ¿°±Ë®£ºFe3+¡¢NH4+¡¢SO42£­¡¢OH£­

B. ¼ÓÈë¹ýÁ¿NaClOÈÜÒº£ºFe2+¡¢Na+¡¢ClO£­¡¢SO42£­

C. ¼ÓÈë¹ýÁ¿NaOHÈÜÒº£ºNa+¡¢AlO2£­¡¢SO42£­¡¢OH£­

D. ¼ÓÈë¹ýÁ¿NaHCO3ÈÜÒº£ºNa+¡¢Al3+¡¢SO42£­¡¢HCO3£­

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸