¡¾ÌâÄ¿¡¿ÁòËáÊÇÈý´ó¹¤ÒµÓÃËáÖ®Ò»£¬ÔÚÒ±½ð¡¢Ê¯Ó͹¤Òµ¡¢ÖÆÒ©¡¢¹ú·ÀµÈ·½Ã涼ҪÓõ½ÁòËá¡£½«ÊÊÁ¿µÄÕáÌÇ·ÅÈëÉÕ±­ÖУ¬¼ÓÉÙÁ¿Ë®°èÔÈ£¬ÔÙ¼ÓÊÊÁ¿Å¨ÁòËᣬѸËÙ½Á°è£¬·Å³ö´óÁ¿µÄÈÈ£¬Í¬Ê±¹Û²ìµ½ÕáÌÇÖð½¥±äºÚ£¬Ìå»ýÅòÕÍ£¬²¢·Å³öÓд̼¤ÐÔÆøζµÄÆøÌå¡£ÊԻشð£º

£¨1£©¼ÓÈëÉÙÁ¿Ë®µÄÔ­ÒòÊÇ_________________________________________________¡£

£¨2£©Éú³ÉµÄºÚÉ«ÎïÖÊÊÇ______________£¨Ð´³ö»¯Ñ§Ê½£©¡£

£¨3£©¡°Ìå»ýÅòÕÍ£¬²¢·Å³öÓд̼¤ÐÔÆøζµÄÆøÌ壨SO2£©¡±£¬Ð´³ö¶ÔÓ¦µÄ»¯Ñ§·½³Ìʽ£º_________________________________________________¡£

£¨4£©ÉÏÊöÏÖÏóÌåÏÖÁËŨÁòËáµÄ_________________£¨ÌîÐòºÅ£©

¢ÙËáÐÔ ¢ÚÎüË®ÐÔ ¢ÛÍÑË®ÐÔ ¢ÜÇ¿Ñõ»¯ÐÔ

£¨5£©Ïò80mLŨÁòËáÖмÓÈë5.6gÍ­£¬¼ÓÈÈÒ»¶Îʱ¼äºóÖÁ²»ÔÙ·´Ó¦ÎªÖ¹£¬ÊµÑé²âµÃ·´Ó¦Öй²ÓÐ1.12L£¨±ê×¼×´¿öÏ£©SO2ÆøÌåÉú³É£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________________________£¬·´Ó¦ÖÐתÒƵç×Ó______mol£¬Í­Ê£Óà_______g£¬¸Ã·´Ó¦ÖÐŨÁòËáÌåÏÖÁË___________ºÍ__________¡£

¡¾´ð°¸¡¿Å¨ÁòËáºÍÉÙÁ¿Ë®»ìºÏʱ»á·Å³ö´óÁ¿µÄÈÈ£¬¼Ó¿ì·´Ó¦ C C+2H2SO4£¨Å¨£©CO2¡ü+2SO2¡ü+2H2O ¢Ú¢Û¢Ü Cu+2H2SO4(Ũ) CuSO4+SO2¡ü+2H2O 0.1 2.8 Ç¿Ñõ»¯ÐÔ ËáÐÔ

¡¾½âÎö¡¿

½«ÊÊÁ¿µÄÕáÌÇ·ÅÈëÉÕ±­ÖУ¬¼ÓÉÙÁ¿Ë®°èÔÈ£¬ÀûÓÃŨÁòËáÓëË®»ìºÏ·Å³ö´óÁ¿µÄÈÈ£¬Î¶ÈÉý¸ß£¬¼Ó¿ì·´Ó¦ËÙÂÊ£¬ÔÙ¼ÓÊÊÁ¿Å¨ÁòËᣬѸËÙ½Á°è£¬·Å³ö´óÁ¿µÄÈÈ£¬Í¬Ê±¹Û²ìµ½ÕáÌÇÖð½¥±äºÚ£¬ÀûÓÃŨÁòËáµÄÍÑË®ÐÔ£¬½«ÕáÌDZäΪÁËÌ¿ºÚ£¬Ì¿ºÚºÍŨÁòËáÔÚ¼ÓÈÈÌõ¼þÏ·¢Éú·´Ó¦Éú³É¶þÑõ»¯ÁòºÍ¶þÑõ»¯Ì¼ÆøÌ壬Ìå»ýÅòÕÍ£¬²¢·Å³öÓд̼¤ÐÔÆøζµÄÆøÌå¡£

¢Å¼ÓÈëÉÙÁ¿Ë®Ö÷ÒªÊÇÀûÓÃË®ÓëŨÁòËá»ìºÏ·Å³ö´óÁ¿µÄÈÈ£¬·Å³öµÄÈÈÁ¿Ê¹Î¶ÈÉý¸ß£¬¼Ó¿ì·´Ó¦ËÙÂÊ£¬¹Ê´ð°¸Îª£ºÅ¨ÁòËáºÍÉÙÁ¿Ë®»ìºÏʱ»á·Å³ö´óÁ¿µÄÈÈ£¬¼Ó¿ì·´Ó¦£»

¢ÆŨÁòËáÓëÕáÌÇ»ìºÏ£¬Å¨ÁòËá¾ßÓÐÍÑË®ÐÔ£¬Ê¹ÕáÌDZäΪ̿ºÚ£¬Òò´Ë¸ÃÉú³ÉµÄºÚÉ«ÎïÖÊÊÇ̼£¬¹Ê´ð°¸Îª£ºC£»

¢ÇÉú³ÉµÄ̼ºÍŨÁòËá»ìºÏÔÚÊÜÈȵÄÌõ¼þÏ·´Ó¦·Å³ö¶þÑõ»¯Ì¼ºÍ¶þÑõ»¯ÁòÆøÌ壬¡°Ìå»ýÅòÕÍ£¬²¢·Å³öÓд̼¤ÐÔÆøζµÄÆøÌ壨SO2£©¡±£¬Ð´³ö¶ÔÓ¦µÄ»¯Ñ§·½³Ìʽ£ºC+2H2SO4£¨Å¨£©CO2¡ü+2SO2¡ü+2H2O£¬¹Ê´ð°¸Îª£ºC+2H2SO4£¨Å¨£©CO2¡ü+2SO2¡ü+2H2O£»

¢ÈÒòŨÁòËá¾ßÓÐÎüË®ÐÔ£¬Å¨ÁòËáÎüˮϡÊͶø·Å³ö´óÁ¿µÄÈÈ£¬Å¨ÁòËá¾ßÓÐÍÑË®ÐÔ£¬¿ÉʹÕáÌÇ̼»¯¶ø±äºÚ£¬Éú³ÉµÄºÚÉ«ÎïÖÊÊÇC£¬Ì¼ºÍŨÁòËá·¢ÉúÑõ»¯»¹Ô­·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍ¶þÑõ»¯ÁòÆøÌ壬ÆäÖжþÑõ»¯Áò¾ßÓд̼¤ÐÔÆøζ£¬Å¨ÁòËáÇ¿Ñõ»¯ÐÔ£¬Òò´ËÌåÏÖÎüË®ÐÔ¡¢ÍÑË®ÐÔ¡¢Ç¿Ñõ»¯ÐÔ£»¹Ê¢Ú¢Û¢ÜÕýÈ·£¬¹Ê´ð°¸Îª£º¢Ú¢Û¢Ü£»

¢ÉÏò80mLŨÁòËáÖмÓÈë5.6gÍ­£¬¼ÓÈÈÒ»¶Îʱ¼äºóÖÁ²»ÔÙ·´Ó¦ÎªÖ¹£¬ÊµÑé²âµÃ·´Ó¦Öй²ÓÐ1.12L£¨±ê×¼×´¿öÏ£©SO2ÆøÌåÉú³É£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCu+2H2SO4(Ũ) CuSO4+SO2¡ü+2H2O£¬¹Êµ«ÎªCu+2H2SO4(Ũ) CuSO4+SO2¡ü+2H2O£»·´Ó¦Éú³ÉÁË1.12L¶þÑõ»¯ÁòÆøÌå¼´0.05 mol£¬·´Ó¦ÖÐתÒƵç×Ó2¡Á0.05 mol =0.1 mol£¬¹Ê´ð°¸Îª0.1 mol£¬Í­ÏûºÄÁË0.05 mol£¬Í­Ê£Óà5.6 ¡ª0.05¡Á64 = 2.4g£¬¹Ê´ð°¸Îª2.4 g£¬Å¨ÁòËáÉú³É¶þÑõ»¯Áò£¬»¯ºÏ¼Û½µµÍ£¬ÌåÏÖÇ¿Ñõ»¯ÐÔ£¬Å¨ÁòËá±äΪÁòËáÍ­£¬ÌåÏÖËáÐÔ£¬¹Ê´ð°¸Îª£ºÇ¿Ñõ»¯ÐÔºÍËáÐÔ£»

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÏÂÁйØÓÚÈÈ»¯Ñ§·´Ó¦µÄÃèÊöÖÐÕýÈ·µÄÊÇ

A. HClºÍNaOH·´Ó¦µÄÖкÍÈÈ¡÷H£½£­57.3 kJ¡¤mol1£¬ÔòH2SO4ºÍCa(OH)2·´Ó¦µÄÖкÍÈÈ¡÷H=2¡Á(£­57.3)kJ¡¤mol1

B. ¼×ÍéµÄ±ê׼ȼÉÕÈȦ¤H£½£­890.3 kJ¡¤mol1£¬ÔòCH4(g)£«2O2(g)£½CO2(g)£«2H2O(g) ¦¤H£¼£­890.3 kJ¡¤mol1

C. ÒÑÖª£º500¡æ¡¢30MPaÏ£¬N2(g)£«3H2(g)2NH3(g) ¦¤H£½£­92.4kJ¡¤mol£­1£»½«1.5 mol H2ºÍ¹ýÁ¿µÄN2ÔÚ´ËÌõ¼þϳä·Ö·´Ó¦£¬·Å³öÈÈÁ¿46.2 kJ

D. CO(g)µÄȼÉÕÈÈÊÇ283.0kJ¡¤mol1£¬Ôò2CO2(g) ===2CO(g)+O2(g)·´Ó¦µÄ¡÷H£½+566.0 kJ¡¤mol1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿·Ö×ÓʽΪC9H18O2µÄõ¥ÔÚËáÐÔÌõ¼þÏÂË®½â£¬ËùµÃµÄôÈËáºÍ´¼ÔÚÏàͬÌõ¼þ϶þÕßÕôÆûµÄÃܶÈÏàͬ£¬·ûºÏ´ËÌõ¼þµÄõ¥µÄͬ·ÖÒì¹¹ÊýĿΪ

A. 2 B. 8 C. 10 D. 16

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿CO¡¢H2¡¢CH3OH¾ùÊÇÇå½àÄÜÔ´¡£

£¨1£©ÒÑÖª²¿·Ö»¯Ñ§¼ü¼üÄÜÊý¾ÝÈçÏ£º

»¯Ñ§¼ü

CO

O=O

C=O

C-O

E/(kJ mol-1)

958.5

497

745

351

2CO(g) +O2(g)==2CO2(g) H1 H2O(g)+CO(g)==H2(g) + CO2(g) H2 = -41 kJmol-1

CH3OH(g)+ 3/2O2(g)==CO2(g)+2H2O(g) H3 = -660kJmol-1

Ôò¡÷H1=_____ kJmol-1£¬·´Ó¦CO(g)+2H2(g)CH3OH(g)µÄ¡÷H=_____ kJmol-1¡£

£¨2£©Ò»¶¨Ìõ¼þÏ£¬ÔÚÈÝ»ýΪ2 LµÄÃܱÕÈÝÆ÷QÖгäÈëa mol COÓëb molH2ºÏ³É¼×´¼£ºCO(g) +2H2(g) CH3OH(g)¡£²âµÃƽºâʱ»ìºÏÆøÌåÖÐCH3OHµÄÌå»ý°Ù·Öº¬Á¿Óëζȡ¢ ѹǿ֮¼äµÄ¹ØϵÈçͼ1Ëùʾ£¬Í¼2±íʾÔÚÒ»¶¨Î¶ÈÏ£¬H2µÄƽºâת»¯ÂÊÓë·´Ó¦¿ªÊ¼Ê±Á½ÖÖ·´Ó¦ÎïµÄͶÁÏÎïÖʵÄÁ¿Ö®±È£¨ÓÃX±íʾ£©¡¢Ñ¹Ç¿Ö®¼äµÄ¹Øϵ¡£

¢ÙѹǿÏàͬʱ£¬Î¶ÈΪT1¡¢T2ʱ£¬·´Ó¦´ïµ½Æ½ºâËùÐèÒªµÄʱ¼ä·Ö±ðΪt1¡¢t2£¬Ôò¶þÕßÖ®¼äµÄÏà¶Ô´óСΪt1___ t2(Ìî¡°>¡±¡¢¡°<¡±¡¢¡°=¡±»ò¡°ÎÞ·¨È·¶¨¡±)¡£

¢ÚP1_____P2(Ìî¡°>¡±¡¢¡°<¡±¡¢¡°=¡±»ò¡°ÎÞ·¨È·¶¨¡±)¡£

¢ÛÈôa =2£¬b=4£¬ÔòѹǿΪP1¡¢Î¶ÈΪT1ʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK=______________¡£

¢ÜÈôÔÚѹǿΪP1¡¢Î¶ÈΪT1ʱ£¬ÏòQÈÝÆ÷ÖÐͬʱ¼ÓÈëµÈÎïÖʵÄÁ¿µÄCO¡¢H2¡¢CH3OHÈýÖÖÆøÌ壬Ôò·´Ó¦¿ªÊ¼Ê±£¬v(CH3OH)Õý_____v(CH3OH)Äæ(Ìî¡°>¡±¡¢¡°<¡±¡¢¡°=¡±»ò¡°ÎÞ·¨È·¶¨¡±)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿µÈÎïÖʵÄÁ¿µÄNa2CO3¡¤mH2OºÍBaCl2¡¤nH2OµÄ»ìºÏÎï3.68g£¬¼ÓÈë×ãÁ¿Ë®½Á°è£¬¾­³ä·Ö·´Ó¦ºó¿ÉµÃ1.97g³Áµí¡£ÔòmºÍnÖµ·Ö±ðÊÇ

A.10ºÍ2B.7ºÍ3C.3ºÍ1D.1ºÍ2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓûÅäÖÆ0.1mol/LµÄNaHCO3ÈÜÒº250mL£¬Íê³ÉÏÂÁв½Ö裺

£¨1£©Óõç×ÓÌìƽ³ÆÈ¡NaHCO3¹ÌÌå______________g£»

£¨2£©½«³ÆºÃµÄNaHCO3¹ÌÌå·ÅÈë______________£¬¼ÓÊÊÁ¿µÄÕôÁóË®½«Æä___________£»

£¨3£©´ýÈÜÒº_____________ºó£¬½«ÈÜÒºÑØ×Å______________ÒÆÈë250mLÈÝÁ¿Æ¿ÖУ¬ÔÚ²Ù×÷¹ý³ÌÖв»ÄÜËðʧÈκÎÒ»µãÒºÌ壬·ñÔò»áʹÈÜÒºµÄŨ¶ÈÆ«_________£¨¸ß»òµÍ£©£»

£¨4£©ÓÃÉÙÁ¿ÕôÁóË®³åÏ´_______________2-3´Î£¬½«Ï´µÓÒºÒÆÈë______________ÖУ»

£¨5£©ÏòÈÝÁ¿Æ¿ÄÚ¼ÓË®ÖÁ¾àÀë¿Ì¶ÈÏß______________ʱ£¬¸ÄÓÃ________________СÐļÓË®ÖÁ___________________£¬Èô¼ÓË®³¬¹ý¿Ì¶ÈÏߣ¬»áÔì³ÉÈÜҺŨ¶ÈÆ«_________£¨¸ß»òµÍ£©£¬Ó¦¸Ã_____________£»

£¨6£©×îºó¸ÇºÃÆ¿¸Ç£¬___________£¬È»ºó½«ÅäºÃµÄÈÜÒºÒÆÈëÊÔ¼ÁÆ¿ÖУ¬ÌùºÃ±êÇ©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿ÓÃÖÊÁ¿·ÖÊýΪ375%µÄŨÑÎËáÃܶÈΪ116 g/cm3ÅäÖÆ250mLŨ¶ÈΪ1mol/LµÄÏ¡ÑÎËá¡£ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©ÅäÖÆÏ¡ÑÎËáʱ£¬Ó¦Ñ¡ÓÃÈÝÁ¿Îª______mLµÄÈÝÁ¿Æ¿£»

£¨2£©¾­¼ÆËãÐèÒª______ mLŨÑÎËᣬÔÚÁ¿È¡Ê±ÒËÑ¡ÓÃÏÂÁÐÁ¿Í²ÖеÄ______¡££¨Ìî×Öĸ£©

A. 5 mL B. 10 mL C. 25 mL D. 50 mL

£¨3£©ÔÚÁ¿È¡Å¨ÑÎËáºó£¬½øÐÐÁËÏÂÁвÙ×÷£º

¢ÙµÈÏ¡ÊͺóµÄÑÎËáζÈÓëÊÒÎÂÒ»Öºó£¬Ñز£Á§°ô½«ÑÎËá×¢ÈëÈÝÁ¿Æ¿ÖС£

¢ÚÍùÈÝÁ¿Æ¿ÖÐСÐļÓÕôÁóË®ÖÁÒºÃæ½Ó½ü»·ÐαêÏß2¡«3 cm´¦£¬¸ÄÓýºÍ·µÎ¹Ü¼ÓÕôÁóË®£¬Ê¹ÈÜÒºµÄ°¼Ãæµ×²¿ÓëÆ¿¾±µÄ»·ÐαêÏßÏàÇС£

¢ÛÔÚÊ¢ÑÎËáµÄÉÕ±­ÖÐ×¢ÈëÕôÁóË®¼¸Ê®ºÁÉý£¬²¢Óò£Á§°ô½Á¶¯£¬Ê¹Æä»ìºÏ¾ùÔÈ¡£

¢ÜÓÃÕôÁóˮϴµÓÉÕ±­ºÍ²£Á§°ô2ÖÁ3´Î£¬²¢½«Ï´µÓҺȫ²¿×¢ÈëÈÝÁ¿Æ¿¡£

ÉÏÊö²Ù×÷ÖУ¬ÕýÈ·µÄ˳ÐòÊÇ£¨ÌîÐòºÅ£©________________________¡£

£¨4£©ÔÚÉÏÊöÅäÖƹý³ÌÖУ¬ÓøոÕÏ´µÓ½à¾»µÄÁ¿Í²À´Á¿È¡Å¨ÑÎËᣬÆäÅäÖƵÄÏ¡ÑÎËáŨ¶ÈÊÇ__________£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±¡¢¡°ÎÞÓ°Ï족£©¡£ÈôδÓÃÕôÁóˮϴµÓÉÕ±­ÄÚ±Ú»òδ½«Ï´µÓҺעÈëÈÝÁ¿Æ¿£¬ÔòÅäÖƵÄÏ¡ÑÎËáŨ¶ÈÊÇ_______£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±¡¢¡°ÎÞÓ°Ï족£©¡£

£¨5£©ÅäÖÆÍêºó£¬Ó¦½«ÈÝÁ¿Æ¿ÖеÄÏ¡ÑÎËáתÒƵ½_______Öдæ·Å£¬²¢ÌùÉϱêÇ©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿¼ºÖªCO2Óë¼îÈÜÒº·¢Éú·´Ó¦Ê±£¬ÒÀ´Î·¢ÉúÈçÏÂÀë×Ó·´Ó¦£ºCO2£«2OH£­¡úCO32£­£«H2O¡¢CO2£«CO32£­£«H2O¡ú2HCO

£¨1£©Ïòº¬0.01molÇâÑõ»¯¸ÆµÄʯ»ÒË®ÖÐͨÈëCO2£¬µÃµ½0.4g°×É«³Áµí£¬ÔòͨÈëCO2µÄÎïÖʵÄÁ¿Îª______mol»ò______mol£®

£¨2£©ÏÖÓÐNaOHºÍBa(OH)2»ìºÏÈÜÒº100mL£¬ÒÑÖª¸ÃÈÜÒºÖÐc(OH£­)=1mol/L£¬ÏòÈÜÒºÖлº»ºÍ¨ÈëCO2ÆøÌ壨ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©£®

¢Ùµ±ÖÁÉÙͨÈëCO2ÆøÌåÌå»ýΪ0.56L£¨±ê׼״̬£©Ê±Éú³ÉµÄ³Áµí×î¶à£¬ÔòÔ­ÈÜÒºÖÐÇâÑõ»¯ÄƺÍÇâÑõ»¯±µµÄÎïÖʵÄÁ¿Å¨¶È¸÷Ϊ¶àÉÙ£¿__________________

¢Úµ±Í¨ÈëµÄCO2ÆøÌå×ÜÌå»ýΪ2.24L£¨±ê׼״̬£©Ê±£¬ÈÜÒºÖÐÒõÀë×Ó£¨OH£­³ýÍ⣩µÄÎïÖʵÄÁ¿Å¨¶ÈÊǶàÉÙ£¿______________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¡¾ÌâÄ¿¡¿´×ËáÊÇÒ»ÖÖ³£¼ûµÄÈõËᣬΪÁËÖ¤Ã÷´×ËáÊÇÈõµç½âÖÊ£¬Ä³Í¬Ñ§¿ªÕ¹ÁËÌâΪ¡°´×ËáÊÇÈõµç½âÖʵÄʵÑéÑо¿¡±µÄ̽¾¿»î¶¯¡£

£¨1£©±¾¿ÎÌâµÄÑо¿¶ÔÏóÊÇCH3COOH£¬Ñо¿Ä¿µÄÊÇ_____£¬Ñо¿·½·¨ÊÇʵÑéÑо¿¡£

£¨2£©¸ÃͬѧÉè¼ÆÁËÈçÏ·½°¸£º

A.ÏÈÅäÖÆÒ»¶¨Á¿µÄ0.10mol/L CH3COOHÈÜÒº£¬È»ºó²âÈÜÒºµÄpH£¬ÈôpH´óÓÚ1£¬Ôò¿ÉÖ¤Ã÷´×ËáΪÈõµç½âÖÊ¡£

B.ÏÈÅäÖÆ0.01mol/LºÍ0.10mol/LµÄCH3COOH£¬·Ö±ð²âËûÃǵÄpH£¬ÈôÁ½ÕßµÄpHÏà²îСÓÚ1¸öµ¥Î»£¬Ôò¿ÉÖ¤Ã÷´×ËáÊÇÈõµç½âÖÊ¡£

C.ÏȲâ0.10mol/LCH3COOHµÄpH£¬È»ºó½«Æä¼ÓÈÈÖÁ100¡æ£¨²»¿¼ÂÇ´×Ëá»Ó·¢£©£¬ÔÙ²âpH£¬Èç¹ûpH±äС£¬Ôò¿ÉÖ¤Ã÷´×ËáÊÇÈõµç½âÖÊ¡£

D.ÅäÖÆÒ»¶¨Á¿µÄCH3COONaÈÜÒº£¬²âÆäpH£¬Èô³£ÎÂÏÂpH´óÓÚ7£¬Ôò¿ÉÖ¤Ã÷´×ËáÊÇÈõµç½âÖÊ¡£

ÄãÈÏΪÉÏÊö·½°¸¿ÉÐеÄÊÇ____£¨¶àÏîÑ¡Ôñ£©

£¨3£©Çë»Ø´ðÒÔÏÂÁ½¸öÎÊÌ⣺

¢ÙÈõµç½âÖʵÄÌصãÊÇ____£¨¶àÏîÑ¡Ôñ£©

A.ÔÚË®ÈÜÒºÖÐÈ«²¿µçÀë B.ÔÚË®ÈÜÒºÖв¿·ÖµçÀë

C.Ë®ÈÜÒº³ÊËáÐÔ D.Ë®ÈÜÒºÖдæÔÚµçÀëƽºâ

¢Úд³ö´×ËáµÄµçÀë·½³Ìʽ£º____¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸