13£®Èç¹û1¸ö·´Ó¦¿ÉÒÔ·Ö¼¸²½½øÐУ¬Ôò¸÷·Ö²½·´Ó¦µÄ·´Ó¦ÈÈÖ®ºÍÓë¸Ã·´Ó¦Ò»²½Íê³ÉʱµÄ·´Ó¦ÈÈÊÇÏàͬµÄ£¬Õâ¸ö¹æÂɳÆΪ¸Ç˹¶¨ÂÉ£®¾Ý´Ë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©±±¾©°ÂÔ˻ᡰÏéÔÆ¡±»ð¾æȼÁÏÊDZûÍ飨C3H8£©£¬ÑÇÌØÀ¼´ó°ÂÔË»á»ð¾æȼÁÏÊDZûÏ©£¨C3H6£©£®±ûÍéÍÑÇâ¿ÉµÃ±ûÏ©£®
ÒÑÖª£º¢ÙCH3CH2CH3£¨g£©¡úCH4£¨g£©+HC¡ÔCH£¨g£©+H2£¨g£©£»¡÷H1=+156.6kJ•mol-1?
¢ÚCH3CH=CH2£¨g£©¡úCH4£¨g£©+HC¡ÔCH£¨g£©£»¡÷H2=+32.4kJ•mol-1?
ÔòÏàͬÌõ¼þÏ£¬±ûÍéÍÑÇâµÃ±ûÏ©µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºC3H8£¨g£©=CH3CH=CH2£¨g£©+H2£¨g£©¡÷H=+124.2 kJ•mol-1
£¨2£©ÒÒÅðÍ飨B2H6£©ÊÇÒ»ÖÖÆø̬¸ßÄÜȼÁÏ£¬0.3molµÄÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5kJÈÈÁ¿£¬ÆäÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£»
ÈôH2O£¨l£©=H2O£¨g£©£»¡÷H=+44kJ/mol£¬Ôò5.6L£¨±ê¿ö£©ÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆø̬ˮʱ£¬·Å³öµÄÈÈÁ¿ÊÇ508.25 kJ£®

·ÖÎö £¨1£©ÒÀ¾ÝÈÈ»¯Ñ§·½³Ìʽ½áºÏ¸Ç˹¶¨ÂɼÆËãµÃµ½£»
£¨2£©¸ù¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨¿ÉÖª£¬ÎïÖʵÄÎïÖʵÄÁ¿Óë·´Ó¦·Å³öµÄÈÈÁ¿³ÉÕý±È£¬²¢×¢Òâ±êÃ÷¸÷ÎïÖʵľۼ¯×´Ì¬À´½â´ð£»
Ïȸù¾Ý¸Ç˹¶¨ÂÉд³ö·½³Ìʽ£¬È»ºó¸ù¾ÝÎïÖʵÄÎïÖʵÄÁ¿Óë·´Ó¦·Å³öµÄÈÈÁ¿³ÉÕý±ÈÀ´½â´ð£®

½â´ð ½â£º£¨1£©¢ÙC3H8£¨g£©=CH4£¨g£©+HC¡ÔCH£¨g£©+H2£¨g£©¡÷H1=+156.6kJ•mol-1
¢ÚCH3CH=CH2£¨g£©=CH4£¨g£©+HC¡ÔCH£¨g £©¡÷H2=+32.4kJ•mol-1
ÒÀ¾Ý¸Ç˹¶¨ÂÉ¢Ù-¢ÚµÃµ½ÈÈ»¯Ñ§·½³ÌʽΪ£ºC3H8£¨g£©=CH3CH=CH2£¨g£©+H2£¨g£©¡÷H=+124.2 kJ•mol-1 £»
¹Ê´ð°¸Îª£ºC3H8£¨g£©=CH3CH=CH2£¨g£©+H2£¨g£©¡÷H=+124.2 kJ•mol-1 £»
£¨2£©0.3molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5KJµÄÈÈÁ¿£¬Ôò1molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö2165KJµÄÈÈÁ¿£¬·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£¬
¸ù¾Ý£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£» ¢Ù
H2O£¨l£©¡úH2O£¨g£©£»¡÷H=+44kJ/moL£¬¢Ú
¢Ù+¢Ú¡Á3µÃ£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨g£©¡÷H=-2033kJ/mol£¬¹Ê5.6L£¨±ê×¼×´¿ö£©¼´0.25molÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆø̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇΪ2033kJ¡Á0.25=508.25kJ£»
¹Ê´ð°¸Îª£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol£»508.25 kJ£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËÈÈ»¯Ñ§·½³ÌʽµÄÊéд£¬¸Ç˹¶¨ÂɵÄÔËÓÃÒÔ¼°ÈÈÁ¿µÄ¼ÆË㣬ÐèҪעÒâµÄÊÇ·´Ó¦ÈȵÄÊýÖµÓ뻯ѧ·½³ÌʽǰÃæµÄϵÊý³ÉÕý±È£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

3£®Ñ§Ï°ÁËÔªËØÖÜÆÚÂɵÄÓйØ֪ʶºó£¬Ä³Í¬Ñ§¸ù¾ÝÔªËطǽðÊôÐÔÓë¶ÔÓ¦×î¸ß¼Ûº¬ÑõËáÖ®¼äµÄ¹Øϵ£¬Ñ¡ÔñÓɶÌÖÜÆÚÔªËØ×é³ÉµÄ»¯ºÏÎïÉè¼ÆÁËÈçͼ1×°ÖÃÀ´Ò»´ÎÐÔÍê³ÉͬÖ÷×åºÍͬÖÜÆÚÔªËطǽðÊôÐÔÇ¿Èõ±È½ÏµÄʵÑéÑо¿£®
£¨1£©Ð´³öÑ¡ÓÃÎïÖʵĻ¯Ñ§Ê½£ºBNa2CO3¡¢CNa2SiO3£®¸ÉÔï¹ÜDµÄ×÷ÓÃÊÇ·Àµ¹Îü
£¨2£©¸Ã×°ÖôæÔÚȱÏÝ£¬ÇëÖ¸³ö£ºÏõËá¾ßÓлӷ¢ÐÔ£¬ÓëÉú³ÉµÄCO2Ò»Æð½øÈë×°ÖÃCÖУ¬ÔòÎÞ·¨Ö¤Ã÷C¡¢SiµÄ·Ç½ðÊôÐÔÇ¿Èõ£¬¸Ä½øµÄ·½·¨ÊÇ£ºÔÚB£¬C¼äÁ¬½ÓÒ»¸ö×°Óб¥ºÍNaHCO3ÈÜÒº µÄÏ´ÆøÆ¿
£¨3£©¸Ä½øºó£¬½øÐÐʵÑ飮¸ù¾ÝʵÑéÄ¿µÄ£¬ÉÕ±­CÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽH2O+CO2+SiO32-¨TH2SiO3¡ý+CO32-£¬
£¨4£©Í¨¹ýʵÑéµÃ³öµÄ½áÂÛ£ºÔªËطǽðÊôÐÔÇ¿Èõ˳ÐòΪN£¾C£¾Si£»´ÓÔ­×ӽṹµÄ½Ç¶È¼òÒª·ÖÎöͬÖÜÆÚÔªËØÐÔÖʵݱäµÄÔ­Òò£ºÍ¬ÖÜÆÚÔªËغËÍâµç×Ó²ãÊýÏàͬ£¬ËæÔ­×ÓÐòÊýÔö´ó£¬ºËµçºÉÊýÔö´ó£¬Ô­×Ӻ˶ԺËÍâµç×ÓµÄÎüÒýÄÜÁ¦Öð½¥ÔöÇ¿£¬Ô­×Óʧȥµç×ÓµÄÄÜÁ¦Öð½¥¼õÈõ£¬µÃµç×ÓÄÜÁ¦Öð½¥ÔöÇ¿£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

4£®¹ýÁ¿µÄÍ­ÓëŨÁòËáÔÚ¼ÓÈÈÌõ¼þϵķ´Ó¦»áÒòÁòËáŨ¶ÈϽµ¶øÍ£Ö¹£®Îª²â¶¨·´Ó¦²ÐÓàÇåÒºÖÐÁòËáµÄŨ¶È£¬Ì½¾¿Ð¡×éͬѧÌá³öµÄÏÂÁÐʵÑé·½°¸£º
¼×·½°¸£ºÓë×ãÁ¿BaCl2ÈÜÒº·´Ó¦£¬³ÆÁ¿Éú³ÉµÄBaSO4ÖÊÁ¿£®
ÒÒ·½°¸£ºÓë×ãÁ¿Ð¿·´Ó¦£¬²âÁ¿Éú³ÉÇâÆøµÄÌå»ý£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Åж¨¼×·½°¸²»¿ÉÐУ¬ÀíÓÉÊǼÓÈë×ãÁ¿µÄBaCl2ÈÜÒºÖ»ÄÜÇó³öÁòËá¸ùÀë×ÓµÄÁ¿£¬¶ø²»ÄÜÇó³öÊ£ÓàÁòËáµÄŨ¶È£®
£¨2£©¼×ͬѧÓÃͼ1ËùʾװÖýøÐÐÒÒ·½°¸µÄʵÑ飮
¢Ù¼ì²é´Ë×°ÖÃÆøÃÜÐԵķ½·¨Á¬½Ó×°Öúó£¬ÏòÁ¿Æø¹ÜÓÒ¶Ëעˮ£¬Ö±µ½×óÓÒÁ½±ß¹Ü×ÓÐγÉÒ»¶ÎÒºÃæ²î£¬Ò»¶Îʱ¼äºóÒºÃæ²î²»±ä£®
¢ÚʹYÐιÜÖеIJÐÓàÇåÒºÓëпÁ£·´Ó¦µÄÕýÈ·²Ù×÷Êǽ«ZnÁ£×ªÒƵ½Ï¡ÁòËáÖУ®²ÐÓàÇåÒºÓëпÁ£»ìºÏºóµÄÏÖÏó±íÃæÎö³ö°µºìÉ«¹ÌÌ壬ÓдóÁ¿ÆøÅÝ£¬Ð¿Á£²¿·ÖÈܽ⣮
¢Û·´Ó¦Íê±Ï£¬Ã¿¼ä¸ô1·ÖÖÓ¶ÁÈ¡ÆøÌåÌå»ý¡¢ÆøÌåÌå»ýÖð½¥¼õС£¬Ö±ÖÁÌå»ý²»±ä£®ÆøÌåÌå»ýÖð´Î¼õСµÄÔ­ÒòÊǸ÷´Ó¦·ÅÈÈ£¬ÆøÌåδÀäÈ´£¨ÅųýÒÇÆ÷ºÍʵÑé²Ù×÷µÄÓ°ÏìÒòËØ£©£®
£¨3£©ÒÒͬѧÄâÑ¡ÓÃͼ2ʵÑé×°ÖÃÍê³ÉʵÑ飺
¢ÙÄãÈÏΪ×î¼òÒ×µÄ×°ÖÃÆäÁ¬½Ó˳ÐòÊÇ£ºA½ÓE½ÓD½ÓG£¨Ìî½Ó¿Ú×Öĸ£¬¿É²»ÌîÂú£©
¢ÚʵÑ鿪ʼʱ£¬ÏÈ´ò¿ª·ÖҺ©¶·ÉϿڵIJ£Á§Èû£¬ÔÙÇáÇá´ò¿ªÆä»îÈû£¬Ò»»á¶ùºó²ÐÓàÇåÒº¾Í²»ÄÜ˳ÀûµÎÈë׶ÐÎÆ¿£®ÆäÔ­ÒòÊÇпÓë²ÐÓàÇåÒºÖÐÁòËáÍ­·´Ó¦Éú³ÉÍ­£¬Í­Ð¿ÐγÉÔ­³Ø£¬Ê¹Ð¿ÓëÁòËá·´Ó¦ËÙÂʼӿ죬·´Ó¦Éú³É´óÁ¿ÆøÌåÇÒ·ÅÈÈ£¬µ¼ÖÂ׶ÐÎÆ¿ÄÚѹǿÔö´ó£®
¢ÛÒÒͬѧµÎÈë׶ÐÎÆ¿ÖеIJÐÓàÇåÒºÌå»ýΪamL£¬²âµÃÁ¿Í²ÖÐË®µÄÌå»ýΪbmL£¨¼ÙÉèÌåϵÖÐÆøÌåΪ±ê×¼×´¿ö£©£¬²ÐÓàÇåÒºÖÐÁòËáµÄÎïÖʵÄÁ¿Å¨¶ÈµÄ±í´ïʽ$\frac{b-a}{22.4a}$mol/L£¨ÓÃa¡¢b±íʾ£©
¢ÜijѧÉúÏëÓÃÅÅË®·¨ÊÕ¼¯ÇâÆø£¬²¢¼ìÑéÆä´¿¶È£¬Ó¦Ñ¡Ôñͼ3Öм¯Æø×°ÖÃB£¨ÌîA»òB£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

1£®ÊµÑéÊÒÐèÒª0.1mol/LNaOHÈÜÒº500mL£¬¸ù¾ÝÕâÖÖÈÜÒºµÄÅäÖÆÇé¿ö»Ø´ðÏÂÁÐÎÊÌ⣮
£¨1£©ÔÚÈçͼËùʾÒÇÆ÷ÖУ¬ÅäÖÆÉÏÊöÈÜÒº¿Ï¶¨²»ÐèÒªµÄÊÇAC£¨ÌîÐòºÅ£©£¬³ýͼÖÐ
ÒÑÓÐÒÇÆ÷Í⣬ÅäÖÆÉÏÊöÈÜÒº»¹ÐèÒªµÄ²£Á§ÒÇÆ÷ÊÇÉÕ±­¡¢²£Á§°ôºÍ500mLÈÝÁ¿Æ¿£®
£¨2£©ÔÚÈÝÁ¿Æ¿µÄʹÓ÷½·¨ÖУ¬ÏÂÁвÙ×÷²»ÕýÈ·µÄÊÇBCD
A£®Ê¹ÓÃÈÝÁ¿Æ¿Ç°¼ìÑéÊÇ·ñ©ˮ
B£®ÈÝÁ¿Æ¿ÓÃˮϴ¾»ºó£¬ÔÙÓôýÅäÈÜҺϴµÓ
C£®ÅäÖÆÈÜҺʱ£¬Èç¹ûÊÔÑùÊǹÌÌ壬°Ñ³ÆºÃµÄ¹ÌÌåÓÃÖ½ÌõСÐĵ¹ÈëÈÝÁ¿Æ¿ÖУ¬»ºÂý¼ÓË®ÖÁ½Ó½ü¿Ì¶ÈÏß1¡«2cm´¦£¬ÓýºÍ·µÎ¹Ü¼ÓÕôÁóË®ÖÁ¿Ì¶ÈÏß
D£®ÅäÖÆÈÜҺʱ£¬ÈôÊÔÑùÊÇÒºÌ壬ÓÃÁ¿Í²È¡ÑùºóÓò£Á§°ôÒýÁ÷µ¹ÈëÈÝÁ¿Æ¿ÖУ¬»ºÂý¼ÓË®ÖÁ¿Ì¶ÈÏß1¡«2cm´¦£¬ÓýºÍ·µÎ¹Ü¼ÓÕôÁóË®ÖÁ¿Ì¶ÈÏß
E£®¸ÇºÃÆ¿Èû£¬ÓÃʳָ¶¥×¡Æ¿Èû£¬ÁíÒ»Ö»ÊÖÍÐסƿµ×£¬°ÑÈÝÁ¿Æ¿·´¸´µ¹×ª¶à´Î£¬Ò¡ÔÈ£®
£¨3£©¸ù¾Ý¼ÆËãÓÃÍÐÅÌÌìƽ³ÆÈ¡µÄÖÊÁ¿Îª2.0g£®ÔÚʵÑéÖÐÆäËû²Ù×÷¾ùÕýÈ·£¬Èô¶¨ÈÝʱÑöÊӿ̶ÈÏߣ¬ÔòËùµÃÈÜҺŨ¶ÈСÓÚ0.1mol/L£¨Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

8£®ÓÃÈçͼװÖÃʵÑ飬A¡¢BÁ½ÉÕ±­·Ö±ðÊ¢·Å200g10%NaOHºÍ×ãÁ¿CuSO4ÈÜÒº£®Í¨µçÒ»¶Îʱ¼äºó£¬c¼«ÉÏÓÐCuÎö³ö£¬ÓÖ²âµÃA±­ÖÐÈÜÒºµÄÖÊÁ¿¼õÉÙ4.5g£¨²»¿¼ÂÇË®µÄÕô·¢£©£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µçÔ´P¼«Îª¸º¼«£»Çë·Ö±ðд³öb¼«ºÍc¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦Ê½£º
b¼«4OH--4e-=2H2O+O2¡ü£»
c¼«Cu2++2e-=Cu
£¨2£©c¼«ÉÏÎö³ö¹ÌÌåÍ­µÄÖÊÁ¿Îª16g
£¨3£©Èô×°ÖÃÖÐÓÃǦÐîµç³Ø×÷µçÔ´£¬ÒÑ֪ǦÐîµç³Ø·Åµçʱ·¢ÉúÈçÏ·´Ó¦£º
¸º¼«£ºPb+SO42-¨TPbSO4+2e-
Õý¼«£ºPbO2+4H++SO42-+2e-¨TPbSO4+2H2O
¼ÙÉèÔÚa¼«ÖƵÃÆøÌå0.050mol£¬Õâʱµç³ØÄÚÏûºÄµÄH2SO4µÄÎïÖʵÄÁ¿ÖÁÉÙÊÇ0.10mol£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

18£®ÔÚ±ê×¼×´¿öÏ£¬ÏÂÁÐÎïÖÊËùÕ¼Ìå»ý×î´óµÄÊÇ£¨¡¡¡¡£©
A£®98 g H2SO4B£®56 g FeC£®44.8 L HClD£®6 g H2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

5£®Ï±íÖУ¬¶Ô³ÂÊö¢ñ¡¢¢òµÄÕýÈ·ÐÔ¼°Á½Õß¼äÊÇ·ñ¾ßÓÐÒò¹û¹ØϵµÄÅж϶¼ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
Ñ¡Ïî³ÂÊö¢ñ³ÂÊö¢òÅжÏ
A̼ËáÄÆÈÜÒº¿ÉÓÃÓÚÖÎÁÆθ²¡Na2CO3¿ÉÓëÑÎËá·´Ó¦¢ñ¶Ô£¬¢ò¶Ô£¬ÓÐ
BÏòNa2O2µÄË®ÈÜÒºÖеÎÈë·Ó̪±äºìÉ«Na2O2ÓëË®·´Ó¦Éú³ÉÇâÑõ»¯ÄÆ¢ñ¶Ô£¬¢ò´í£¬ÎÞ
C½ðÊôÄƱ£´æÔÚúÓÍÖУ¬ÒÔ¸ô¾ø¿ÕÆø³£ÎÂÏ£¬½ðÊôÄÆÔÚ¿ÕÆøÖлáÉú³É¹ýÑõ»¯ÄÆ¢ñ¶Ô£¬¢ò¶Ô£¬ÓÐ
D¹ýÑõ»¯ÄÆ¿ÉÓÃÓÚº½ÌìÔ±µÄ¹©Ñõ¼ÁNa2O2ÄܺÍCO2ºÍH2O·´Ó¦Éú³ÉO2¢ñ¶Ô£¬¢ò¶Ô£¬ÓÐ
A£®AB£®BC£®CD£®D

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

2£®£¨1£©ÏÂÁÐÎïÖÊ£º¢ÙNa ¢ÚÑÎËᠢ۾ƾ« ¢ÜCO2 ¢ÝNH3 ¢ÞCu£¨OH£©2¢ßBa£¨OH£©2
¢à±ù´×Ëᣨ´¿¾»µÄ´×Ëᣩ  ¢áÕáÌÇ ¢âNaClÈÜÒº⑪BaSO4£®
ÆäÖÐÊôÓÚµç½âÖʢޢߢà⑪£¬£¨ÌîÐòºÅ£¬ÏÂͬ£©ÊôÓڷǵç½âÖʢۢܢݢᣮ
£¨2£©ÒÑ֪Ũ¶ÈΪ0.01mol•L-1µÄHClÈÜÒºVmL£¬¼ÓˮϡÊ͵½2VmL£¬È¡³ö10mL£¬ÔòÕâ10mLÈÜÒºÖÐc£¨H+£©0.005mol/L£®
£¨3£©Ð´³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ£º
¢ÙÏòä廯ÑÇÌúÈÜÒºÖÐͨÈë×ãÁ¿µÄÂÈÆø£º2Fe2++4Br-+3Cl2=2Fe3++2Br2+6Cl-£®
¢ÚÏòÁòËáÂÁ¼ØÈÜÒºÖеÎÈëÇâÑõ»¯±µÈÜҺʹÁòËá¸ùÀë×ÓÇ¡ºÃÍêÈ«³Áµí£ºAl3++2SO42-+2Ba2++4OH-=2BaSO4¡ý+AlO2-+2H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

3£®£¨1£©ÊµÑéÊÒÖƱ¸ÇâÑõ»¯Ìú½ºÌ廯ѧ·½³Ìʽ£ºFeCl3+3H2O$\frac{\underline{\;\;¡÷\;\;}}{\;}$Fe£¨OH£©3£¨½ºÌ壩+3HCl£®
£¨2£©ÄÜÖ¤Ã÷Na2SO3ÈÜÒºÖдæÔÚSO32-+H2O?HSO3-+OH-Ë®½âƽºâµÄÊÂʵÊÇC£¨ÌîÑ¡Ïî×Öĸ£©£®
A£®µÎÈë·Ó̪ÈÜÒº±äºì£¬ÔÙ¼ÓÈëH2SO4ÈÜÒººóºìÉ«ÍËÈ¥
B£®µÎÈë·Ó̪ÈÜÒº±äºì£¬ÔÙ¼ÓÈëÂÈË®ºóºìÉ«ÍËÈ¥
C£®µÎÈë·Ó̪ÈÜÒº±äºì£¬ÔÙ¼ÓÈëBaCl2ÈÜÒººó²úÉú³ÁµíÇÒºìÉ«ÍÊÈ¥
£¨3£©ÏÖÓпÉÄæ·´Ó¦£º2A£¨g£©+2B£¨g£©?C£¨g£©+3D£¨s£©£¬ÔÚÃܱÕÈÝÆ÷µÄÈÝ»ý¡¢Î¶ȶ¼ÏàͬµÄÌõ¼þÏ£¬·Ö±ð´ÓÒÔÏÂÁ½Ìõ;¾¶½¨Á¢Æ½ºâ£º¢ñ£®A¡¢BµÄÆðʼÎïÖʵÄÁ¿¾ùΪ2mol£¬¢ò£®C¡¢DµÄÆðʼÎïÖʵÄÁ¿·Ö±ðΪ2molºÍ6mol£®ÒÔÏÂ˵·¨ÖÐÕýÈ·µÄÊÇC£¨ÌîÑ¡Ïî×Öĸ£©£®
A£®¢ñ¡¢¢òÁ½Ìõ;¾¶×îÖմﵽƽºâʱ£¬ÌåϵÄÚ»ìºÏÆøÌåµÄ°Ù·Ö×é³ÉÏàͬ
B£®´ïµ½Æ½ºâʱ£¬Í¾¾¶¢òËùµÃ»ìºÏÆøÃܶÈΪ;¾¶¢ñËùµÃ»ìºÏÆøÃܶȵÄ2±¶
C£®´ïµ½Æ½ºâʱ£¬Í¾¾¶¢òÖÐCµÄƽºâŨ¶È´óÓÚ;¾¶¢ñÖÐCµÄƽºâŨ¶ÈµÄ2±¶
£¨4£©ÔÚºãÈݾøÈÈ£¨²»ÓëÍâ½ç½»»»ÄÜÁ¿£©Ìõ¼þϽøÐÐ2A £¨g£©+B£¨g£©?2C£¨g£©+D£¨s£©·´Ó¦£¬°´Ï±íÊý¾ÝͶÁÏ£¬·´Ó¦´ïµ½Æ½ºâ״̬£¬²âµÃÌåϵѹǿÉý¸ß£®¼òÊö¸Ã·´Ó¦µÄƽºâ³£ÊýÓëζȵı仯¹Øϵ£º¸Ã·´Ó¦µÄƽºâ³£ÊýKËæζȵÄÉý¸ß¶ø¼õС£®
ÎïÖÊABCD
ÆðʼͶÁÏ/mol2120

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸