¡¾ÌâÄ¿¡¿Ã¸Ö¸¾ßÓÐÉúÎï´ß»¯¹¦Äܵĸ߷Ö×ÓÎïÖÊ£®ÔÚøµÄ´ß»¯·´Ó¦ÌåϵÖУ¬·´Ó¦Îï·Ö×Ó±»³ÆΪµ×Îµ×Îïͨ¹ýøµÄ´ß»¯×ª»¯ÎªÆäËû·Ö×Ó£®¼¸ºõËùÓеÄϸ°û»î¶¯½ø³Ì¶¼ÐèҪøµÄ²ÎÓ룬ÒÔÌá¸ßЧÂÊ£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ã¸Ñõ»¯Ã¸Óë·ÓÀàµ×ÎïÔÚϸ°ûÖÐÄÜʵÏÖ·ÖÀà´æ·Å£¬ÊÇÒòΪϸ°ûÄÚ¾ßÓРϵͳ£¬×é³É¸ÃϵͳµÄ½á¹¹¾ßÓеŦÄÜÌØÐÔÊÇ £®²èҶϸ°ûÖÐÒ²´æÔÚÖÚ¶àÖÖÀàµÄ·ÓÀàÎïÖÊÓë·ÓÑõ»¯Ã¸£®Â̲èÖÆÈ¡¹ý³ÌÖбØÐëÏȽøÐÐÈȹø¸ßγ´ÖÆ£¬ÕâÒ»¹ý³ÌµÄÄ¿µÄÊÇ £®

£¨2£©²èÊ÷µÄRubiconøÔÚCO2Ũ¶È½Ï¸ßʱ£¬¸Ãø´ß»¯C2ÓëCO2·´Ó¦£¬RubiconøµÄ´æÔÚ³¡ËùΪ £»¸Ãø¾ßÓС°Á½ÃæÐÔ¡±£¬ÔÚO2Ũ¶È½Ï¸ßʱ£¬¸Ãø´ß»¯C2ÓëO2·´Ó¦£¬²úÎᆳһϵÁб仯ºóµ½ÏßÁ£ÌåÖлá²úÉúCO2£¬Æä¡°Á½ÃæÐÔ¡±ÓëøµÄ £¨ÌØÐÔ£©Ïàì¶Ü£®

£¨3£©ÈçͼÖÐÇúÏß±íʾ½«Ã¸ÔÚ²»Í¬Î¶Èϱ£ÎÂ×ã¹»³¤Ê±¼ä£¬ÔÙÔÚø»îÐÔ×î¸ßµÄζÈϲâÆä²ÐÓàø»îÐÔ£¬ÓÉͼ¿ÉµÃ³ö£º £®ÎªÁËÑéÖ¤ÕâÒ»½áÂÛ£¬ÇëÒÔÄ͸ßεÄÏËάËØ·Ö½âøΪʵÑé²ÄÁÏ£¬±È½ÏÔÚµÍκÍ×îÊÊζÈÏ´¢´æ¶Ôø»îÐÔµÄÓ°Ï죬д³öʵÑéÉè¼Æ˼·£º £®

¡¾´ð°¸¡¿£¨1£©ÉúÎïĤ Ñ¡Ôñ͸¹ýÐÔ ¸ßÎÂʹ·ÓÑõ»¯Ã¸Ê§»î

£¨2£©Ò¶ÂÌÌå»ùÖÊ ×¨Ò»ÐÔ

£¨3£©Ã¸ÔڽϵÍζÈÏ´¢´æ»îÐÔÊÜÓ°ÏìС£¬ÔڽϸßζÈÏ´¢´æ»îÐÔÊÜÓ°Ïì´ó

½«Ä͸ßεÄÏËάËØ·Ö½âø·Ö±ðÔÚµÍκÍ×îÊÊζÈÏ´¢´æ×ã¹»³¤Ê±¼äºó£¬ÔÙÔÚ×îÊÊζÈϲâÁ¿Æä»îÐÔ

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©Ã¸Ñõ»¯Ã¸Óë·ÓÀàµ×ÎïÔÚϸ°ûÖÐÄÜʵÏÖ·ÖÀà´æ·Å£¬ÊÇÒòΪϸ°ûÄÚ¾ßÓÐÉúÎïĤϵͳ£¬ÉúÎïĤµÄ½á¹¹µÄ¹¦ÄÜÌØÐÔÊÇÑ¡Ôñ͸¹ýÐÔ£®Â̲èÖÆÈ¡¹ý³ÌÖбØÐëÏȽøÐÐÈȹø¸ßγ´ÖÆ£¬ÕâÒ»¹ý³ÌµÄÄ¿µÄÊǸßÎÂʹ·ÓÑõ»¯Ã¸Ê§»î£®

£¨2£©²èÊ÷µÄRubiconøÔÚCO2Ũ¶È½Ï¸ßʱ£¬¸Ãø´ß»¯C2ÓëCO2·´Ó¦£¬¸Ã·´Ó¦ÊôÓÚ°µ·´Ó¦ÖеÄÎïÖʱ仯£¬Òò´ËRubiconøµÄ´æÔÚ³¡ËùΪҶÂÌÌå»ùÖÊ£»¸Ãø¾ßÓС°Á½ÃæÐÔ¡±£¬ÔÚO2Ũ¶È½Ï¸ßʱ£¬¸Ãø´ß»¯C2ÓëO2·´Ó¦£¬²úÎᆳһϵÁб仯ºóµ½ÏßÁ£ÌåÖлá²úÉúCO2£¬Æä¡°Á½ÃæÐÔ¡±ÓëøµÄרһÐÔÏàì¶Ü£®

£¨3£©ÈçͼÖÐÇúÏß±íʾ½«Ã¸ÔÚ²»Í¬Î¶Èϱ£ÎÂ×ã¹»³¤Ê±¼ä£¬ÔÙÔÚø»îÐÔ×î¸ßµÄζÈϲâÆä²ÐÓàø»îÐÔ£¬ÓÉͼ¿ÉµÃ³öøÔڽϵÍζÈÏ´¢´æ»îÐÔÊÜÓ°ÏìС£¬ÔڽϸßζÈÏ´¢´æ»îÐÔÊÜÓ°Ïì´ó£®ÎªÁËÑéÖ¤ÕâÒ»½áÂÛ£¬¿ÉÒÔ½«Ä͸ßεÄÏËάËØ·Ö½âø·Ö±ðÔÚµÍκÍ×îÊÊζÈÏ´¢´æ×ã¹»³¤Ê±¼äºó£¬ÔÙÔÚ×îÊÊζÈϲâÁ¿Æä»îÐÔ£®

¹Ê´ð°¸Îª£º£¨1£©ÉúÎïĤ Ñ¡Ôñ͸¹ýÐÔ ¸ßÎÂʹ·ÓÑõ»¯Ã¸Ê§»î

£¨2£©Ò¶ÂÌÌå»ùÖÊ ×¨Ò»ÐÔ

£¨3£©Ã¸ÔڽϵÍζÈÏ´¢´æ»îÐÔÊÜÓ°ÏìС£¬ÔڽϸßζÈÏ´¢´æ»îÐÔÊÜÓ°Ïì´ó

½«Ä͸ßεÄÏËάËØ·Ö½âø·Ö±ðÔÚµÍκÍ×îÊÊζÈÏ´¢´æ×ã¹»³¤Ê±¼äºó£¬ÔÙÔÚ×îÊÊζÈϲâÁ¿Æä»îÐÔ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ä³¿ÎÌâ×éÒÔÄÉÃ×Fe2O3×÷Ϊµç¼«²ÄÁÏÖƱ¸ï®Àë×Óµç³Ø£¨ÁíÒ»¼«Îª½ðÊô﮺ÍʯīµÄ¸´ºÏ²ÄÁÏ£©£¬Í¨¹ýÔÚÊÒÎÂÌõ¼þ϶Ôï®Àë×Óµç³Ø½øÐÐÑ­»·³ä·Åµç£¬³É¹¦µØʵÏÖÁ˶ԽøÐԵĿÉÄæµ÷¿Ø£®Èçͼ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨ £©

A. ¸Ãµç³Ø¿ÉÒÔÓÃNaOHÈÜҺΪµç½âÖÊÈÜÒº

B. ·Åµçʱµç³ØÕý¼«µÄµç¼«·´Ó¦Ê½ÎªFe2O3+6Li++6e-¨T3Li2O+2Fe

C. ³äµçʱ£¬Fe×÷ΪÒõ¼«£¬µç³Ø²»±»´ÅÌúÎüÒý

D. ´ÅÌúµÄÖ÷Òª³É·ÖÊÇFe3O4£¬Æä¼È¿ÉÎüÒýÌú£¬Ò²¿ÉÎüÒýFe2O3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÒÑÖª1molSiO2º¬ÓÐ4mol Si¡ªO¡£ÓйؼüÄÜÊý¾ÝÈç±í£¬¾§Ìå¹èÔÚÑõÆøÖÐȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£ºSi(s)£«O2(g)£½SiO2(s)£»¦¤H£½£­989.2kJ¡¤mol£­1£¬Ôò±íÖÐxµÄֵΪ

»¯Ñ§¼ü

Si¡ªO

O£½O

Si¡ªSi

¼üÄÜ/kJ¡¤mol£­1

x

498.8

176

A. 460 B. 920 C. 1165.2 D. 423.3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÈýÂÈÑõÁ×(POCl3)Ë×ÃûÁ×õ£ÂÈ¡¢ÑõÂÈ»¯Á×£¬ÊÇÖØÒªµÄ»ù´¡»¯¹¤Ô­ÁÏ£¬¿ÉÓÃÓÚÖÆȡȾÁÏ¡¢Å©Ò©£¬ÓлúºÏ³ÉµÄÂÈ»¯¼Á¡¢´ß»¯¼ÁºÍ×èȼ¼ÁµÈ¡£Ä³»¯Ñ§ÊµÑéС×éÄ£ÄâPCl3Ö±½ÓÑõ»¯·¨ÖƱ¸POCl3,ÆäʵÑé×°ÖÃÉè¼ÆÈçÏ£º

ÓйØÎïÖʵIJ¿·ÖÐÔÖÊÈçÏÂ±í£º

ÈÛµã/¡æ

·Ðµã/¡æ

ÆäËû

PCl3

-112

75.5

ÓöË®Éú³ÉH3PO3ºÍHCl£¬ÓöO2Éú³ÉPOCl3

POCl3

2

105.3

ÓöË®Éú³ÉH3PO4ºÍHCl£¬ÄÜÈÜÓÚPCl3

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ÒÇÆ÷aµÄÃû³ÆÊÇ__________£¬×°ÖÃAÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ__________________¡£

£¨2£©BÖÐËù×°ÊÔ¼ÁΪ__________£¬B×°ÖõÄ×÷Óóý¹Û²ìO2µÄÁ÷ËÙÖ®Í⣬»¹ÓÐ_____________________¡£

£¨3£©C×°ÖÿØÖÆ·´Ó¦ÔÚ60¡æ~65¡æ½øÐУ¬Æä¿ØεÄÖ÷ҪĿµÄÊÇ__________¡£

£¨4£©ÊµÑéÖƵõÄPOCl3´Ö²úÆ·Öг£º¬ÓÐPCl3£¬¿ÉÓÃ__________·½·¨Ìá´¿¡£

£¨5£©Í¨¹ý·ð¶û¹þµÂ·¨¿ÉÒԲⶨÈýÂÈÑõÁײúÆ·ÖÐClÔªËغ¬Á¿£¬ÊµÑé²½ÖèÈçÏ£º

¢ñ£®È¡ag²úÆ·ÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈë×ãÁ¿NaOHÈÜÒº£¬´ýÍêÈ«·´Ó¦ºó¼ÓÏ¡ÏõËáÖÁËáÐÔ¡£

¢ò£®Ïò׶ÐÎÆ¿ÖмÓÈë0.1000mol¡¤L-1µÄAgNO3ÈÜÒº40.00mL,ʹCl-ÍêÈ«³Áµí¡£

¢ó£®ÏòÆäÖмÓÈë2mLÏõ»ù±½£¬ÓÃÁ¦Ò¡¶¯£¬Ê¹³Áµí±íÃæ±»ÓлúÎ︲¸Ç£¬ÒÔ·ÀÖ¹ÔڵμÓNH4SCNʱ£¬½«AgCl³Áµíת»¯ÎªAgSCN³Áµí¡£

¢ô£®¼ÓÈëָʾ¼Á£¬ÓÃcmol¡¤L-1NH4SCNÈÜÒºµÎ¶¨¹ýÁ¿Ag+ÖÁÖյ㣬¼ÇÏÂËùÓÃÌå»ýVmL¡£

ÒÑÖª£ºKsp(AgCl)=3.2¡Á10-10,Ksp(AgSCN)=2¡Á10-12

¢ÙµÎ¶¨Ñ¡ÓõÄָʾ¼ÁÊÇ__________ (Ìî±êºÅ)¡£

a£®FeCl2 b£®¼×»ù³È c£®µí·Û d£®NH4Fe(SO4)2

¢ÚClÔªËصÄÖÊÁ¿°Ù·Öº¬Á¿Îª__________ (ÁгöËãʽ)¡£

¢ÛÔÚ²½Öè¢óÖУ¬ÈôÎÞ¼ÓÈëÏõ»ù±½µÄ²Ù×÷£¬Ëù²âCl-ÔªËغ¬Á¿½«»á__________(Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°²»±ä¡±)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¶ÌÖÜÆÚÔªËØA¡¢B¡¢C¡¢D,Ô­×ÓÐòÊýÒÀ´ÎÔö´ó£¬AµÄ»ù̬ԭ×ÓµÄL²ãµç×ÓÊÇK²ãµç×ÓµÄÁ½±¶£»BµÄ¼Ûµç×Ó²ãÖеÄδ³É¶Ôµç×ÓÓÐ3¸ö£»CÓëBͬ×壻DµÄ×î¸ß¼Ûº¬ÑõËáΪËáÐÔ×îÇ¿µÄÎÞ»úº¬ÑõËá¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©CµÄ»ù̬ԭ×ӵĵç×ÓÅŲ¼Ê½Îª___________£»DµÄ×î¸ß¼Ûº¬ÑõËáËáÐÔ±ÈÆäµÍÁ½¼ÛµÄº¬ÑõËáËáÐÔÇ¿µÄÔ­ÒòÊÇ_____________________________________________¡£

£¨2£©ÔÓ»¯¹ìµÀ·ÖΪµÈÐԺͲ»µÈÐÔÔÓ»¯£¬²»µÈÐÔÔÓ»¯Ê±ÔÚÔÓ»¯¹ìµÀÖÐÓв»²Î¼Ó³É¼üµÄ¹Âµç×ӶԵĴæÔÚ¡£A¡¢B¡¢C¶¼·Ö±ðÄÜÓëDÐγÉÖÐÐÄÔ­×ÓÔÓ»¯·½Ê½¾ùΪ___________µÄ¹²¼Û»¯ºÏÎïX¡¢Y¡¢Z¡£ÆäÖУ¬ÊôÓÚ²»µÈÐÔÔÓ»¯µÄÊÇ___________ (д»¯Ñ§Ê½)¡£

£¨3£©±È½ÏY¡¢ZµÄÈ۷еãY______Z(Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±)£¬²¢½âÊÍÀíÓÉ_____________________¡£

£¨4£©DÔªËØÄÜÓëCuÐγÉ×Ø»ÆÉ«¹ÌÌ壬¼ÓË®ÈܽⲢϡÊ͹ý³ÌÖУ¬ÈÜÒºÑÕÉ«ÓÉÂÌÉ«Öð½¥×ª»¯ÎªÀ¶É«£¬ÏÔÂÌÉ«Àë×ÓÊÇ_________________£¬ÏÔÀ¶É«Àë×ÓÖеÄÅäλԭ×Ó________________¡£

£¨5£©AºÍBÄÜÐγɶàÖֽṹµÄ¾§Ìå¡£ÆäÖÐÒ»ÖÖÀàËÆʯīµÄ½á¹¹£¬Æä½á¹¹ÈçÏÂͼËùʾ(ͼ1Ϊ¾§Ìå½á¹¹£¬Í¼2ΪÇÐƬ²ã×´½á¹¹)£¬Æ仯ѧʽΪ__________¡£ÊµÑé²âµÃ´Ë¾§Ìå½á¹¹ÊôÓÚÁù·½¾§Ïµ,¾§°û½á¹¹¼ûͼ3¡£ÒÑ֪ͼʾԭ×Ó¶¼°üº¬ÔÚ¾§°ûÄÚ£¬¾§°û²ÎÊýa=0.64nm£¬c=0.24nm¡£Æ侧ÌåÃܶÈΪ__________g/cm3(ÒÑÖª£º2=1.414,ºó=1.732,½á¹û¾«È·µ½Ð¡ÊýµãºóµÚ2λ¡£)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿Ï±íÖÐÎïÖʵķÖÀà×éºÏÍêÈ«ÕýÈ·µÄÊÇ

񅧏

A

B

C

D

Ç¿µç½âÖÊ

KNO3

H2SO4

BaSO4

HClO4

Èõµç½âÖÊ

HClO4

CaCO3

HClO

NH3¡¤H2O

·Çµç½âÖÊ

SO2

CS2

H2O

C2H5OH

A. A B. B C. C D. D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÈçͼÊÇÒ»¸ö»¯Ñ§¹ý³ÌµÄʾÒâͼ:

Çë»Ø´ð:

£¨1£©Í¼Öм׳ØÊÇ______×°Öã¬ÆäÖÐOH£­ÒÆÏò_________¼«(Ìî¡°CH3OH¡±»ò¡°O2¡±)¡£

£¨2£©Ð´³öͨÈëCH3OHµç¼«µÄµç¼«·´Ó¦Ê½:_______________________________¡£

£¨3£©ÏòÒÒ³ØÁ½µç¼«¸½½üµÎ¼ÓÊÊÁ¿×ÏɫʯÈïÊÔÒº£¬¸½½ü±äºìµÄµç¼«Îª_____¼«(Ìî¡°A¡±»ò¡°B¡±)£¬

²¢Ð´³ö´Ëµç¼«µÄµç¼«·´Ó¦Ê½:______________________¡£

£¨4£©ÒÒ³ØÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________¡£

£¨5£©µ±ÒÒ³ØÖÐB(Ag)¼«ÖÊÁ¿Ôö¼Ó5.40 gʱ£¬ÒÒ³Øc(H+)ÊÇ_______(ÈôÒÒ³ØÖÐÈÜҺΪ500 mL)£»´Ëʱ±û³Øijµç¼«Îö³ö1.60 gij½ðÊô£¬Ôò±ûÖеÄijÑÎÈÜÒº¿ÉÄÜÊÇ_________(ÌîÐòºÅ)¡£

A£®MgSO4 B£®CuSO4 C£®NaCl D£®AgNO3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁÐÓйØpH±ä»¯µÄÅжÏÖУ¬ÕýÈ·µÄÊÇ£¨ £©

A. ËæζȵÄÉý¸ß£¬´¿Ë®µÄKw¼õС

B. ËæζȵÄÉý¸ß£¬´¿Ë®µÄpHÔö´ó

C. ÐÂÖÆÂÈË®¾­¹âÕÕÒ»¶Îʱ¼äºó£¬pH¼õС

D. ÇâÑõ»¯ÄÆÈÜÒº¾ÃÖÃÓÚ¿ÕÆøÖУ¬pHÔö´ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÖÓÃ0.1000 mol¡¤L£­1KMnO4ËáÐÔÈÜÒºµÎ¶¨Î´ÖªÅ¨¶ÈµÄÎÞÉ«H2C2O4ÈÜÒº£¬·´Ó¦Àë×Ó·½³ÌʽÊÇ£º2MnO4£­£«5H2C2O4£«6H+ = 2Mn2+£«10CO2¡ü£«8H2O

Ìî¿ÕÍê³ÉÎÊÌ⣺

£¨1£©¸ÃµÎ¶¨ÊµÑéËùÐèµÄ²£Á§ÒÇÆ÷ÓÐ______________¡££¨Ìî×Öĸ£©

A£®ËáʽµÎ¶¨¹ÜB£®¼îʽµÎ¶¨¹Ü C£®Á¿Í² D£®×¶ÐÎÆ¿ E£®Ìú¼Ų̈F£®µÎ¶¨¹Ü¼ÐG£®ÉÕ±­H£®°×Ö½ I£®Â©¶·

£¨2£©²»ÓÃ________(Ìî¡°Ëᡱ»ò¡°¼î¡±)ʽµÎ¶¨¹ÜÊ¢·Å¸ßÃÌËá¼ØÈÜÒº¡£ÊÔ·ÖÎöÔ­Òò___________________________________________¡£

£¨3£©µÎ¶¨ÖÕµãµÄÏÖÏóΪ___________________________________¡£

£¨4£©ÈôµÎ¶¨¿ªÊ¼ºÍ½áÊøʱ£¬µÎ¶¨¹ÜÖеÄÒºÃæÈçͼËùʾ£¬ÔòÆðʼ¶ÁÊýΪ________mL£¬ÖÕµã¶ÁÊýΪ________mL¡£

£¨5£©Ä³Ñ§Éú¸ù¾Ý3´ÎʵÑé·Ö±ð¼Ç¼ÓйØÊý¾ÝÈçÏÂ±í£º

µÎ¶¨

´ÎÊý

´ý²âH2C2O4ÈÜÒºµÄÌå»ý/mL

0.1000 mol/L KMnO4µÄÌå»ý£¨mL£©

µÎ¶¨Ç°¿Ì¶È

µÎ¶¨ºó¿Ì¶È

ÈÜÒºÌå»ý/mL

µÚÒ»´Î

25.00

0.00

26.11

26.11

µÚ¶þ´Î

25.00

1.56

30.30

28.74

µÚÈý´Î

25.00

0.22

26.31

26.09

ÒÀ¾ÝÉϱíÊý¾ÝÁÐʽ¼ÆËã¸ÃH2C2O4ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ_______________¡£

£¨6£©ÏÂÁвÙ×÷ÖпÉÄÜʹ²â¶¨½á¹ûÆ«µÍµÄÊÇ___________(Ìî×Öĸ)¡£

A£®ËáʽµÎ¶¨¹ÜδÓñê×¼ÒºÈóÏ´¾ÍÖ±½Ó×¢ÈëKMnO4±ê×¼Òº

B£®µÎ¶¨Ç°Ê¢·Å²ÝËáÈÜÒºµÄ׶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»ºóûÓиÉÔï

C£®ËáʽµÎ¶¨¹Ü¼â×첿·ÖÔڵζ¨Ç°Ã»ÓÐÆøÅÝ£¬µÎ¶¨ºóÓÐÆøÅÝ

D£®¶ÁÈ¡KMnO4±ê׼Һʱ£¬¿ªÊ¼ÑöÊÓ¶ÁÊý£¬µÎ¶¨½áÊøʱ¸©ÊÓ¶ÁÊý

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸